Post by: elemis on October 03, 2010, 02:40:16 pm
Post by: ruby92 on October 08, 2010, 09:35:37 am
Post by: elemis on October 08, 2010, 09:47:15 am
Can someone please explain in detail how to derive the equation for centerpital acceleration?
http://dev.physicslab.org/Document.aspx?doctype=3&filename=CircularMotion_CentripetalAcceleration.xml
http://www.youtube.com/watch?v=YRBRarbMCyE <-------- watch video first,
Post by: ruby92 on October 09, 2010, 11:05:36 am
I have two more doubts aswell atm.
from circular motion:
At an airshow an aircraft diving at the speed of 170ms-1 pulls out of the dive by moving in the arc of a circle at the bottom of the dive.
a) Calculate the minimum radius of this circle if the centerpital acceleration of the aircraft is not to exceed 5 times the acceleration of freefall.
From gravitaitonal fields:
The radius of the moons orbit is 3.84*10^8, and its period is 27.4 days.Use keplers law to calculate the period of the orbit of a satellite orbiting the earth just above the earths surface.
Post by: S.M.A.T on October 09, 2010, 11:40:01 am
from circular motion:
At an airshow an aircraft diving at the speed of 170ms-1 pulls out of the dive by moving in the arc of a circle at the bottom of the dive.
a) Calculate the minimum radius of this circle if the centerpital acceleration of the aircraft is not to exceed 5 times the acceleration of freefall.
a=5g ms^-2
a=(v^2)/r
r=(v^2)/a
r=(170^2)/5g
r=(5780/g)m
Post by: ruby92 on October 09, 2010, 02:41:35 pm
Post by: ashish on October 09, 2010, 03:42:06 pm
m/j 2008 paper 4 q1 b
the centripetal force is provided by the frictional force..
note w=omega
W=weight
Fc= Frictional force
M(w^2)r =0.72W
M(w^2)(35/100) =0.72Mg
M cancel out
w^2=(0.72*9.81)/0.35
w=4.49 rad/s
frequency of rotation = w/2pi
=0.7146
in 60 sec no. of rotation = 0.7146*60
=43
Post by: ruby92 on October 09, 2010, 05:15:07 pm
also could someone please help me with oct/nov 2009 paper 41 q1 bii
Post by: ashish on October 09, 2010, 05:55:21 pm
Gravitational force = centripetal force
(GMm)/x^2 = Mw^2 x (w=omega)
m is eliminated on both side
GM= w^2 x^3-----------1
we know that g is Gravitational field strength on earth surface
g=GM/R^2
BY making GM subject of formula
GM = gR^2------------
now
now replace GM by gR^2 in 1
gR^2 =w^2 x^3
:D
Post by: Ghost Of Highbury on October 10, 2010, 06:39:44 am
Doubt : (i) (direction)
2) For a force F of magnitude 150N, determine the force in S1.
I didn't understand the diagram very well. How exactly will the system move if a force is applied to the end of the lever. Also, how are equal forces produced in the strings? Could someone explain please?
Post by: ashish on October 10, 2010, 07:47:04 am
Question Attached.
Doubt : (i) (direction)
2) For a force F of magnitude 150N, determine the force in S1.
I didn't understand the diagram very well. How exactly will the system move if a force is applied to the end of the lever. Also, how are equal forces produced in the strings? Could someone explain please?
if the string would no have been there then the disc would have rotated upon itself!
just like opening nuts with a Cylinder Opener Key
2) the moment produce will be equal to the couple ( the two string will have opposite but equal forces)
moment = (30/100)* 150
=45Nm
couple is perpendicular distance between the 2 forces * magnitude of one force
45=(12/100)*F
F=375N
sorry but i am getting problem while uploading pictures :(
the tension in the string would be toward the disc
Post by: ashish on October 10, 2010, 08:00:25 am
thank you:)
I have two more doubts aswell atm.
From gravitaitonal fields:
The radius of the moons orbit is 3.84*10^8, and its period is 27.4 days.Use keplers law to calculate the period of the orbit of a satellite orbiting the earth just above the earths surface.
Keplers law T^2 is direcly proportional to R^3
therefore T^2/R^3 = constant
(27.4*24*60*60)^2 / (3.84*10^8)^3 =9.8977*10^-14
let the period be T
T^2 / R^3 = 9.8977*10^-14
T^2 = (9.8977*10^-14)* (6.4*10^6)^3
T = 5093.74 sec
T=1h24min
is that right?
Post by: Ghost Of Highbury on October 10, 2010, 08:00:51 am
if the string would no have been there then the disc would have rotated upon itself!
just like opening nuts with a Cylinder Opener Key
2) the moment produce will be equal to the couple ( the two string will have opposite but equal forces)
moment = (30/100)* 150
=45Nm
couple is perpendicular distance between the 2 forces * magnitude of one force
45=(12/100)*F
F=375N
sorry but i am getting problem while uploading pictures :(
the tension in the string would be toward the disc
k thanks a lot, + rep.
Post by: ashish on October 10, 2010, 08:03:19 am
k thanks a lot, + rep.
thanks mate :)
Post by: Deadly_king on October 10, 2010, 08:08:22 am
Keplers law T^2 is direcly proportional to R^3
therefore T^2/R^3 = constant
(27.4*24*60*60)^2 / (3.84*10^8)^3 =9.8977*10^-14
let the period be T
T^2 / R^3 = 9.8977*10^-14
T^2 = (9.8977*10^-14)* (6.4*10^6)^3
T = 5093.74 sec
T=1h24min
is that right?
Good job pal :)
keep it up ;)
+rep
Post by: ashish on October 10, 2010, 08:12:05 am
Post by: Ghost Of Highbury on October 10, 2010, 10:53:08 am
Post by: Deadly_king on October 10, 2010, 11:07:53 am
Can u explain the steps too please? Thank you20.
You should first know that points at the same level/height experience same pressure and that pressure due to a liquid is heg.
Back to the diagram, let's take it at the level x from below.
Since in both limbs, points at same level experience same pressure :
Patm + xeqg = Patm + 2xepg
Therefore simplify this equation to obtain ;
ep/eq = 1/2
Answer is A
12.
Since the forces are in equilibrium, the resultant force should be zero. For this, the forces should make a complete cycle back to the same original point.
Only triangles R and S show this possiblity. Hence answer is C
Hope it helps :)
Post by: Ghost Of Highbury on October 10, 2010, 12:01:31 pm
20.
You should first know that points at the same level/height experience same pressure and that pressure due to a liquid is heg.
Back to the diagram, let's take it at the level x from below.
Since in both limbs, points at same level experience same pressure :
Patm + xeqg = Patm + 2xepg
Therefore simplify this equation to obtain ;
ep/eq = 1/2
Answer is A
12.
Since the forces are in equilibrium, the resultant force should be zero. For this, the forces should make a complete cycle back to the same original point.
Only triangles R and S show this possiblity. Hence answer is C
Hope it helps :)
Thank you. +rep
Post by: Deadly_king on October 10, 2010, 12:08:47 pm
Thank you. +repAnytime dude :)
Post by: Deadly_king on October 10, 2010, 12:15:38 pm
One doubt, which point have u considered?Height x above the ground.
It will be 2x for the left limb but x in the right one. Got me?
Post by: ruby92 on October 10, 2010, 04:19:25 pm
Keplers law T^2 is direcly proportional to R^3Thanks:)
therefore T^2/R^3 = constant
(27.4*24*60*60)^2 / (3.84*10^8)^3 =9.8977*10^-14
let the period be T
T^2 / R^3 = 9.8977*10^-14
T^2 = (9.8977*10^-14)* (6.4*10^6)^3
T = 5093.74 sec
T=1h24min
is that right?
Post by: ruby92 on October 10, 2010, 04:38:23 pm
why is it that the normal reaction force=force of gravity - centerpital force.
shouldnt it be plus?
Also m/j 2007 paper 4 q1 bi
Post by: Ghost Of Highbury on October 10, 2010, 04:54:51 pm
The load on each of the arrangements is L.
For each arrangement , complete the table by determining
(i) the total extension in terms of e,
(ii) the spring constant in terms of k.
Post by: Deadly_king on October 11, 2010, 05:15:34 am
i need help in this question
The load on each of the arrangements is L.
For each arrangement , complete the table by determining
(i) the total extension in terms of e,
(ii) the spring constant in terms of k.
Hmm......Ashish......your spring constants are ok. But am afraid the extensions are not :-[
Anyway dude, you should know that for spring constant,
When springs are in parallel, their net spring constants are added together. ( Same as for a capacitor and opposite to that of resistors)
When they are in series they are calculated as shown KR = [1/k + 1/k]-1 or 1/KR = 1/k + 1/k
For extension,
When in series, net extension is the sum of the extension of each spring.
When in parallel, it is calculated as shown eR = [1/e + 1/e]-1 or 1/eR = 1/e + 1/e
Let's back to the question now :
1. Total extension will be the sum of extension by each springs when they are connected in series. Each spring extends by e
Hence total extension will be 2e
Spring constant will be [1/k + 1/k]-1 = k/2
2. Since the two springs are connected in parallel, total extension will be e/2 and net spring constant will be 2k
eR = [1/e + 1/e]-1 = e/2
KR = k + k = 2k
3. Total extension : 3e/2
Net spring constant : 2k/3
Use the values you obtained in part(ii)
Total extension = e/2 + e = 3e/2
For the two springs in parallel, net spring constant will be 2k. Hence spring constant of the whole system will be [1/2k + 1/k]-1 = 2k/3
NOTE : kR and eR are the resultant spring constants and resultant extensions respectively.
Post by: Deadly_king on October 11, 2010, 05:44:47 am
Physics A2 oct/nov 2008 paper 4 q 1aiii
why is it that the normal reaction force=force of gravity - centerpital force.
shouldnt it be plus?
Also m/j 2007 paper 4 q1 bi
Nov 08 p4 No 1
a)(iii) The normal reaction exerted by a planet on a mass is equal to the weight of the mass.
Normally for an object undergoing circular motion,
Resultant towards centre = Centripetal force
Resultant force towards centre of the planet will be : Gravitational force - Weight
Hence
Gravitational force - Weight = Centripetal force ---> Weight = GMm/R2 - mR
Since Weight = Normal reaction = GMm/R2 - mR
Jun 07 p4 No 1
b)(i) Gravitational potential energy is given by the formula = -GMm/R
In the question you are asked to find the change in gravitational potential.
Let ^Ep = Change in gravitational potential energy
Uf = Final gravitational potential energy = -GMm/3R
Ui = Initial gravitational potential energy = -GMm/2R
Hence ^Ep = Uf - Ui
^Ep = (-GMm/3R) - (-GMm/2R)
Simplify this equation and you'll get your required solution which is ^Ep = GMm/6R
Post by: ashish on October 11, 2010, 11:26:10 am
Hmm......Ashish......your spring constants are ok. But am afraid the extensions are not :-[
Anyway dude, you should know that for spring constant,
When springs are in parallel, their net spring constants are added together. ( Same as for a capacitor and opposite to that of resistors)
When they are in series they are calculated as shown KR = [1/k + 1/k]-1 or 1/KR = 1/k + 1/k
For extension,
When in series, net extension is the sum of the extension of each spring.
When in parallel, it is calculated as shown eR = [1/e + 1/e]-1 or 1/eR = 1/e + 1/e
Let's back to the question now :
1. Total extension will be the sum of extension by each springs when they are connected in series. Each spring extends by e
Hence total extension will be 2e
Spring constant will be [1/k + 1/k]-1 = k/2
2. Since the two springs are connected in parallel, total extension will be e/2 and net spring constant will be 2k
eR = [1/e + 1/e]-1 = e/2
KR = k + k = 2k
3. Total extension : 3e/2
Net spring constant : 2k/3
Use the values you obtained in part(ii)
Total extension = e/2 + e = 3e/2
For the two springs in parallel, net spring constant will be 2k. Hence spring constant of the whole system will be [1/2k + 1/k]-1 = 2k/3
NOTE : kR and eR are the resultant spring constants and resultant extensions respectively.
the answer i gave were in terms of k an L :(
Post by: Deadly_king on October 11, 2010, 05:10:51 pm
the answer i gave were in terms of k an L :(
Yeah.......the question asked in terms of e ;)
Post by: ashish on October 12, 2010, 02:34:28 pm
Post by: WARRIOR on October 12, 2010, 03:07:47 pm
Post by: Deadly_king on October 12, 2010, 05:40:23 pm
a simple test if anyone has the time to do it , good for basics practise
Thanks for your questions buddy ;)
Post by: Vin on October 14, 2010, 02:16:13 pm
Pg. 111, Q.3 From the book International A/AS Level Physics by Chris Mee, Mike Crundell and others.
A uniform plank of weight 120 N rests on two stools as shown in Fig. 4.39. A weight of 80 N is placed on the plank, midway between the stools. Calculate :
(a) the force acting on the stool at A,
(b) the force acting on the stool at B.
Thanks a lot!
Post by: Vin on October 14, 2010, 02:35:11 pm
Post by: Deadly_king on October 14, 2010, 03:37:55 pm
Oh, By the way the answers are (a) 80 N and (b) 120 N. Just in case. ;)
Are you sure about the answers dude?
Because am getting (a) 88N and (b) 112N
I think there must be some kind of misprint in the book ;)
Post by: lana on October 14, 2010, 09:14:37 pm
thanks =]
Post by: WARRIOR on October 15, 2010, 06:32:47 am
Thanks for your questions buddy ;)well if u have time can u just solve them and give me the mark scheme? jsut the answers ! caz the whole test would be pointless. i have my answers but i want to make sure with someone else!
Post by: Deadly_king on October 15, 2010, 07:16:02 am
well if u have time can u just solve them and give me the mark scheme? jsut the answers ! caz the whole test would be pointless. i have my answers but i want to make sure with someone else!Sorry dude.......I won't have time now since am in exams :-\
Will be glad to help you after that ;)
Post by: ashish on October 15, 2010, 10:30:37 am
Are you sure about the answers dude?
Because am getting (a) 88N and (b) 112N
I think there must be some kind of misprint in the book ;)
well me too i got 88 and 112
Post by: elemis on October 15, 2010, 10:42:09 am
Or vice versa.
Post by: Deadly_king on October 15, 2010, 10:43:58 am
One reason for the discrepancy maybe that you are using 9.8 for g instead of 10.
Or vice versa.
Sorry br0 but g is not used in this question since weight has already been given ;)
Post by: elemis on October 15, 2010, 10:52:42 am
Sorry br0 but g is not used in this question since weight has already been given ;)
I havent even looked at the question closely :P
Post by: Deadly_king on October 15, 2010, 11:01:48 am
I havent even looked at the question closely :P
If you have time, try it out!
Then let us know your answers ;)
Post by: Ivo on October 15, 2010, 08:15:11 pm
Three forces are applied to the point O as shown in the diagram. Calculate: a) The component in directions OX and OY respectively of each of the forces; b) The resultant force acting at O.
Answers I got for part a) (couldn't do b) -
For 6N force: Vertical component (VC) = 4.2N, Horizontal component (HC) = 4.2N
For 8N force: VC = 5.7N, HC = 5.7N
For 14N force: VC = 0N, HC = 14N
I don't know if answers are right, please check kindly. Also, please explain part b). Many thanks! :D
Post by: S.M.A.T on October 15, 2010, 09:26:29 pm
Please could someone help me with this question; diagram attached.Here :)................sorry for low visibility :-[
Three forces are applied to the point O as shown in the diagram. Calculate: a) The component in directions OX and OY respectively of each of the forces; b) The resultant force acting at O.
Answers I got for part a) (couldn't do b) -
For 6N force: Vertical component (VC) = 4.2N, Horizontal component (HC) = 4.2N
For 8N force: VC = 5.7N, HC = 5.7N
For 14N force: VC = 0N, HC = 14N
I don't know if answers are right, please check kindly. Also, please explain part b). Many thanks! :D
Post by: Ivo on October 15, 2010, 10:43:09 pm
Here :)................sorry for low visibility :-[
Ahh, a very clever way of doing it! Thanks! +rep for you!!! ;)
Post by: Ivo on October 15, 2010, 10:45:09 pm
The diagram attached shows a mass of 2.5 kg initially at rest on a rough inclined plane. The mass is now released and acquires a velocity of 4.0ms-1 at P, the base of the incline. Find (a) the work done against friction, (b) the (average) friction force.
I got 20J, 5N, but I did I completely different approach (using GPE) and got dissimilar answers: 17.5J, 4.4N. Which is right - By the way for
Thanks in advance!
Post by: S.M.A.T on October 15, 2010, 11:45:32 pm
Post by: Deadly_king on October 16, 2010, 05:00:30 am
What about this question, there seems to be two ways of doing it, getting two different solutions. I'm not quite sure whether the 1.5 m is required for the calculation, (maybe for GPE?), or do we use the acceleration formula:
The diagram attached shows a mass of 2.5 kg initially at rest on a rough inclined plane. The mass is now released and acquires a velocity of 4.0ms-1 at P, the base of the incline. Find (a) the work done against friction, (b) the (average) friction force.
I got 20J, 5N, but I did I completely different approach (using GPE) and got dissimilar answers: 17.5J, 4.4N. Which is right - By the way for, can we use this formula if we are working out the friction force (ie. the force against the acceleration), or is this only valid for resultant force? I'm confused.
Thanks in advance!
Do you mind if I ask how you got 20J and 5N ???
The only possible answers am getting are 17.5J and 4.375N.
You can use both methods and I assure you, you'll be getting same answers. I guess you must have done some little mistakes somewhere ;)
I'll elaborate on both methods to help you find your errors :
1. Loss in P.E = Gain in K.E + Work done against friction
mgh = 0.5mv2 + Work done against friction
Hence Work done against friction = mgh - 0.5mv2 = 2.5(10)(1.5) - 0.5(2.5)(4)2 = 17.5J
Now you can find the average frictional force = 17.5/4 = 4.375N
2. First you need to find the angle of inclination(x) using the distances provided -----> sin x = 1.5/4
Now you need to find the acceleration using v2 = u2 + 2aS
42 = 0 + 2a(4) -----> a = 2m/s2
Using Newton's 2nd law of motion :
Resultant force = ma
mgsin x - Friction = ma
Hence Friction = mgsin x - ma = 2.5(10)(1.5/4) - 2.5(2) = 4.375N
Now, Work done against friction = 4.375 x 4 = 17.5J
Hope it helps :)
Post by: Ivo on October 16, 2010, 01:49:58 pm
Do you mind if I ask how you got 20J and 5N ???
The only possible answers am getting are 17.5J and 4.375N.
You can use both methods and I assure you, you'll be getting same answers. I guess you must have done some little mistakes somewhere ;)
I'll elaborate on both methods to help you find your errors :
1. Loss in P.E = Gain in K.E + Work done against friction
mgh = 0.5mv2 + Work done against friction
Hence Work done against friction = mgh - 0.5mv2 = 2.5(10)(1.5) - 0.5(2.5)(4)2 = 17.5J
Now you can find the average frictional force = 17.5/4 = 4.375N
2. First you need to find the angle of inclination(x) using the distances provided -----> sin x = 1.5/4
Now you need to find the acceleration using v2 = u2 + 2aS
42 = 0 + 2a(4) -----> a = 2m/s2
Using Newton's 2nd law of motion :
Resultant force = ma
mgsin x - Friction = ma
Hence Friction = mgsin x - ma = 2.5(10)(1.5/4) - 2.5(2) = 4.375N
Now, Work done against friction = 4.375 x 4 = 17.5J
Hope it helps :)
Thanks for your help. I understand the 1st method, but I'm not sure about the 2nd. How did you get mgsin x? I assume you're taking angle x is the point at P on the diagram - so for the angle:
But then how do we work out the acclerating force (how did you get your value)? I'm assuming that the weight of 2.5kg mass is acting vertically downwards, so it is 25N. So I take the acclerating force to be:
Many thanks once again in advance!
Post by: Freaked12 on October 16, 2010, 02:40:53 pm
and distance travelled
find acceleration
2 ms-2
calculate the force of weight acting downwards
thats coming 3/8
this force minus friction gives you ma
you know m and a
equate and find friction
and then multiply it by distance, 4m
and you'll get work done
Post by: Ivo on October 16, 2010, 02:57:03 pm
you know initial and final velocities
and distance travelled
find acceleration
2 ms-2
calculate the force of weight acting downwards
thats coming 3/8
this force minus friction gives you ma
you know m and a
equate and find friction
and then multiply it by distance, 4m
and you'll get work done
How did you get 3/8? Surely force acting downards is 25N, no?
Post by: Freaked12 on October 16, 2010, 03:05:47 pm
thats 3/8 into 25
thats it
FsinQ
sinQ = 3/8
Post by: Ivo on October 16, 2010, 03:22:23 pm
sorry
thats 3/8 into 25
thats it
FsinQ
sinQ = 3/8
I agree that
Thanks in advance!
Post by: Deadly_king on October 16, 2010, 05:14:45 pm
Thanks for your help. I understand the 1st method, but I'm not sure about the 2nd. How did you get mgsin x? I assume you're taking angle x is the point at P on the diagram - so for the angle:
But then how do we work out the acclerating force (how did you get your value)? I'm assuming that the weight of 2.5kg mass is acting vertically downwards, so it is 25N. So I take the acclerating force to be:, where F is the accelerating force to be found?
Many thanks once again in advance!
Indeed the weight acts downwards and is equal to 25N. But for this question you need not the weight but the component of the weight along the plane,i.e the inclined plane.
If you draw a straight line vertically downwards from the centre of the mass, the latter will form a right angled triangle with the horizontal line. From this you find the angle of inclination which is sin x = 1.5/4.
Now you can find the component of the weight downwards along the plane which will be 25sinx or 25(1.5/4)
NOTE : Angle x which is equal to the angle of inclination is also the angle between the vertical and the perpendicular line of the plane from the centre of mass.
Try to draw the triangle and the normal, you'll see that the angle I mean is equal to x and also how I got 25sin x as component of the weight downwards but along the plane.
Hope you understand what am trying to say ;)
Post by: Ivo on October 16, 2010, 11:02:59 pm
Indeed the weight acts downwards and is equal to 25N. But for this question you need not the weight but the component of the weight along the plane,i.e the inclined plane.
If you draw a straight line vertically downwards from the centre of the mass, the latter will form a right angled triangle with the horizontal line. From this you find the angle of inclination which is sin x = 1.5/4.
Now you can find the component of the weight downwards along the plane which will be 25sinx or 25(1.5/4)
NOTE : Angle x which is equal to the angle of inclination is also the angle between the vertical and the perpendicular line of the plane from the centre of mass.
Try to draw the triangle and the normal, you'll see that the angle I mean is equal to x and also how I got 25sin x as component of the weight downwards but along the plane.
Hope you understand what am trying to say ;)
Thanks, for your help. Great explanation, now I get it! + rep for you! (By the way, which method would you advise?)
Post by: Deadly_king on October 17, 2010, 05:17:21 am
Thanks, for your help. Great explanation, now I get it! + rep for you! (By the way, which method would you advise?)
You're welcome :)
Hmm.....that will depend on you. Use the method you find easier.
But the question ask for the work done first, so i believe the first method would be shorter ;)
Post by: Ivo on October 17, 2010, 09:35:38 am
My (wrong) answers:
a) For equlibrium to be achieved, forces are balanced, zero resultant.
b) i) 141N
ii) 131N
iii) No idea! Difference is not 0N, I'm stuck!
Many thanks in advance!
Post by: S.M.A.T on October 17, 2010, 12:33:10 pm
b iii)because the frictional force in the bed is equal to the horizontal force exerted by the rope so resultant force is zero
Post by: ashish on October 17, 2010, 12:42:25 pm
OK, I think this is slightly harder than the previous ones that I am struggling with. I know that the tension is the same all round the system, but I don't know which directions each one are (could someone draw the arrows to show them, thanks!)
My (wrong) answers:
a) For equlibrium to be achieved, forces are balanced, zero resultant.
b) i) 141N
ii) 131N
iii) No idea! Difference is not 0N, I'm stuck!
Many thanks in advance!
your answers are not wrong
i got the same
look at the picture
there is a constant tension(10*8) in the string throughout the system i took g as ten
when the string has inclined it will cause an additional force
additional horizontal force will be 80cos40 =61.3
total force will be =61.3+80
=141 N
additional vertical force will be 80sin40 = 51.42
total force will be = 51.42+80
=131.42 N
Post by: Deadly_king on October 17, 2010, 01:14:11 pm
I got the same answer like U in b) i) ii) ....what is the correct answer ??? ???
b iii)because the frictional force in the bed is equal to the horizontal force exerted by the rope so resultant force is zero
Do you mind to explain how you got your answers ???
b)(iii) It's not necessary that resultant is zero. The frictional force between the bed and the body of the skier may be greater or equal to the horizontal force.
@ ashish : Weight cannot be the answer since the latter acts vertically downwards ans will have no horizontal component ;)
Post by: ashish on October 17, 2010, 01:18:32 pm
Do you mind to explain how you got your answers ???
b)(iii) It's not necessary that resultant is zero. The frictional force between the bed and the body of the skier may be greater or equal to the horizontal force.
@ ashish : Weight cannot be the answer since the latter acts vertically downwards ans will have no horizontal component ;)
yeah you are right but are my answers right?
Post by: Deadly_king on October 17, 2010, 01:52:15 pm
yeah you are right but are my answers right?
Am not sure about it myself :-\
Let horizontal component of the tension be TH and the vertical component to be TV while T is the tension in the string.
Since the system is in equilibrium, TV = 8(9.81) -----> T x sin 40 = 8(9.81)
Therefore the tension in the string is 8(9.81)/sin 40 = 122 N
Horizontal component will be TH = T x cos 40 = 122 x cos 40 = 93.5 N
I have a feeling I made a mistake somewhere but can't really find out where :-[
Can anyone confirm the answer ???
Post by: nid404 on October 17, 2010, 02:15:26 pm
3rd one I'm not sure of either :-\
Post by: Deadly_king on October 17, 2010, 02:34:21 pm
I got the same answers as Deadly king for the first and second part.
3rd one I'm not sure of either :-\
Alright then.
For the third one, am pretty sure both answers(mine and Asif) will be accepted.
But I'll prefer mine since the question has not stated that the system is in limiting equilibrium. If that was the case, then asif's answer would have been best. ;)
Post by: ashish on October 17, 2010, 02:44:17 pm
Am not sure about it myself :-\
Let horizontal component of the tension be TH and the vertical component to be TV while T is the tension in the string.
Since the system is in equilibrium, TV = 8(9.81) -----> T x sin 40 = 8(9.81)
Therefore the tension in the string is 8(9.81)/sin 40 = 122 N
Horizontal component will be TH = T x cos 40 = 122 x cos 40 = 93.5 N
I have a feeling I made a mistake somewhere but can't really find out where :-[
Can anyone confirm the answer ???
i have a doubt, the 122 N will be the tension in the inclined string which will cause the upward lift?
Post by: Deadly_king on October 17, 2010, 02:52:12 pm
i have a doubt, the 122 N will be the tension in the inclined string which will cause the upward lift?
I don't think the upward lift is applied here.
The tension along the whole piece of string will be the same. (122 N)
Post by: lana on October 17, 2010, 09:38:11 pm
do you have any advice on how to prepare for the practical paper 3 ?? and could someone please explain to me an easy way to find uncertainties ??
thank you...
Post by: ashish on October 18, 2010, 03:43:54 am
I don't think the upward lift is applied here.
The tension along the whole piece of string will be the same. (122 N)
hey deadly i kept on thinking on it ,the tension in the string cannot be 122N throughout it has to be 9.81*8 !
Post by: S.M.A.T on October 18, 2010, 11:09:21 am
Do you mind to explain how you got your answers ???
b)(iii) It's not necessary that resultant is zero. The frictional force between the bed and the body of the skier may be greater or equal to the horizontal force.
@ ashish : Weight cannot be the answer since the latter acts vertically downwards ans will have no horizontal component ;)
DK :D frictional force cannot be greater than the horizontal force,if it was greater than the body will start to move to the left ::) .But u can still say that this is not the maximum frictional force that the bed can apply on the body because the question did not said that the body is in limiting equilibrium so if horizontal force increase than the frictional force will increase.
Since the body did not move(i.e. in equilibrium) that means frictional force=horizontal force
I hope u understood what i said :)
Post by: nid404 on October 18, 2010, 11:13:16 am
Post by: S.M.A.T on October 18, 2010, 11:20:03 am
Limiting friction is greater than the horizontal force. Friction can take any value from 0 to uR.
yes this is what i meant to say :D
So when an object is in equilibrium,the frictional force will take a value equal to the horizontal force ;)
Post by: nid404 on October 18, 2010, 11:21:53 am
yes this is what i meant to say :D
So when an object is in equilibrium,the frictional force will take a value equal to the horizontal force ;)
Bingo :P
Post by: S.M.A.T on October 18, 2010, 11:45:13 am
OK, I think this is slightly harder than the previous ones that I am struggling with. I know that the tension is the same all round the system, but I don't know which directions each one are (could someone draw the arrows to show them, thanks!)DK here is the working :)
My (wrong) answers:
a) For equlibrium to be achieved, forces are balanced, zero resultant.
b) i) 141N
ii) 131N
iii) No idea! Difference is not 0N, I'm stuck!
Many thanks in advance!
b)i and ii
Post by: Deadly_king on October 18, 2010, 12:45:30 pm
yes this is what i meant to say :D
So when an object is in equilibrium,the frictional force will take a value equal to the horizontal force ;)
Understood you perfectly br0 ;)
I had the same thinking in mind but i think i misplaced my words. When I said frictional force is greater than the tension, I was referring to the maximum constant value that the friction may reach ;)
By the way good job :D
Post by: ashish on October 18, 2010, 02:42:22 pm
DK here is the working :)
b)i and ii
nice work thought i posted the same working check my post , lol the answer was simple it was just that i didn't concentrate .... EXAMMMMMMMMMMMMMMMMMMs ss :-\
Post by: S.M.A.T on October 18, 2010, 04:14:33 pm
nice work thought i posted the same working check my post , lol the answer was simple it was just that i didn't concentrate .... EXAMMMMMMMMMMMMMMMMMMs ss :-\
SORRY i didn't notice. :)...+rep for your work ;)
Post by: lana on October 19, 2010, 06:39:29 pm
thanks =]
Post by: Ivo on October 19, 2010, 07:05:53 pm
yes this is what i meant to say :D
So when an object is in equilibrium,the frictional force will take a value equal to the horizontal force ;)
Just to make sure, so can I say for part iii) That because the system is in equlibrium, there is no resultant force acting on the skier. Therefore, the friction force exerted by the bed is said to be equal to the net horizontal forces exerted by the tension of the rope?
What I'm not sure about is should I say 'net', and should I say 'taking into account of horizontal components'.
Also, I'm not sure about the directions of Tension that should be on the question. Here's my attempt, please could someone correct me. I'm not sure if there should be a Tension force on the horizontal right at the top of the system.
Post by: Deadly_king on October 20, 2010, 05:23:37 am
could someone please explain to me to estimating uncertainties ??
thanks =]
Try this :
http://www.astarmathsandphysics.com/a_level_physics_notes/principles_dimensions_units_error_analysis_etc/a_level_physics_notes_error%20analysis%281%29.html
If ever you need some explanation from us, post a question!
It's easier to explain when answering to an appropriate question ;)
Post by: Deadly_king on October 20, 2010, 05:36:31 am
Just to make sure, so can I say for part iii) That because the system is in equlibrium, there is no resultant force acting on the skier. Therefore, the friction force exerted by the bed is said to be equal to the net horizontal forces exerted by the tension of the rope?
What I'm not sure about is should I say 'net', and should I say 'taking into account of horizontal components'.
Also, I'm not sure about the directions of Tension that should be on the question. Here's my attempt, please could someone correct me. I'm not sure if there should be a Tension force on the horizontal right at the top of the system.
Yeah..........the word net fits exactly here. Your answer is right ;)
For the tensions, I think all of them should be directed towards the pulleys. Hence the horizontals will have tensions in both directions.
But am not 100% confident about that :-[
Post by: lana on October 20, 2010, 02:16:28 pm
Thank you sooo much =]
Try this :
http://www.astarmathsandphysics.com/a_level_physics_notes/principles_dimensions_units_error_analysis_etc/a_level_physics_notes_error%20analysis%281%29.html
If ever you need some explanation from us, post a question!
It's easier to explain when answering to an appropriate question ;)
Post by: Deadly_king on October 20, 2010, 03:24:22 pm
Thank you sooo much =]
You're welcome dear :)
Post by: Ivo on October 20, 2010, 06:12:29 pm
(http://img101.imageshack.us/img101/2651/diagramk.jpg)
Post by: Ivo on October 20, 2010, 06:21:30 pm
Hmm, don't really know how to start this!
(http://img101.imageshack.us/img101/2651/diagramk.jpg)
Here are my answers - I doubt my abilities though!
a) 5600N
b) 1600N
c) 1600N
Post by: ashish on October 21, 2010, 04:26:34 am
Here are my answers - I doubt my abilities though!
a) 5600N
b) 1600N
c) 1600N
for part a
F=ma
=(3000+4000)*0.8
=5600N
B)
total force is 5600 N
net force acting on 1st trailer will be 5600- ( 3000*0.8 )
=3200N
c)5600-((3000+2000)*0.8 )
=1600N
Post by: TJ-56 on October 22, 2010, 12:01:55 pm
(http://img201.imageshack.us/img201/5643/unfi.jpg)
http://img201.imageshack.us/img201/5643/unfi.jpg
Post by: $!$RatJumper$!$ on October 22, 2010, 03:51:18 pm
someone please explain
(http://img201.imageshack.us/img201/5643/unfi.jpg)
http://img201.imageshack.us/img201/5643/unfi.jpg
This should answer your question. If it isn't clear let me know :)
(http://i53.tinypic.com/jawhgp.jpg)
http://i53.tinypic.com/jawhgp.jpg
Post by: TJ-56 on October 23, 2010, 11:22:38 am
Post by: TJ-56 on October 23, 2010, 11:26:00 am
june 05 q13 paper 1
thanx in advance
Post by: Deadly_king on October 23, 2010, 12:48:27 pm
(http://img254.imageshack.us/img254/9700/s0513.png)
june 05 q13 paper 1
thanx in advance
Moment is defined as the product of the force to the perpendicular distance from the pivot.
Total clockwise moments = 20(3)
Total anti-clockwise moment = 5(2) + 10(2)
Resultant moment = 20(3) - [5(2) + 10(2)] = 30Nm
NOTE : For the 10N force, the perpendicular distance from P is 2m since the line of action of the force is horizontal.
Hope it helps :)
Post by: missbeautiful789 on October 23, 2010, 12:55:33 pm
Force x Perpendicular distance from the pivot
resultant torque= sum of clokwise moment - sum of anti-clockwise moments
Clokwise moments:
the 20N has a clockwise moment
20 x perp distance
20X3
clockwise moment=60 Nm
Anti-clockwise moments:
the 10N and 5N have a anti-clockwise moment, so
10N x its perp distance from pivot + 5N x its per distance from pivot
10X2 + 5X2
anticlockwise moment = 30 Nm
so resulant torque = 60 - 30
= 30 Nm
Post by: Deadly_king on October 23, 2010, 01:01:10 pm
The turning effect of a Force is given by
Force x Perpendicular distance from the pivot
resultant torque= sum of clokwise moment - sum of anti-clockwise moments
Clokwise moments:
the 20N has a clockwise moment
20 x perp distance
20X3
clockwise moment=60 Nm
Anti-clockwise moments:
the 10N and 5N have a anti-clockwise moment, so
10N x its perp distance from pivot + 5N x its per distance from pivot
10X2 + 5X2
anticlockwise moment = 30 Nm
so resulant torque = 60 - 30
= 30 Nm
Good job dear :)
Very informative explanation..........much appreciated :D
+rep after i spread the love ;)
Post by: TJ-56 on October 23, 2010, 01:41:50 pm
can anyone help with this one
(http://img231.imageshack.us/img231/6720/q36j.png)
answers A
pls explain
Post by: SpongeBob on October 23, 2010, 02:13:18 pm
Current flowing in the circuit= 2/7.5 = 4/15 A
Current flowing in each loop= 2/15
Voltage across each resistor= 5 X 2/15= 2/3 V
Across X (2 resistors) V= 4/3 V
Across Y (1 resistor) V= 2/3 V
Potential difference= Potential across X - Potential across Y= 4/3 - 2/3 = 2/3 V
Post by: TJ-56 on October 23, 2010, 03:32:39 pm
Got another question
(http://img109.imageshack.us/img109/3395/s02q28.png)
The answer is C pls explain i tried everything, couldnt get to C
thanx in advance
Post by: missbeautiful789 on October 23, 2010, 04:56:51 pm
so thats suggests that the phase difference of the two waves at that point is 180 degrees, which is the same as half wavelength
so path difference=0.5 lamda
path difference = distance S2 - distance S 1 = 0.5 lamda
so ans is C
you need to understand that at ponit X. 1 wave is at its max and 1 is at its min to make a resulting displacement of zero.
that happens at a phase diff of 180 degrees, assuming that the both have same amplitude.
you might have said ans is D, but that is only correct if it is a maxima
Post by: TJ-56 on October 23, 2010, 06:36:44 pm
(http://img227.imageshack.us/img227/968/s08q14.png)
the answers A
Can you please explain
Post by: missbeautiful789 on October 23, 2010, 07:13:24 pm
This should answer your question. If it isn't clear let me know :)
(http://i53.tinypic.com/jawhgp.jpg)
http://i53.tinypic.com/jawhgp.jpg
You made a mistake RatJumper. T1 seconds after the ball is released the ball travels a distance of x, so you cant go and say that u=0 for the journey between time interval between T2 and T1, because it would have picked up speed when it reached point M1
This is how i did it:
(http://i54.tinypic.com/zxtfzm.jpg)
<a href="http://tinypic.com?ref=zxtfzm" target="_blank"><img src="http://i54.tinypic.com/zxtfzm.jpg" border="0" alt="Image and video hosting by TinyPic">[/url]
Distance to cover h given by area of trapesium
Distance=(1/2)(a+b)(h)
h=(1/2)(aT1+aT2)(T2-T1)
h=(1/2)a(T1+T2)(T2-T1) difference of squares so>>
h=(1/2)a(T12-T22)
2h/(T12-T22)=a
Post by: $!$RatJumper$!$ on October 23, 2010, 07:22:38 pm
Post by: TJ-56 on October 23, 2010, 09:41:06 pm
thanx alot for the answer
(http://img227.imageshack.us/img227/968/s08q14.png)
the answers A
Can you please explain
can anyone help me with this ?
Post by: Deadly_king on October 24, 2010, 10:42:34 am
can anyone help me with this ?
Moment of a force is defined as the turning effect obtained by the product of force to the perpendicular distance from a particular point.
Taking the pivot at the top of the ladder where it meets the wall.
Clockwise moments = Anti-clockwise moments
Wa + Fh = W(2a)
NOTE : The perpendicular distance from the pivot to force F is h since the force is horizontal. The other force F being at the pivot has no component ;)
Post by: TJ-56 on October 24, 2010, 02:50:30 pm
Moment of a force is defined as the turning effect obtained by the product of force to the perpendicular distance from a particular point.
Taking the pivot at the top of the ladder where it meets the wall.
Clockwise moments = Anti-clockwise moments
Wa + Fh = W(2a)
NOTE : The perpendicular distance from the pivot to force F is h since the force is horizontal. The other force F being at the pivot has no component ;)
Thank you, +rep
Post by: Deadly_king on October 24, 2010, 03:24:57 pm
Thank you, +rep
Anytime dude ;)
Post by: WARRIOR on October 30, 2010, 05:45:02 pm
A block of mass m1=3kg rests on a frictionless horizontal surface.A second block of mass m2 = 2kg from a cord of negligble mass that runs over a frictionless pulley & then is connected to the first block . The blocks are released from rest
-----------
l
l
l
An idea of how it looks ^
QUESTION
(a) Find acc of the two block after they are releaased
(b)what is the velocity of the first block after 1.2s after the release of the blocks, assuming the first block does not run out of room on the table and the second block does not land on the floor?
(c)How far has block 1 moved during 1.2s interval?
(d) what is the displacement of the blocks from their initial positions .4 seconds after they are release?
--------------------------------------------------------------------------------------------------------------
A rope is attached from a truck to a car of 1400kg.The rope will break if the tension is greater than 2500N.Neglecting friction , what is the max possible acceleration of the truck if the rope does not break.Should the driver of the truck be concerned that the top might break
--------------------------------------------------------------------
In a phys lab a glider is released from rest on a frictionless air track inclined at an angle.If the glider has gained a speed of 25 cm/s in travelling 50cm from starting point , what is the angle of inclination of track
Post by: $!$RatJumper$!$ on October 30, 2010, 06:16:19 pm
Post by: WARRIOR on October 30, 2010, 06:20:44 pm
Could you tell which paper these questions are from? it would make it easier.
its questions from my school book :/
Post by: missbeautiful789 on October 30, 2010, 09:19:21 pm
Resultant force = mass x acceleration
The only force on m1 is the tension on the string. Its weight is being "neutralised" by the reaction force of the table, so forget about its weight.
so
T=3xa
For mass M2:
Its hanging, so the string is pulling it upward and its weight is acting downward. So its RF= T-mg
assuming g=10
T-20=2xa
T=2xa+20
T is the same so we can say:
3a=2a+20
a=20
B) final velocity=initial velocity + acceleration x time
v=u+at
from rest, so u=o
V=at
v=20x1.2
V=24
c) S=ut+0.5xaxt2
s=0.5x20x1.22
s=14.4
d) s=0.5x20x0.42
s=1.6
______________________________________________
2500=1400xa
a=1.79
_________________________________________
resolved force of weight down plane= mg sin@
v2=u2+2aS
625=2xax50
a=6.25
RF=ma so
mg sin@=mx6.25
mx10xsin@=mx6.25
sin@=0.625
@=38.7 degrees
Looks like homework..don't forget the units
Post by: WARRIOR on October 31, 2010, 03:12:00 am
For mass m1:yes it was infact hw!
Resultant force = mass x acceleration
The only force on m1 is the tension on the string. Its weight is being "neutralised" by the reaction force of the table, so forget about its weight.
so
T=3xa
For mass M2:
Its hanging, so the string is pulling it upward and its weight is acting downward. So its RF= T-mg
assuming g=10
T-20=2xa
T=2xa+20
T is the same so we can say:
3a=2a+20
a=20
B) final velocity=initial velocity + acceleration x time
v=u+at
from rest, so u=o
V=at
v=20x1.2
V=24
c) S=ut+0.5xaxt2
s=0.5x20x1.22
s=14.4
d) s=0.5x20x0.42
s=1.6
______________________________________________
2500=1400xa
a=1.79
_________________________________________
resolved force of weight down plane= mg sin@
v2=u2+2aS
625=2xax50
a=6.25
RF=ma so
mg sin@=mx6.25
mx10xsin@=mx6.25
sin@=0.625
@=38.7 degrees
Looks like homework..don't forget the units
thanks ! + rep
that last one was pretty hard for me! even with your answers inftont of me
thanks again
Post by: WARRIOR on October 31, 2010, 01:27:43 pm
i have a tension test and a vector and projectiles test
if anyone has any exams or notes or question and can be kind enough to post them..CHEERS!
Post by: Deadly_king on October 31, 2010, 03:01:39 pm
For mass m1:
Resultant force = mass x acceleration
The only force on m1 is the tension on the string. Its weight is being "neutralised" by the reaction force of the table, so forget about its weight.
so
T=3xa
For mass M2:
Its hanging, so the string is pulling it upward and its weight is acting downward. So its RF= T-mg
assuming g=10
T-20=2xa
T=2xa+20
T is the same so we can say:
3a=2a+20
a=20
B) final velocity=initial velocity + acceleration x time
v=u+at
from rest, so u=o
V=at
v=20x1.2
V=24
c) S=ut+0.5xaxt2
s=0.5x20x1.22
s=14.4
d) s=0.5x20x0.42
s=1.6
She made just a small mistake there.
Actually the tension is upwards towards the pulley, but the acceleration is downwards due to the weight of M2.
Hence the equation should have been 20 - T = 2a
Now you can solve it with the first equation T = 3a and you'll be getting the acceleration to be 4ms-2
b) All the other answers were affected since the acceleration was not correct. Otherwise the method was perfect ;)
v = u + at ----> v = 0 + 4(1.2) = 4.8ms-1
c) v2 = u2 + 2aS ----> 4.82= 0 + 2(4)S
Hence the distance S is found to be 2.88m
d) Displacement is given by S = ut + 0.5at2
S = 0 + 0.5(4)(0.4)2 ---> S = 0.32m
Post by: missbeautiful789 on October 31, 2010, 04:52:27 pm
Thanks for that DK
+rep
Post by: Deadly_king on October 31, 2010, 05:04:57 pm
An acceleration of 20m/s!! down :o, that should've raised a few flags. Its at falling twice g ::)
Thanks for that DK
+rep
Hehe.............i guess you were not very attentive. Otherwise you would have noticed your mistake right away ;)
No problem dear............just doing my job :)
Post by: WARRIOR on November 02, 2010, 04:04:47 pm
!
I need some notes on resolving vectors along wth questions and answers !
i have a test tomorrow !
sorry for the short notice~
Post by: Dania on November 04, 2010, 09:26:50 pm
Refer to the attached pdf.
Post by: Deadly_king on November 05, 2010, 05:20:59 am
Can someone please help me on this question on stationary waves. It's no. 4 (c)
Refer to the attached pdf.
Sound with lowest frequency is produced => An antinode is found at the end of the tube and it is the only one.
Hence there should be a node at the piston. Distance between a node and the consecutive antinode is given to be lamda/4
Lambda/4 = 45 cm -----> lambda is given to be 180cm = 1.8m
V = f x lambda
Therefore f = v/lambda = 330/1.8 = 180 Hz
Frequency is given to the nearest tenth since v = 330 and lambda=1.8 (nearest 10)
Hope it helps :)
Post by: Dania on November 05, 2010, 11:24:14 am
Also, there's something I always struggle to grasp. There are questions where they give a diagram with a wave on a graph & tell you that there's another wave, giving you the phase angle of the latter one. I don't know if you get me. But what I'm trying to say is that how would I make the necessary calculations to enable me to draw the second wave? Like the time difference, etc. Would you happen to have a link that could help me?
Ref. October/November 2002, Paper 2, Q. (b) (i).
Thanks in advance.
Post by: ruby92 on November 05, 2010, 12:58:01 pm
q2 part c(iii)
Post by: ashish on November 05, 2010, 01:53:32 pm
M/j 2010 physics p41
q2 part c(iii)
I will give you a tip try it
Internal energy = PEs + KEs
since ideal gas does not have intermolecular forces of attraction PE =o
Internal energy= KEs
= number of gas molecules * KE of one atom
Post by: TJ-56 on November 05, 2010, 09:38:09 pm
the second one is q5 (c) why isn't it a straight line from origin, why does it have a curve as it says on the mark scheme?
thanks so much for any help in advance
Post by: Deadly_king on November 06, 2010, 05:09:04 am
Thanks :)
Also, there's something I always struggle to grasp. There are questions where they give a diagram with a wave on a graph & tell you that there's another wave, giving you the phase angle of the latter one. I don't know if you get me. But what I'm trying to say is that how would I make the necessary calculations to enable me to draw the second wave? Like the time difference, etc. Would you happen to have a link that could help me?
Ref. October/November 2002, Paper 2, Q. (b) (i).
Thanks in advance.
Yeah, I understand what you mean. Once it was rather complicated for me too. ;)
You should know this formula => Phase angle = 2(pie)
This formula applies when you have a graph of x against wavelength(lambda). Since in this case the graph is x against time(t), we can change the formula to => Phase angle = 2(pie)
It has been said that both waves have the same waveform, which implies same wavelength and same period. The only difference is that one would lead the other.
Take 2(pie) as 360o since the phase angle has been given in degrees and not radians.
Hence 60o = 2(180)
Therefore the new wave will lag behind by 0.5s, i.e it will have its first maximum at 0.5s and the first minimum at 2s.
Hope you get it. :)
Am sorry but I don't have any specific sites for this topic, but here (https://studentforums.biz/revison-notes/useful-websites/) are some links which a member has been so kind to look for us. :D
Post by: Deadly_king on November 06, 2010, 06:14:06 am
Can sumone help me with diagrams in nov 2003 qp2, the first one is q3 (c) (i) just wanna make sure if its right and explain if you can pls,
the second one is q5 (c) why isn't it a straight line from origin, why does it have a curve as it says on the mark scheme?
thanks so much for any help in advance
Nov 03 p2
3.(c)(i) In both strings the tension is directed towards the pulley. But I don't know how to explain it. :-[
5.(c) Hmm...........first you should realize that it is a graph of velocity against distance from A.
Hence the gradient will be velocity/distance which is also equal to 1/time.
Therefore as the electron moves from A to B, time is increasing.
Since gradient = 1/time, an increase in time will lead to a decrease in gradient.
But velocity of the electron will be increasing due to the force of attraction from plate B and the force of repulsion from plate A.
Thus the graph should indicate an increase in speed while the gradient will be decreasing. This is why a curve is drawn instead of a straight line. ;)
Hope it helps :)
Post by: $!$RatJumper$!$ on November 06, 2010, 08:28:18 am
Yeah, I understand what you mean. Once it was rather complicated for me too. ;)
You should know this formula => Phase angle = 2(pie)/lambda
This formula applies when you have a graph of x against wavelength(lambda). Since in this case the graph is x against time(t), we can change the formula to => Phase angle = 2(pie)/T, where T is the period of oscillation and
It has been said that both waves have the same waveform, which implies same wavelength and same period. The only difference is that one would lead the other.
Take 2(pie) as 360o since the phase angle has been given in degrees and not radians.
Hence 60o = 2(180)
Therefore the new wave will lag behind by 0.5s, i.e it will have its first maximum at 0.5s and the first minimum at 2s.
Hope you get it. :)
Am sorry but I don't have any specific sites for this topic, but here (https://studentforums.biz/revison-notes/useful-websites/) are some links which a member has been so kind to look for us. :D
Excellent info. Thank you! +rep
Post by: $!$RatJumper$!$ on November 06, 2010, 08:32:00 am
D*n*(lambda) / d
D*(lambda) / d
Post by: Dania on November 06, 2010, 09:45:21 am
Post by: Deadly_king on November 06, 2010, 10:05:10 am
Do you mind drawing it out for me and uploading the pic? I get what you're saying. But I'm finding it hard to draw it. Thanks.
Am sorry but I don't have a scanner :-[
Nevertheless I'll see what i can do for you.
Post by: Deadly_king on November 06, 2010, 10:19:00 am
What is the difference between the equations:
D*n*(lambda) / d
D*(lambda) / d
D(lambda)/d : It is the distance between the centres of two consecutive bright fringes or the distance between the centres of two consecutive dark fringes.
Dn(lamda)/d : It is the distance between the centres of two bright fringes which separated by n wavelengths or the distance between the centres of two dark fringes which are separated by n wavelengths.
But am not sure about the second one since it is only rarely used. :-[
Post by: Dania on November 06, 2010, 10:40:57 am
Am sorry but I don't have a scanner :-[
Nevertheless I'll see what i can do for you.
You could take a picture with your phone, it doesn't have to be immaculate.
Post by: Dania on November 06, 2010, 10:42:40 am
The mark scheme answer is so vague. I don't get it.
Post by: TJ-56 on November 06, 2010, 10:50:35 am
Nov 03 p2Thanx for the help!
3.(c)(i) In both strings the tension is directed towards the pulley. But I don't know how to explain it. :-[
5.(c) Hmm...........first you should realize that it is a graph of velocity against distance from A.
Hence the gradient will be velocity/distance which is also equal to 1/time.
Therefore as the electron moves from A to B, time is increasing.
Since gradient = 1/time, an increase in time will lead to a decrease in gradient.
But velocity of the electron will be increasing due to the force of attraction from plate B and the force of repulsion from plate A.
Thus the graph should indicate an increase in speed while the gradient will be decreasing. This is why a curve is drawn instead of a straight line. ;)
Hope it helps :)
+rep
Post by: Hypernova on November 06, 2010, 11:01:01 am
Do you mind drawing it out for me and uploading the pic? I get what you're saying. But I'm finding it hard to draw it. Thanks.
Here.
<a href="http://tinypic.com?ref=2wpqs09" target="_blank"><img src="http://i51.tinypic.com/2wpqs09.jpg" border="0" alt="Image and video hosting by TinyPic">[/url]
(http://i51.tinypic.com/2wpqs09.jpg)
Post by: Hypernova on November 06, 2010, 11:13:51 am
Help!
The mark scheme answer is so vague. I don't get it.
Understand that the weight of the body may be considered to act at its center of gravity. Weight is a force, so if it acts about a pivot it will cause a turning effect.
We know that the carboard comes to rest, therefore we conclude that there is no turning effect.
The only time there is no turning effect is when the weight acts through the pivot.
Think about it like this...moment=F*d...so if d=0, then there is no moment.
The only force on it is weight, and weight acts through the pivot, then say there is no resulting moments.
Post by: Deadly_king on November 06, 2010, 11:21:00 am
Understand that the weight of the body may be considered to act at its center of gravity. Weight is a force, so if it acts about a pivot it will cause a turning effect.
We know that the carboard comes to rest, therefore we conclude that there is no turning effect.
The only time there is no turning effect is when the weight acts through the pivot.
Think about it like this...moment=F*d...so if d=0, then there is no moment.
The only force on it is weight, and weight acts through the pivot, then say there is no resulting moments.
Thank you very much dear member for your help :)
It's a real pleasure to have you here. :D
+rep
Post by: Deadly_king on November 06, 2010, 11:44:34 am
Here.
<a href="http://tinypic.com?ref=2wpqs09" target="_blank"><img src="http://i51.tinypic.com/2wpqs09.jpg" border="0" alt="Image and video hosting by TinyPic">[/url]
(http://i51.tinypic.com/2wpqs09.jpg)
That's exactly what I meant Dania. ;)
Post by: Dania on November 06, 2010, 01:30:38 pm
Understand that the weight of the body may be considered to act at its center of gravity. Weight is a force, so if it acts about a pivot it will cause a turning effect.
We know that the carboard comes to rest, therefore we conclude that there is no turning effect.
The only time there is no turning effect is when the weight acts through the pivot.
Think about it like this...moment=F*d...so if d=0, then there is no moment.
The only force on it is weight, and weight acts through the pivot, then say there is no resulting moments.
Thank You! That was very helpful.
Post by: Dania on November 06, 2010, 05:10:56 pm
Post by: Hypernova on November 07, 2010, 07:37:53 am
Okay, so I seem to be stuck on (b) [Refer to the attachment]. Would I use the equation "Phase angle = 2(pie)x/lambda"?
I don't know the DK's method, hopefully, he'll explain it to you.
This is what i do.
First find the amplitude of the second waveform.
We know that the intensity of the second wave is doulde the first,
So, 2I1=I2
Intesity is proportional to amplitude2
I1=K x 52 I2= K x a2
K=I1/52 K= I2/a2
now equate the two Ks
I1/52 = I2/a2
using 2I1=I2, we say: 2= I2/I1
I1/52 = I2/a2
I2/I1 = a2/25
2 = a2/25
50 = a2
a = 7
Now draw the graph.
See I just know how to draw the graph. If the Wave has a hase difference of 180 degress, then it is completely out of phase,
If it is 90, then a 0.25 out of phase.
In other words. Look at the graph they gave you, 0.6 is one wavelength. So from 0 to O.6 is 360 degrees. so at 0.3, the phase angle of that point is 180 degrees. The new wave, when it starts from 0, will be in phase with the wave at this point.
So my wave looks like the one attached.
Post by: $!$RatJumper$!$ on November 07, 2010, 08:34:57 am
Post by: $!$RatJumper$!$ on November 07, 2010, 10:13:03 am
Torque = Force * Perp Distance between two forces
Post by: Hypernova on November 07, 2010, 12:30:46 pm
a 'moment' is called 'torque' when it involves a couple.
Look at this
Torque = Force * Perp Distance between two forces
Torque= Force x (Perp Distance between pivot x 2)
Torque= 2 x Force x Perp Distance between pivot
Moment = Force1 x Perp Distance between pivot + Force2 x Perp Distance between pivot (Since F1=F2)
Moment= 2 x Force x Perp Distance between pivot
There is no difference bettween them, just the definition.
Post by: $!$RatJumper$!$ on November 07, 2010, 12:51:42 pm
Torque and moment is essentially the same, both being the turning effect of a force
a 'moment' is called 'torque' when it involves a couple.
Look at this
Torque = Force * Perp Distance between two forces
Torque= Force x (Perp Distance between pivot x 2)
Torque= 2 x Force x Perp Distance between pivot
Moment = Force1 x Perp Distance between pivot + Force2 x Perp Distance between pivot (Since F1=F2)
Moment= 2 x Force x Perp Distance between pivot
There is no difference bettween them, just the definition.
Thank you :) that makes a lot of sense :)
Another question, S04 Q 4 (b)(i), what is the direction of the force?
Post by: Hypernova on November 07, 2010, 01:02:07 pm
Thank you :) that makes a lot of sense :)
Another question, S04 Q 4 (b)(i), what is the direction of the force?
Upward
Use Newton's Third law: To every action there is always an equal and opposite reaction: or the forces of two bodies on each other are always equal and are directed in opposite directions.
Basically, The Ball exerts a downward force on the ground, so the ground will exert an equal force but opposite in direction (upward) on the ball.
Post by: $!$RatJumper$!$ on November 07, 2010, 01:04:05 pm
Post by: Hypernova on November 07, 2010, 01:10:40 pm
gr8 :) thankx +rep
I assumed that you were you ok with finding its magnitude. Should I do it?
Post by: $!$RatJumper$!$ on November 07, 2010, 01:11:15 pm
Post by: TJ-56 on November 07, 2010, 01:18:10 pm
I have a question, regarding a drawing again pls
November 2005 paper 2 q8 (b) (ii) (last question)
can you draw it for me pls? the mark scheme is rly confusing me
thank you so much
Post by: $!$RatJumper$!$ on November 07, 2010, 01:29:08 pm
Im finding it difficult to 'spread the love' deadly king, i will +rep the moment i can do so,
I have a question, regarding a drawing again pls
November 2005 paper 2 q8 (b) (ii) (last question)
can you draw it for me pls? the mark scheme is rly confusing me
thank you so much
The equation for Gravitational Potential Energy is: mgh. Now it says that a ball is falling vertically through the air. Which means, that it was initially dropped mid air and is thus falling towards the ground.
From this we can conclude that since it is falling towards the ground, its height above the ground is decreasing as it falls closer. We also know that the mass "m" of the ball and the gravitational feild strength "g" are constant and do not change. So the only thing changing is the value of "h" in the equation. Since it is decreasing and everything else staying constant, the value for the Gravitational Potential Energy will decrease as a result.
This is why the mark scheme says negative gradient because the value for Gravitational Potential Energy is decreasing. Since it was at its maximum height when it was dropped, it will have maximum Gravitational Potential Energy thus the line starts at the maximum y intercept above Ek and ends at h0 on the x-axis.
Post by: TJ-56 on November 07, 2010, 01:32:25 pm
The equation for Gravitational Potential Energy is: mgh. Now it says that a ball is falling vertically through the air. Which means, that it was initially dropped mid air and is thus falling towards the ground.thanx +rep that cleared it up
From this we can conclude that since it is falling towards the ground, its height above the ground is decreasing as it falls closer. We also know that the mass "m" of the ball and the gravitational feild strength "g" are constant and do not change. So the only thing changing is the value of "h" in the equation. Since it is decreasing and everything else staying constant, the value for the Gravitational Potential Energy will decrease as a result.
This is why the mark scheme says negative gradient because the value for Gravitational Potential Energy is decreasing. Since it was at its maximum height when it was dropped, it will have maximum Gravitational Potential Energy thus the line starts at the y intercept and ends at h0.
just one last point, why does it start from above the maximum kinetic energy as stated on the ms?
Post by: $!$RatJumper$!$ on November 07, 2010, 01:39:33 pm
It starts above because it has the maximum possible gravitational energy when it is at its maximum point just as it is released. The point is, that you need to start your negative sloping line right at the top of the y-axis just to show the examiner that you understand it is at its maximum energy initially. It has to have nothing to do with the kinetic energy because they are separate things. Yes, they are related, but not in such a way that they need to start at the same point.
Post by: TJ-56 on November 07, 2010, 01:51:15 pm
Thankx for the +rep :)thanx alot, i finally got it!
It starts above because it has the maximum possible gravitational energy when it is at its maximum point just as it is released. The point is, that you need to start your negative sloping line right at the top of the y-axis just to show the examiner that you understand it is at its maximum energy initially. It has to have nothing to do with the kinetic energy because they are separate things. Yes, they are related, but not in such a way that they need to start at the same point.
Post by: $!$RatJumper$!$ on November 07, 2010, 02:01:05 pm
Post by: WARRIOR on November 07, 2010, 02:24:34 pm
Post by: $!$RatJumper$!$ on November 07, 2010, 02:36:40 pm
Post by: $!$RatJumper$!$ on November 07, 2010, 02:39:48 pm
Thankx
Post by: Deadly_king on November 07, 2010, 03:42:21 pm
What will happen to the water level if someone was sitting in a boat floating on a very small pond and then he take the anchor out of the boat and drop it into the water? ( assuming the anchor reached the bottom of the sea and the boat stops )
The level of water will rise.
But we may not note this rise due to the large surface area covered by the pond, even if the pond is very small. ;)
The volume displaced by the anchor upon sinking may be negligible compared to the total volume of water in the pond.
Post by: TJ-56 on November 07, 2010, 04:40:49 pm
W04 Q1 (b)(ii) pleaseIf the percentage uncertainty of the diameter is (0.02/0.50)*100% = 4%
Area is pi* r^2 , so to calculate the uncertainty of a squared value, we multiply it by the power, and here the power is 2, so 4% *2 = 8%
Hope that was helpful
Post by: TJ-56 on November 07, 2010, 04:44:53 pm
And may someone please put the values of all the frequencies & wavelengths in the electromagnetic spectrum.
Thankx
Post by: Deadly_king on November 07, 2010, 04:48:20 pm
If the percentage uncertainty of the diameter is (0.02/0.50)*100% = 4%
Area is pi* r^2 , so to calculate the uncertainty of a squared value, we multiply it by 2m so 8%
Better?
Good job dude :)
You may look here (http://hyperphysics.phy-astr.gsu.edu/hbase/hframe.html) as well for the electromagnetic spectrum. (click on electromagnetic spectrum in the index on your right ) ;)
Post by: SpongeBob on November 07, 2010, 05:02:55 pm
Post by: ruby92 on November 07, 2010, 05:32:29 pm
I will give you a tip try it
Internal energy = PEs + KEs
since ideal gas does not have intermolecular forces of attraction PE =o
Internal energy= KEs
= number of gas molecules * KE of one atom
thanks :)
Post by: TJ-56 on November 07, 2010, 06:57:52 pm
thanx for the help
Post by: $!$RatJumper$!$ on November 07, 2010, 08:03:31 pm
Can someone solve s07 qp2 q4 (d) both parts, literally driving me crazy, and if u can pls post the vector diagram,
thanx for the help
Hope this helps :)
Post by: TJ-56 on November 07, 2010, 08:59:40 pm
+rep
Post by: Hypernova on November 07, 2010, 09:12:12 pm
The level of water will rise.
But we may not note this rise due to the large surface area covered by the pond, even if the pond is very small. ;)
The volume displaced by the anchor upon sinking may be negligible compared to the total volume of water in the pond.
wait a sec.
First the anchor is on the boat. It exerts a downward force on the boat, forcing the boat down, so displacing alot of water.
When the anchor is released, less of the ship is in the water therefore less water diplacement. The volume diplaced by the anchor is small.
The higher the water displacement, the higher the water level.
Since water displacement decreased the water level should also decrease.
Imagine yourself in a bathtub. Imagine that the anchor is on you, you will go deeper into the water, the water will rise. when the anchor is released, less of you will be in the water, so the water level will decrease.
Post by: astarmathsandphysics on November 07, 2010, 10:22:15 pm
If anyone could help me with this? (:
Need part i) of the question. :D
Post by: astarmathsandphysics on November 07, 2010, 10:26:59 pm
If the anchor hits the bottom then archimedes principle only applies to the boat. The weight of water displaced equals the weight of the boat only, so less water is displaced and the water level falls.
Post by: WARRIOR on November 08, 2010, 03:29:27 am
Post by: birchy33 on November 08, 2010, 03:53:12 am
Question number 5a and 5b. Sorry i don't know how to link it :( but help would be appreciated because I'm just stuck for all money on how to approach it. Cheers guys.
Post by: Deadly_king on November 08, 2010, 03:59:19 am
First the anchor is on the boat. It exerts a downward force on the boat, forcing the boat down, so displacing alot of water.
When the anchor is released, less of the ship is in the water therefore less water diplacement. The volume diplaced by the anchor is small.
The higher the water displacement, the higher the water level. Since water displacement decreased the water level should also decrease.
Imagine yourself in a bathtub. Imagine that the anchor is on you, you will go deeper into the water, the water will rise. when the anchor is released, less of you will be in the water, so the water level will decrease.
If the anchor doesnt reach the bottom the overall water level will not change cos the weight of water displaced equals weight of boat + weight of anchor.
If the anchor hits the bottom then archimedes principle only applies to the boat. The weight of water displaced equals the weight of the boat only, so less water is displaced and the water level falls.
That's right. Sorry for the confusion Kimo :-[
Post by: Deadly_king on November 08, 2010, 06:32:30 am
Hi I'm New to the forum and need some help with this question. Its from May/June 2009 paper 1st variant.
Question number 5a and 5b. Sorry i don't know how to link it :( but help would be appreciated because I'm just stuck for all money on how to approach it. Cheers guys.
Welcome to the forum.
I'll advice you to introduce yourself in the Introduction thread. ;)
For your question, is it paper 2 or paper 4 ???
Post by: $!$RatJumper$!$ on November 08, 2010, 06:33:50 am
Post by: $!$RatJumper$!$ on November 08, 2010, 07:19:15 am
Q3 (c)- what is the direction?
Q4 (a)(ii)- why doesnt it work when I find the area under of the triangle for the first 0.3 sec then times it by 2
Q5 (c)(ii)- please explain
Thank you
Post by: birchy33 on November 08, 2010, 08:07:41 am
Post by: astarmathsandphysics on November 08, 2010, 09:05:10 am
Post by: astarmathsandphysics on November 08, 2010, 09:22:18 am
Q4 (a)(ii)- why doesnt it work when I find the area under of the triangle for the first 0.3 sec then times it by 2
Q5 (c)(ii)- please explain
3c)36 degrees below the horizontal, at right angles to the upper string
4aii) It should work the way you did it. I get about 2*1/2*0.3*0.10 =0.03 by APPROXIMATING THE slope by a line
5vii)amplitude of A=sqrt(1) and amplitude of B=sqrt(4/9)=2/3
1-2/3=1/3 then 1/3^2=1/9 . Subtract since there are 180 degrees out of phase
Post by: $!$RatJumper$!$ on November 08, 2010, 11:36:49 am
Q3 (c)- what is the direction?
Q4 (a)(ii)- why doesnt it work when I find the area under of the triangle for the first 0.3 sec then times it by 2
Q5 (c)(ii)- please explain
3c)36 degrees below the horizontal, at right angles to the upper string
4aii) It should work the way you did it. I get about 2*1/2*0.3*0.10 =0.03 by APPROXIMATING THE slope by a line
5vii)amplitude of A=sqrt(1) and amplitude of B=sqrt(4/9)=2/3
1-2/3=1/3 then 1/3^2=1/9 . Subtract since there are 180 degrees out of phase
i really dont understand 3 and 5.. please could you elaborate :)
and fir 4, the mark scheme says horizontal.. doesnt that mean 90 degrees? so why is it 36?
thankx a lot
Post by: TJ-56 on November 08, 2010, 12:01:48 pm
i really dont understand 3 and 5.. please could you elaborate :)For q5 (c):
and fir 4, the mark scheme says horizontal.. doesnt that mean 90 degrees? so why is it 36?
thankx a lot
To find the resultant intensity, you have to find the resultant amplitude first, since they are 180 degrees out of phase, find the difference between the amplitudes of 2 peaks (or troughs) which is (-) 1, so I(resultant)/I=1(sqr)/3(sqr) simplified gives the resultant intensity of 1/9.
(you could also use the intensity 4/9 * I but with amplitude of 2 squared)
Hope that was helpful.
Post by: $!$RatJumper$!$ on November 08, 2010, 12:06:32 pm
Post by: TJ-56 on November 08, 2010, 12:09:46 pm
ok thankx that makes sense now :) and jus a query, when finding resultant amplitudes, dont we ADD the amplitudes. like in this question, i think they added them in order to get their resultant amplitude as 0.0001That is correct, we add them.
Take for example the 1st peak on wave A, wave A's amp is 3, and wave B's amp is -2 at that time, so 3 + (-2) is 1 x 10^-4
Post by: $!$RatJumper$!$ on November 08, 2010, 12:11:15 pm
Post by: TJ-56 on November 08, 2010, 12:18:28 pm
ok :) so is it a rule that we should ALWAYS ADD amplitudes to find the resultant?I think so yeah, theres another question regarding this topic on w02 qp2 q5 b iii , so, positive
Post by: $!$RatJumper$!$ on November 08, 2010, 12:32:24 pm
umm and could you also explain why 3(c) direction is 35 below the horizontal. isnt it just 90 degrees?
Post by: $!$RatJumper$!$ on November 08, 2010, 12:34:06 pm
Q4 (a)(ii)- why doesnt it work when I find the area under of the triangle for the first 0.3 sec then times it by 2
4aii) It should work the way you did it. I get about 2*1/2*0.3*0.10 =0.03 by APPROXIMATING THE slope by a line
why did u use 0.1
Post by: TJ-56 on November 08, 2010, 12:51:38 pm
Thankx for all ur help +repI think you just need to state that the direction is horizontal, but its not 90 degrees, its 90 degrees when its in the center (i.e. the weight acts through the pivot), but no need since they dont want the angle
umm and could you also explain why 3(c) direction is 35 below the horizontal. isnt it just 90 degrees?
Post by: $!$RatJumper$!$ on November 08, 2010, 01:04:05 pm
Post by: TJ-56 on November 08, 2010, 01:17:25 pm
ok.. and can u explain 4(a)(ii) .. why should we use 0.1Ok i didnt understand the calculation that much, but heres what i did, ur way, calc till 0.3 seconds then multiplied the value by 2:
Divided the area into 2 triangles 2 squares and 1 trapezium (thats alot but its the only way it matched the ms value): so respectively:
triangles : 0.1*4*0.5 + 0.1*2.9*0.5
squares: .2*4 + 0.1*2
trapezium: 0.1 * (0.9+2 / 2)
Multiply the sum by 2, then by 10^-2 (in metres) gives us 0.0298 which is 0.030 rounded to 2 sf
Post by: $!$RatJumper$!$ on November 08, 2010, 01:40:11 pm
Post by: TJ-56 on November 08, 2010, 01:45:37 pm
thank you very much. +repAnytime :)
thanx for the rep
Post by: $!$RatJumper$!$ on November 08, 2010, 01:55:19 pm
Post by: astarmathsandphysics on November 08, 2010, 02:46:13 pm
which paper
Post by: Deadly_king on November 08, 2010, 02:54:58 pm
S06 Q6 please explain
which paper
I think it should be paper 2 since he is sitting for AS. ;)
Post by: $!$RatJumper$!$ on November 08, 2010, 03:07:03 pm
Post by: joel on November 09, 2010, 03:41:48 am
paper found on freeexampapers.com
Post by: ashish on November 09, 2010, 07:11:42 am
How do you do Question No. 3, part (a), from May/June 2001 Paper 2 (CIE)
paper found on freeexampapers.com
a) assuming no energy is loss
loss in PE = gain in KE
mgh=1/2 mv2
m cancel out
(10*0.13*2)1/2=v
v=1.6 ms-1
b) since momentum is conserved
m*1.6 = m*v + m*0.9
m cancel out
1.6=v+0.9
v=0.7
c) inelastic collision since KE has not been conserved
Post by: $!$RatJumper$!$ on November 09, 2010, 09:11:28 am
Please explain.
Post by: TJ-56 on November 09, 2010, 11:01:27 am
S06 Q6 P2 and W07 Q 5(a) P2
Please explain.
W07 Q 5(a) P2
phi/360=AB/lambda
(60/360) * lambda = AB
AB = 1/6 * 6
= 1
So it leads or lags behind (question didnt state which) by 1
see attachment
S06 Q6 P2
What part didnt u understand exactly?
Post by: Deadly_king on November 09, 2010, 11:09:48 am
W07 Q 5(a) P2
phi/360=AB/lambda
(60/360) * lambda = AB
AB = 1/6 * 6
= 1
So it leads or lags behind (question didnt state which) by 1
see attachment
S06 Q6 P2
What part didnt u understand exactly?
Good work :)
He said he need the whole question. Do you mind to take care of it?
I've to help others.
Post by: $!$RatJumper$!$ on November 09, 2010, 11:28:31 am
W07 Q 5(a) P2
phi/360=AB/lambda
(60/360) * lambda = AB
AB = 1/6 * 6
= 1
So it leads or lags behind (question didnt state which) by 1
see attachment
S06 Q6 P2
What part didnt u understand exactly?
Thankx that makes sense about the phase diff. But dont we need to do anything about the amplitude since it mentions its intensity is half? even the mark scheme says something about amplitude.
And if you could do all parts on q6 it would be wonderful :)
thankx
Post by: TJ-56 on November 09, 2010, 11:30:21 am
Good work :)
He said he need the whole question. Do you mind to take care of it?
I've to help others.
(b) 32.4 cm is the distance between 1 nodes, i.e. half the distance of a complete wavelength. v=f * lambda
frequencey is stated, 512 Hz
Lambda= 32.4*2 *10^-2 (im meters)
so 512*64.8 * 10^-2 = 330 m\s
(c) the exact distance of the antinode is half 32.4
So 32.4/2 = 16.2
So the antinode is 16.2 cm above the surface of water in the tube on the left.
As 15.7 cm is the length of the column of air INSIDE the tube, then 16.2-15.7 is the length of air above the where the antinode is located.
Hope that was helpful
Post by: TJ-56 on November 09, 2010, 11:32:17 am
Thankx that makes sense about the phase diff. But dont we need to do anything about the amplitude since it mentions its intensity is half? even the mark scheme says something about amplitude.
And if you could do all parts on q6 it would be wonderful :)
thankx
oh sorry my bad, i will post the correct answer with the amplitude part in a few min
Post by: TJ-56 on November 09, 2010, 11:43:34 am
Thankx that makes sense about the phase diff. But dont we need to do anything about the amplitude since it mentions its intensity is half? even the mark scheme says something about amplitude.
And if you could do all parts on q6 it would be wonderful :)
thankx
say the maximum amp of the given wave is 2 (taking each 5 small blocks as 1)
then we have:
I(new)/I = A(new)²/A²
0.5I/I = A(new)²/(2A)²
Making A(new) ² the subject gives us = sqr root(2A²)
= sqr root(2) * A
which is 1.4 (the new amplitude)
Post by: $!$RatJumper$!$ on November 09, 2010, 12:23:08 pm
say the maximum amp of the given wave is 2 (taking each 5 small blocks as 1)
then we have:
I(new)/I = A(new)²/A²
0.5I/I = A(new)²/(2A)²
Making A(new) ² the subject gives us = sqr root(2A²)
= sqr root(2) * A
which is 1.4 (the new amplitude)
Thank you so much :) i finally understand how to do it :) just to make sure can u please draw it out again? thank you. and im cheking out the other answers for my other question now. thankx again! +rep
Post by: $!$RatJumper$!$ on November 09, 2010, 12:28:28 pm
(b) 32.4 cm is the distance between 1 nodes, i.e. half the distance of a complete wavelength. v=f * lambda
frequencey is stated, 512 Hz
Lambda= 32.4*2 *10^-2 (im meters)
so 512*64.8 * 10^-2 = 330 m\s
(c) the exact distance of the antinode is half 32.4
So 32.4/2 = 16.2
So the antinode is 16.2 cm above the surface of water in the tube on the left.
As 15.7 cm is the length of the column of air INSIDE the tube, then 16.2-15.7 is the length of air above the where the antinode is located.
Hope that was helpful
that was very helpful thankx! though just a question.. why did you only draw half a wavelength in the tube? are there certain rules as in how many wavelengths to draw when drawing stationary waves
Post by: TJ-56 on November 09, 2010, 12:34:15 pm
Thank you so much :) i finally understand how to do it :) just to make sure can u please draw it out again? thank you. and im cheking out the other answers for my other question now. thankx again! +repThere u go
Post by: $!$RatJumper$!$ on November 09, 2010, 12:47:39 pm
Post by: TJ-56 on November 09, 2010, 12:53:57 pm
gr8 stuff :) thankx.. could you please answer my other question aswell about the stationary wavesI didnt understand ur question, the qp though implies that only half a wavelength should be drawn, (difference between 2 loud sounds)
Post by: $!$RatJumper$!$ on November 09, 2010, 01:08:34 pm
I didnt understand ur question, the qp though implies that only half a wavelength should be drawn, (difference between 2 loud sounds)
it just simply says sketch the form of the stationery wave set up in the tube. no mention of only half a wavelength
Post by: TJ-56 on November 09, 2010, 01:19:19 pm
can someone state 2 acceptable experiments?
for the first one i drew a light source and an obstacle, where light is diffracted and an eye at the other end, which observes the diffracted light (and cannot see the source)
for the second one i drew a person shouting, with an obstacle in front of him, and an observer behind the obstacle , where the observer can hear the shouting after being diffracted
Are these correct responses?
thanx in advance
Post by: TJ-56 on November 09, 2010, 01:29:29 pm
it just simply says sketch the form of the stationery wave set up in the tube. no mention of only half a wavelengthOk what do u mean by certain rules?
The tube on the left show an anti node at the top,
To find the NEXT antinode, half a wavelength is drawn,
if we were to draw a complete wavelength instead, 3 anti nodes will result, but they want the NEXT antinode, which is formed by drawing half a wavelength
sorry but i dont know how to explain it more than that,
did that cover ur question?
maybe i didnt fully understand ur que
Post by: $!$RatJumper$!$ on November 09, 2010, 01:48:37 pm
Post by: $!$RatJumper$!$ on November 09, 2010, 01:56:11 pm
w08 q6 (b)
can someone state 2 acceptable experiments?
for the first one i drew a light source and an obstacle, where light is diffracted and an eye at the other end, which observes the diffracted light (and cannot see the source)
for the second one i drew a person shouting, with an obstacle in front of him, and an observer behind the obstacle , where the observer can hear the shouting after being diffracted
Are these correct responses?
thanx in advance
I drew a light source with a cardboard with a slit in it to act as the place where the diffraction will occur. Then a screen where the diffracted light will be seen.
In your case I dont think we can use an eye as a suitable detection media. We will need to use some sort of equipment in order for us to interpret/observe the diffracted light.
For the second one i simply drew a sound source with a small obstacle and had a microphone and cathode ray oscilloscope on the other end to see the wave pattern on the CRO.
I think your second one is fine.
You may want to check the mark scheme to see what cambridge are looking for exactly in your answers
Post by: TJ-56 on November 09, 2010, 02:01:40 pm
I drew a light source with a cardboard with a slit in it to act as the place where the diffraction will occur. Then a screen where the diffracted light will be seen.
In your case I dont think we can use an eye as a suitable detection media. We will need to use some sort of equipment in order for us to interpret/observe the diffracted light.
For the second one i simply drew a sound source with a small obstacle and had a microphone and cathode ray oscilloscope on the other end to see the wave pattern on the CRO.
I think your second one is fine.
You may want to check the mark scheme to see what cambridge are looking for exactly in your answers
ok thanx, just wanted to make sure
Post by: lana on November 09, 2010, 02:58:03 pm
thanks =]
Post by: $!$RatJumper$!$ on November 09, 2010, 03:31:27 pm
may june 03 question 3 b(i) please
thanks =]
The circumference of a circle is known to be 2*pie*r OR d*pie
In this case we will use d*pie since they have given us the diameter as 3.
So the actual circumference is 3*pie. But then it says that it is turned by 6.5 degrees. Now to find the amount it moved in terms of its circumference, we use: (6.5/360)*(3*pie) = 0.17 cm
P.S we divided it by 360 because we need to express 6.5 in terms of its angle of deflection.
Hope that helped :)
Post by: lana on November 09, 2010, 03:37:26 pm
+rep
Post by: $!$RatJumper$!$ on November 09, 2010, 09:12:16 pm
thankx :)
Post by: astarmathsandphysics on November 09, 2010, 11:27:35 pm
a) s1 closed so no current or power
b) s1 and s2 closed so voltage acros A is 240 so power is 1.5KW
c)1.5W+1.5KW since voltage across ABC is 240V, 0V and 240V resp
d)voltage across A and B is 120V so using P=V^2/R power in each element is 1/4 of 1.5KW=375 W and total power is 750W
e)voltage across A and B is 120V so using P=V^2/R power in each element is 1/4 of 1.5KW=375 W and total power is 750W +1.5KW in C=2.25KW
Can you upload S09 P21 Q5b cos I don't have it :-[
Post by: $!$RatJumper$!$ on November 09, 2010, 11:52:39 pm
s09 p2
a)s1 closed no no current or power
b) s1 and s2 closed so voltage acros A is 240 so power is 1,5KW
c)1.5W+1.5KW since voltage across ABC is 240V, 0V and 240V resp
d)voltage across A and B is 120V so using P=V^2/R power in each element is 1/4 of 1.5KW=375 W and total power is 750W
e)voltage across A and B is 120V so using P=V^2/R power in each element is 1/4 of 1.5KW=375 W and total power is 750W +1.5KW in C=2.25KW
Can you upload S09 P21 Q5b cos I dont have it
thankx for that :)
ok iv attached s09 here..
and could you also help me with s10 P22 Q4c
Post by: ashish on November 10, 2010, 04:00:49 am
i have done it but a bit like mechanic question
here it is
8.0a = T
2.0a=20-T-6
after solving it i got 1.4ms-2
there is another method of working it! can anyone please help me with the other method !
Post by: $!$RatJumper$!$ on November 10, 2010, 06:41:53 am
NOV 2008 p1 no.11
i have done it but a bit like mechanic question
here it is
8.0a = T
2.0a=20-T-6
after solving it i got 1.4ms-2
there is another method of working it! can anyone please help me with the other method !
This is how i would have done it. But i do not get the answer as 1.4ms-2. Can someone tell me what im doing wrong. Thankx
The 2kg mass hanging would be the force that pulls the 8kg mass. Since it is hanging gravity would act on it making it have a force of 2*9.81 = 19.62N
Since frictional forces are 6N, we do, 19.62-6 = 13.62N
Thats the resultant force. Now we use F=ma to get its acceleration.
13.62/8 = 1.7 ms-2
Post by: Deadly_king on November 10, 2010, 11:56:45 am
This is how i would have done it. But i do not get the answer as 1.4ms-2. Can someone tell me what im doing wrong. Thankx
The 2kg mass hanging would be the force that pulls the 8kg mass. Since it is hanging gravity would act on it making it have a force of 2*9.81 = 19.62N
Since frictional forces are 6N, we do, 19.62-6 = 13.62N
Thats the resultant force. Now we use F=ma to get its acceleration.
13.62/8 = 1.7 ms-2
You cannot take the acceleration to be 9.81 since that would not be the case due to the tension in the string which is linked to the other mass.
Using Newton's 2nd law of motion for the 2kg mass => W - T = 2a
W : weight of the object.(2 x 9.81)
NOTE : Newton's 2nd law applies only to the resultant force. Here the resultant is W - T.
So you have to do it the way ashish did it. You need to find two equations involving the two variables you're to find; acceleration and tension.
Then you need to solve them simultaneously to get the required solution which is a = 1.4ms-2.
Post by: Deadly_king on November 10, 2010, 12:34:13 pm
and could you also help me with s10 P22 Q4c
Jun 10 p22
4.(c) Sound with lowest frequency is produced => An antinode is found at the end of the tube and it is the only one.
Hence there should be a node at the piston. Distance between a node and the consecutive antinode is given to be lamda/4
Lambda/4 = 45 cm -----> lambda is given to be 180cm = 1.8m
V = f x lambda
Therefore f = v/lambda = 330/1.8 = 180 Hz
Frequency is given to the nearest tenth since v = 330 and lambda=1.8 (nearest 10)
Hope it helps :)
Post by: Arthur Bon Zavi on November 10, 2010, 01:13:29 pm
Post by: Deadly_king on November 10, 2010, 01:31:30 pm
+rep Deadly_king.
Hehe...........thank you Mr the Ancestor ;D
Post by: Dania on November 10, 2010, 02:29:21 pm
Look at the paper, for the full question. [Question no. 4]
Thanks in advance.
Post by: $!$RatJumper$!$ on November 10, 2010, 03:04:23 pm
Jun 10 p22
4.(c) Sound with lowest frequency is produced => An antinode is found at the end of the tube and it is the only one.
Hence there should be a node at the piston. Distance between a node and the consecutive antinode is given to be lamda/4
Lambda/4 = 45 cm -----> lambda is given to be 180cm = 1.8m
V = f x lambda
Therefore f = v/lambda = 330/1.8 = 180 Hz
Frequency is given to the nearest tenth since v = 330 and lambda=1.8 (nearest 10)
Hope it helps :)
Thank you DK :)
Another one, please explain
S09 P21 Q5b
W09 P22 Q5b
Thank you :)
Post by: Deadly_king on November 10, 2010, 03:12:56 pm
Refer to the attachments
Look at the paper, for the full question. [Question no. 4]
Thanks in advance.
I assume that you got the answers for part (a)(ii).
You should then use the two answers namely R =
As you can note, you have been given R = 0.44,
But according to the equation you cannot find it since you don't have L and A. However you can use the 1st equation to find it. ;)
1st equation : R =
Now you replace it in the second equation.
E = W/e x L/A. Since L/A = R/
Hope it helps :)
Post by: Deadly_king on November 10, 2010, 03:43:14 pm
W09 P22 Q5b
Thank you :)
Nov 09 p22
(b) There is no calculation here.
You should have guessed that by the number of marks allocated for the question.
It said state; which means it's pretty obvious from the drawing.
First thing : You should realize that both X and Y are equidistant from a node. Therefore the phase difference should be pie rad or 180o.
Furthermore it can also be noted that X and Y are in antiphase ( One is above while the other one is below). If Y was 1/8L to the right of P, then Y would have been at a phase difference of 180o but in phase. ;)
Hope it helps :)
Post by: TJ-56 on November 10, 2010, 04:12:32 pm
thank you
Post by: $!$RatJumper$!$ on November 10, 2010, 04:16:02 pm
Nov 09 p22
(b) There is no calculation here.
You should have guessed that by the number of marks allocated for the question.
It said state; which means it's pretty obvious from the drawing.
First thing : You should realize that both X and Y are equidistant from a node. Therefore the phase difference should be pie rad or 180o.
Furthermore it can also be noted that X and Y are in antiphase ( One is above while the other one is below). If Y was 1/8L to the right of P, then Y would have been at a phase difference of 180o but in phase. ;)
Hope it helps :)
Hmm im still confused. What i thought was that since X was 1/8th of the wavelength away from the start and Y was 1/8th of the wavelength away from the end, they would each be 45 degrees away from the start and end. Thus 360-(45*2) = 270 degrees.
So what if point X was at the end to the left (by the oscillator) and point Y was at the end to the right (by P), what would the phase diff be then?
:(
Post by: Dania on November 10, 2010, 04:41:49 pm
I assume that you got the answers for part (a)(ii).
You should then use the two answers namely R =L/A and E = WL/Ae.
As you can note, you have been given R = 0.44,
But according to the equation you cannot find it since you don't have L and A. However you can use the 1st equation to find it. ;)
1st equation : R =
Now you replace it in the second equation.
E = W/e x L/A. Since L/A = R/
Hope it helps :)
Thanks! Got it.
Post by: Deadly_king on November 10, 2010, 04:47:01 pm
Hmm im still confused. What i thought was that since X was 1/8th of the wavelength away from the start and Y was 1/8th of the wavelength away from the end, they would each be 45 degrees away from the start and end. Thus 360-(45*2) = 270 degrees.
So what if point X was at the end to the left (by the oscillator) and point Y was at the end to the right (by P), what would the phase diff be then?
:(
Ooh..........i understand what you did. But we don't measure phase difference like this. :-[
If the wave is placed on a graph paper, then the x-axis would be length and not angle. This is why you can't use length as angle.
Phase difference is described by the positions of the wave at the respective points. Whether at X or at Y, the waves undergo similar but opposite motion. This is why they are said to be in anti-phase.
For your question, The phase angle would still be 180o since even then both points will undergo similar motions but will be in phase as they are not opposite.
Post by: $!$RatJumper$!$ on November 10, 2010, 04:54:00 pm
Ooh..........i understand what you did. But we don't measure phase difference like this. :-[
If the wave is placed on a graph paper, then the x-axis would be length and not angle. This is why you can't use length as angle.
Phase difference is described by the positions of the wave at the respective points. Whether at X or at Y, the waves undergo similar but opposite motion. This is why they are said to be in anti-phase.
For your question, The phase angle would still be 180o since even then both points will undergo similar motions but will be in phase as they are not opposite.
Alright i get it now thank you :) +rep
Can you please do S09 P21 Q5b
Thanks
Post by: Deadly_king on November 10, 2010, 04:56:54 pm
Can sumone pls explain s10 P2 Q7 (b) (ii)
thank you
7.(b)(ii)
1. 1.1 MeV is supplied in the form of kinetic energy of the alpha-particle.
2. First you should convert 1.1 MeV into joules. => K.E = (1.1 x 106) x (1.6 x 10-19) = 1.76 x 10-13 J
Now you equate K.E = 0.5mv2 = 1.76 x 10-13
Mass of an alpha-particle is 4u = 4(1.67 x 10-27)kg
Hence speed of alpha-particle is found to be the square root of 2(1.76 x 10-13)/4(1.67 x 10-27)
v = 7.3 x 106 ms-1
Hope it helps :)
Post by: Dania on November 10, 2010, 05:03:57 pm
The whole question.
Post by: $!$RatJumper$!$ on November 10, 2010, 05:15:05 pm
Help! Please. :)
The whole question.
Haha looks like the same question is getting to all of us :P I also need help on that :P
Post by: TJ-56 on November 10, 2010, 09:56:29 pm
Haha looks like the same question is getting to all of us :P I also need help on that :P5(a)
1. The waves should meet in antiphase at M. ( phase difference = pie)
2. Sources must emit waves having same amplitude.
(b)
So the wavelength ranges from 330/4000 to 330/1000 (v/f) which is 8.25 to 33 cm
The distance from S2 to M is 128 (root(100sqr +80sqr))
So the phase difference is 128-100 which is 28
using the equation
phi/2pi = AB/lambda
as phi must be pi rad, 1/2 = AB/lambda
AB= 28
so making lambda the subject: lambda= (28*2)/n , where n could be 1,3,5,7 (to be out of phase)
so trying out 1,3,5,7 give us 56, 18.7, 11.2, and 8 respectively, as 18.7 and 11.2 are the only 2 values inside the range i calculated at the beginning, then the number is 2
hope that was helpful
Post by: TJ-56 on November 10, 2010, 09:57:48 pm
7.(b)(ii)
1. 1.1 MeV is supplied in the form of kinetic energy of the alpha-particle.
2. First you should convert 1.1 MeV into joules. => K.E = (1.1 x 106) x (1.6 x 10-19) = 1.76 x 10-13 J
Now you equate K.E = 0.5mv2 = 1.76 x 10-13
Mass of an alpha-particle is 4u = 4(1.67 x 10-27)kg
Hence speed of alpha-particle is found to be the square root of 2(1.76 x 10-13)/4(1.67 x 10-27)
v = 7.3 x 106 ms-1
Hope it helps :)
Thanx for the answer +rep
But can u explain further part 1 of the question
thanx in advance
Post by: thecandydoll on November 11, 2010, 02:47:35 am
Post by: ashish on November 11, 2010, 04:02:33 am
O/N 2002 Q3 C :(
according to me the time in contact is 0.15 ( from the graph)
F= change in momentum/time
= 0.35/0.15
=2.33N
Post by: $!$RatJumper$!$ on November 11, 2010, 05:12:06 am
5 (b)
So the wavelength ranges from 330/4000 to 330/1000 (v/f) which is 8.25 to 33 cm
The distance from S2 to M is 128 (root(100sqr +80sqr))
So the phase difference is 128-100 which is 28
using the equation
phi/2pi = AB/lambda
as phi must be pi rad, 1/2 = AB/lambda
AB= 28
so making lambda the subject: lambda= (28*2)/n , where n could be 1,3,5,7 (to be out of phase)
so trying out 1,3,5,7 give us 56, 18.7, 11.2, and 8 respectively, as 18.7 and 11.2 are the only 2 values inside the range i calculated at the beginning, then the number is 2
hope that was helpful
What is this equation?? I've never heard about it. What is it used to calculate and when should we use it?
By the way, whats phi?
phi/2pi = AB/lambda
Thankx
Post by: Deadly_king on November 11, 2010, 05:23:44 am
What is this equation?? I've never heard about it. What is it used to calculate and when should we use it?
By the way, whats phi?
phi/2pi = AB/lambda
Thankx
It's the equation of phase difference (phi)
Phase difference = 2(pie)
Take AB as
phi/2(pie) = AB/lambda ;)
Post by: $!$RatJumper$!$ on November 11, 2010, 05:35:30 am
It's the equation of phase difference (phi)
Phase difference = 2(pie)
Take AB as
phi/2(pie) = AB/lambda ;)
Makes sense. Thank you DK and TJ-56 :)
Post by: $!$RatJumper$!$ on November 11, 2010, 05:40:27 am
5(a)
1. The waves should meet in antiphase at M. ( phase difference = pie)
2. Sources must emit waves having same amplitude.
(b)
So the wavelength ranges from 330/4000 to 330/1000 (v/f) which is 8.25 to 33 cm
The distance from S2 to M is 128 (root(100sqr +80sqr))
So the phase difference is 128-100 which is 28
using the equation
phi/2pi = AB/lambda
as phi must be pi rad, 1/2 = AB/lambda
AB= 28
so making lambda the subject: lambda= (28*2)/n , where n could be 1,3,5,7 (to be out of phase)
so trying out 1,3,5,7 give us 56, 18.7, 11.2, and 8 respectively, as 18.7 and 11.2 are the only 2 values inside the range i calculated at the beginning, then the number is 2
hope that was helpful
One other thing i dont get. Where did n suddenly come into this equation from? And how can we take the phase diff to be pie. We dont know if they are completely out of phase. Yes they are out of phase by 28m, but doesnt pie rad mean they are completely out of phase by exactly half a wavelength?
Post by: Deadly_king on November 11, 2010, 05:42:37 am
Thanx for the answer +rep
But can u explain further part 1 of the question
thanx in advance
1. From the equation you can note that mass is conserved.
Total initial mass : 18 and total final mass : 18 -----> ( Add the mass numbers of the different elements and particles taking part in the reaction )
Now you're asked how Energy is conserved.
Energy emitted = delta(m) x c2 = 1.1 MeV
Therefore for total energy to be conserved,
Mass energy of products + energy emitted = Initial mass energy of reactants.
The energy emitted is released in the form of kinetic energy of the alpha-particle. ;)
Hope you understand :)
Post by: Deadly_king on November 11, 2010, 05:49:57 am
One other thing i dont get. Where did n suddenly come into this equation from? And how can we take the phase diff to be pie. We dont know if they are completely out of phase. Yes they are out of phase by 28m, but doesnt pie rad mean they are completely out of phase by exactly half a wavelength?
It is said that the microphone M detects maxima and minima of intensity of sound.
Maxima and Minima takes place only when the two waves meet in anti-phase. ;)
The n indicates the number of wavelengths at which the two waves will meet. It may be any number, but in this case since it is in anti-phase it has to be an odd number.
Phase difference = 2(pie)
Post by: $!$RatJumper$!$ on November 11, 2010, 05:57:02 am
It is said that the microphone M detects maxima and minima of intensity of sound.
Maxima and Minima takes place only when the two waves meet in anti-phase. ;)
The n indicates the number of wavelengths at which the two waves will meet. It may be any number, but in this case since it is in anti-phase it has to be an odd number.
Phase difference = 2(pie)
ok i get the rad part. But for the n part, isnt it meant to be n+1/2 ? why do you say it should be an odd number?
Post by: Deadly_king on November 11, 2010, 06:01:21 am
ok i get the rad part. But for the n part, isnt it meant to be n+1/2 ? why do you say it should be an odd number?
That's right. :D
When it is (n+1)/2, it usually is an odd number. ;)
Post by: birchy33 on November 11, 2010, 06:03:45 am
Post by: TJ-56 on November 11, 2010, 07:20:30 am
1. Form the equation you can note that mass is conserved.
Total initial mass : 18 and total final mass : 18 -----> ( Add the mass numbers of the different elements and particles taking part in the reaction )
Now you're asked how Energy is conserved.
Energy emitted = delta(m) x c2 = 1.1 MeV
Therefore for total energy to be conserved,
Mass energy of products + energy emitted = Initial mass energy of reactants.
The energy emitted is released in the form of kinetic energy of the alpha-particle. ;)
Hope you understand :)
Oh ok.
Thanks alot dk.
Post by: TJ-56 on November 11, 2010, 09:21:45 am
HELP! How do you figure out questions where they ask for the distance between molecules and all that sort of stuff. Like the distance between solid and liquid?
Well the distance between solid particles and liquid particles are quite similar, but its rather the movement of those particles that differ the most. In gases, the separation of particles are much greater than that of solids and liquids (approximately 10 molecular diameters).
Hope that was helpful and covered your question completely.
Post by: $!$RatJumper$!$ on November 11, 2010, 09:28:41 am
That's right. :D
When it is (n+1)/2, it usually is an odd number. ;)
haha no i mean (n+1/2)lambda
Post by: birchy33 on November 11, 2010, 09:34:01 am
Well the distance between solid particles and liquid particles are quite similar, but its rather the movement of those particles that differ the most. In gases, the separation of particles are much greater than that of solids and liquids (approximately 10 molecular diameters).
Hope that was helpful and covered your question completely.
Hmm I mean in like when you are asked to find the distance between molecules or the RATIO between distance between molecules in a solid compared to a liquid or gas.
Post by: astarmathsandphysics on November 11, 2010, 09:45:11 am
Post by: Deadly_king on November 11, 2010, 09:48:45 am
Oh ok.
Thanks alot dk.
Anytime buddy :)
haha no i mean (n+1/2)lambda
Oops.........i mistyped it.
The equation is (2n + 1)/2 or n + 1/2.
This equation is only used when the waves meet in anti-phase, i.e there will be maximum destructive interference.
Post by: $!$RatJumper$!$ on November 11, 2010, 10:15:49 am
Ok can you please have a look at S10 P21 Q4b
So, i use y = (D*lambda) / d to find d. Then I did 1 / d to find the answer. I got it right, but then looked at the MS and saw they used d*sin(theta) = n*lambda to find d. Then they did 1 / d to get the answer.
What im confused about is that when are we meant to know what formula to use where?
Post by: Deadly_king on November 11, 2010, 10:36:33 am
Alright that makes sense now :) thankx
Ok can you please have a look at S10 P21 Q4b
So, i use y = (D*lambda) / d to find d. Then I did 1 / d to find the answer. I got it right, but then looked at the MS and saw they used d*sin(theta) = n*lambda to find d. Then they did 1 / d to get the answer.
What im confused about is that when are we meant to know what formula to use where?
The formula you used can only be used when n = 1 that is only one wavelength is involved. It is normally used to find the fringe width which is the distance between the centres of two consecutive bright fringes or that between two dark fringes.
The formula you used proved to be good here since was actually 1. But if that was not the case you would not have been able to find the correct answer. But it's upto you to choose the method you find easier and more appropriate. The essential thing is that you get the required answer. :D
Post by: $!$RatJumper$!$ on November 11, 2010, 10:41:57 am
The formula you used can only be used when n = 1 that is only one wavelength is involved. It is normally used to find the fringe width which is the distance between the centres of two consecutive bright fringes or that between two dark fringes.
The formula you used proved to be good here since was actually 1. But if that was not the case you would not have been able to find the correct answer. But it's upto you to choose the method you find easier and more appropriate. The essential thing is that you get the required answer. :D
I see. I just called my friend now and he says that d*sin(theta) = n*lambda can only be used for diffraction gratings and y = (D*lambda) / d
can only be used for double slit experiments. Is this true?
Post by: TJ-56 on November 11, 2010, 10:48:07 am
I see. I just called my friend now and he says that d*sin(theta) = n*lambda can only be used for diffraction gratings and y = (D*lambda) / d
can only be used for double slit experiments. Is this true?
Yup. That is correct.
Post by: $!$RatJumper$!$ on November 11, 2010, 10:53:51 am
Its really confusing me because for example,
In W03 Q4(b)(ii), they ask for the angle where the first image is seen. It also says the incident light is at right angles.
But in W06 Q4(b)(i), they ask for the numbers where is it visible so why do we suddenly use a value of 90 for theta in the equation: d*sin(theta) = n*lambda. Wouldnt we need to use the angle like used in W03?
Post by: TJ-56 on November 11, 2010, 11:03:28 am
In d*sin(theta) = n*lambda, what does the theta actually stand for?
Its really confusing me because for example,
In W03 Q4(b)(ii), they ask for the angle where the first image is seen. It also says the incident light is at right angles.
But in W06 Q4(b)(i), they ask for the numbers where is it visible so why do we suddenly use a value of 90 for theta in the equation: d*sin(theta) = n*lambda. Wouldnt we need to use the angle like used in W03?
To find the maximum number n of the visible image; we use the eequation you stated,
which is n*lambda=d* sin (theta)
Arranged with n the subject: n = [d* sin (theta)]/ lambda
So to find the max value of n, we need the max value of sin theta, that is when theta is 90 degrees.
Post by: thecandydoll on November 11, 2010, 11:23:13 am
How is the other V= 4.2?
and for deltap/deltat (how do they find T)
there is one more question similar in 2002/oct/nov how to find the time? when finding force acting on speher?
Post by: TJ-56 on November 11, 2010, 11:35:16 am
May/June 2009 Variant 2 Q2C.
How is the other V= 4.2?
and for deltap/deltat (how do they find T)
there is one more question similar in 2002/oct/nov how to find the time? when finding force acting on speher?
for the 2002 paper
see attachment for the required time
for the 2009 paper
Reading carefully, theyre asking about only the first 3.5 seconds, so following 3.5 s gives a velocity of 4.2 m/s
Hope that helped
Post by: $!$RatJumper$!$ on November 11, 2010, 12:57:49 pm
To find the maximum number n of the visible image; we use the eequation you stated,
which is n*lambda=d* sin (theta)
Arranged with n the subject: n = [d* sin (theta)]/ lambda
So to find the max value of n, we need the max value of sin theta, that is when theta is 90 degrees.
Thank you :)
Post by: TJ-56 on November 11, 2010, 01:04:42 pm
Thank you :)
Don't mention it. :)
Post by: TJ-56 on November 11, 2010, 04:07:08 pm
W01 Q3 (c) (i) The correct directions please. ( no need for explanation.)
W09 Q7 P21 (a) Can you post how to correctly point out the angle of deviation.
S10 P21 Q5 (b) (ii) 2. (decreasing)
Thanx for any help!
Post by: Deadly_king on November 11, 2010, 05:10:23 pm
Someone please help with the following, kinda silly but i just want to make sure my answers are correct.
W01 Q3 (c) (i) The correct directions please. ( no need for explanation.)
W09 Q7 P21 (a) Can you post how to correctly point out the angle of deviation.
S10 P21 Q5 (b) (ii) 2. (decreasing)
Thanx for any help!
Sorry but I won't have time to deal with the 1st question. :-[
Nov 09 P21
7. (a) After you have drawn the two directions, draw straight horizontal line from the particles A and B respectively. Then label each angle of deviation a and b for the specific particles.
Now you just have to mention that b < a
Jun 10 p21
b)(ii)2. Actually it should have been an arrow showing the decrease from positively charged sphere to the earthed metal plate or from the latter to the negatively charged sphere. Indicate the right direction using an arrow. ;)
Post by: Dania on November 11, 2010, 05:20:35 pm
Please explain.
Post by: TJ-56 on November 11, 2010, 05:22:03 pm
Sorry but I won't have time to deal with the 1st question. :-[
Nov 09 P21
7. (a) After you have drawn the two directions, draw straight horizontal line from the particles A and B respectively. Then label each angle of deviation a and b for the specific particles.
Now you just have to mention that b < a
Jun 10 p21
b)(ii)2. Actually it should have been an arrow showing the decrease from positively charged sphere to the earthed metal plate or from the latter to the negatively charged sphere. Indicate the right direction using an arrow. ;)
Thanx dk
i got another,
but i understand if u wont have the time to answer it
S10 QP23 Q3 C (i)
Post by: Deadly_king on November 11, 2010, 05:30:30 pm
Thanx dk
i got another,
but i understand if u wont have the time to answer it
S10 QP23 Q3 C (i)
No problem. ;)
I assume you got the previous answers.
Change in K.E + Useful work done = Total k.E in overcoming friction + Change in P.E
1240 + 600 = Total k.E in overcoming friction + 1170
Total k.E in overcoming friction = 670 J
Post by: TJ-56 on November 11, 2010, 05:33:28 pm
thanx
Post by: Deadly_king on November 11, 2010, 05:36:02 pm
On which formula is the one you stated based on?
thanx
Just the principle of conservation of energy. ;)
Post by: TJ-56 on November 11, 2010, 05:55:06 pm
and
WISH YOU ALL THE BEST OF LUCK FOR TOMORROW!
Post by: $!$RatJumper$!$ on November 11, 2010, 08:12:19 pm
And is speed = distance / time only for horizontal movement?
Post by: Dania on November 11, 2010, 08:25:25 pm
The only difference is that acceleration would be taken as -9.81 ms-2 if the object is going up. If it moving horizontally, the acceleration can be calculated using v2=u2 + 2as or any other motion equation regarding acceleration.
You should remember that speed is a scalar, and velocity is vector. In many questions, they will state whether it is speed or velocity. If they say it is speed, you should use the equation Speed=Distance/Time to find what it is you're looking for.
I hope I was helpful :)
Post by: Deadly_king on November 12, 2010, 04:12:31 am
Is s = ut + (1/2 * a * t2) and v2 = u2 + (2 * a * s) only for vertical movement of an object?
And is speed = distance / time only for horizontal movement?
No, they can be used for either vertical or horizontal.
The only difference is that acceleration would be taken as -9.81 ms-2 if the object is going up. If it moving horizontally, the acceleration can be calculated using v2=u2 + 2as or any other motion equation regarding acceleration.
You should remember that speed is a scalar, and velocity is vector. In many questions, they will state whether it is speed or velocity. If they say it is speed, you should use the equation Speed=Distance/Time to find what it is you're looking for.
I hope I was helpful :)
Very good answer. :D
I would just like to add that for any of the equations of motion to be valid, acceleration should be constant. ;)
Post by: $!$RatJumper$!$ on November 12, 2010, 04:39:06 am
No, they can be used for either vertical or horizontal.
The only difference is that acceleration would be taken as -9.81 ms-2 if the object is going up. If it moving horizontally, the acceleration can be calculated using v2=u2 + 2as or any other motion equation regarding acceleration.
You should remember that speed is a scalar, and velocity is vector. In many questions, they will state whether it is speed or velocity. If they say it is speed, you should use the equation Speed=Distance/Time to find what it is you're looking for.
I hope I was helpful :)
Thank you! Best of luck for today :)
Post by: Deadly_king on November 12, 2010, 04:46:08 am
Thank you! Best of luck for today :)
Good luck to you too buddy ;)
Hope we are blessed with an easy paper ;D
Post by: $!$RatJumper$!$ on November 12, 2010, 04:49:22 am
Good luck to everyone here too! You all have been amazing in helping out. I've learnt so much here and cleared so many doubts :)
Post by: ashish on November 12, 2010, 08:57:04 am
Refer to the attachment.
Please explain.
TO the Examiner sir a voltmeter with infinite resistance doesn't exist XD
it's pretty easy
3.6V is for the 2000ohm resistance
V= IR
3.6/2000=I
I=1.8*10-3A
let resistance of the thermistor be T
R=((1/5000)+(1/T))-1
R=5000T/T+5000
V=IR
2.4=1.8*10-3 5000T/T+5000
when solving for T
you will get 1818.2 ohm
Post by: Deadly_king on November 12, 2010, 10:59:32 am
TO the Examiner sir a voltmeter with infinite resistance doesn't exist XD
To ashish..........it sure does not exist in practice, but it does on theory. ;D
But each time you make use of a voltmeter in a circuit, you should assume it has infinite resistance so that it draws no current. ;)
Post by: ashish on November 12, 2010, 02:01:39 pm
To ashish..........it sure does not exist in practice, but it does on theory. ;D
But each time you make use of a voltmeter in a circuit, you should assume it has infinite resistance so that it draws no current. ;)
yeah but i even argued with a teacher saying him that i will believe him only if he shows me an ideal voltmeter
and he smiled at me :D xD
Post by: Deadly_king on November 12, 2010, 02:12:12 pm
yeah but i even argued with a teacher saying him that i will believe him only if he shows me an ideal voltmeter
and he smiled at me :D xD
Hahaha..................very smart way to baffle the teacher. :P
Post by: thecandydoll on November 13, 2010, 07:02:25 am
q18
Post by: Hypernova on November 13, 2010, 08:01:28 am
Oct/Nov 2004. MCQ Q20.
q18
q18
Your looking for the work done against the 9.0kN force.
Work done by a force = Force x distance moved in the direction of the force
The body has not gained speed, this means that the work done by the 9.0 kN is equal to the work opposing it.
If the body had gained speed, then the work done by the 9.0 kN is greater than the work done opposing by an amount equal to the gain in kinetic energy
the 9.0 kN is equal to the work opposing it.
so the work done by the 9kN is equal to the workdone by friction and by its weight.
Work done by 9kN = Work done by friction + delta Potential energy
9000x40 = Work done by friction + 20000x12
Work done by friction (heat dissipated) = 120000 J
Ans is A
Post by: elemis on November 13, 2010, 08:10:45 am
Post by: $!$RatJumper$!$ on November 13, 2010, 08:48:14 am
Though i have a doubt, the equation is:
Change in K.E + Useful work done = Total k.E in overcoming friction + Change in P.E
Why didnt you use the Change in K.E in this calculation?
Post by: ashish on November 13, 2010, 08:50:56 am
Oct/Nov 2004. MCQ Q20.
q18
Q20
Pressurep at depth 2x= atm+2xDPg
on the right arm pressure =atm+xDQg
they are at equilibrium
atm+2xDPg = atm+ xDQg
simplifying on both side you get
2DP= DQ
DP/DQ= 1/2
note D = density
Post by: Hypernova on November 13, 2010, 08:54:09 am
Thank you for that. It was really useful!
Though i have a doubt, the equation is:
Change in K.E + Useful work done = Total k.E in overcoming friction + Change in P.E
Why didnt you use the Change in K.E in this calculation?
Change in KE is zero, since no chnage in velocity
using your equation
0+9000x40 = Work done by friction + 20000x12
its the same thing
Post by: $!$RatJumper$!$ on November 13, 2010, 08:56:55 am
Change in KE is zero, since no chnage in velocity
using your equation
0+9000x40 = Work done by friction + 20000x12
its the same thing
Thanks :)
Q20
Pressurep at depth 2x= atm+2xDPg
on the right arm pressure =atm+xDQg
they are at equilibrium
atm+2xDPg = atm+ xDQg
simplifying on both side you get
2DP= DQ
DP/DQ= 1/2
note D = density
Please explain this equation u used. We have not been taught it before.. :(
Thank you
Post by: ashish on November 13, 2010, 09:11:04 am
Thanks :)
Please explain this equation u used. We have not been taught it before.. :(
Thank you
pressure in a liquid = h(rho)g
row means density which i represented by D
h=height
pressure at the same level(height) in a liquid is equal
pressure in left arm at a depth of 2x = atmospheric pressure (atm) + pressure by liquid
= atm + 2xDpg
pressure in the right arm (in the diagram the liquid has height x) = atm + xDQg
since pressure in a liquid i at same level(height) is equal
atm + 2xDpg = atm + xDQg
Dp= density of liquid P
DQ= density of liquid Q
hope it helps
Post by: $!$RatJumper$!$ on November 13, 2010, 09:50:08 am
pressure in a liquid = h(rho)g
row means density which i represented by D
h=height
pressure at the same level(height) in a liquid is equal
pressure in left arm at a depth of 2x = atmospheric pressure (atm) + pressure by liquid
= atm + 2xDpg
pressure in the right arm (in the diagram the liquid has height x) = atm + xDQg
since pressure in a liquid i at same level(height) is equal
atm + 2xDpg = atm + xDQg
Dp= density of liquid P
DQ= density of liquid Q
hope it helps
Thank you so much! :) +rep
can u please explain S10 P11 Q16
Post by: Deadly_king on November 13, 2010, 10:16:14 am
can u please explain S10 P11 Q16
Efficiency = Output / Input
Energy input : Force x distance : F
Energy output : P.E + K.E
P.E = mgh where h =
P.E :
K.E = 0
Therefore efficiency =
Hence answer is D
Post by: $!$RatJumper$!$ on November 13, 2010, 10:37:48 am
Efficiency = Output / Input
Energy input : Force x distance : F
Energy output : P.E + K.E
P.E = mgh where h =x sin(alpha)
P.E :sin(alpha)
K.E = 0
Therefore efficiency =sin(alpha) / F
Hence answer is D
What always confuses me is the fact that:
1) How to know which values to take as the Energy input and Energy output
2) Why did you ADD K.E and P.E in this case? Isnt the equation: Change in K.E + Useful work done = Total k.E in overcoming friction + Change in P.E
I dont see how they will be added.
3) The definition of internal energy is sum of K.E and P.E but the equation stated above says otherwise.
Im SO confused :(
Post by: Deadly_king on November 13, 2010, 10:50:43 am
What always confuses me is the fact that:
1) How to know which values to take as the Energy input and Energy output
2) Why did you ADD K.E and P.E in this case? Isnt the equation: Change in K.E + Useful work done = Total k.E in overcoming friction + Change in P.E
I dont see how they will be added.
3) The definition of internal energy is sum of K.E and P.E but the equation stated above says otherwise.
Im SO confused :(
Energy input is the energy required to move the car from its initial position to its destination.
Energy output is the energy the car possesses once it is at its destination.
Change on K.E is zero since velocity is constant which implies that change in velocity is zero.
I added K.E and P.E since these are the two forms of energies that the car possesses once at the top.
3. Internal energy does not apply here. ;)
Post by: $!$RatJumper$!$ on November 13, 2010, 10:52:46 am
Energy input is the energy required to move the car from its initial position to its destination.
Energy output is the energy the car possesses once it is at its destination.
Change on K.E is zero since velocity is constant which implies that change in velocity is zero.
I added K.E and P.E since these are the two forms of energies that the car possesses once at the top.
3. Internal energy does not apply here. ;)
I see :) that makes it a lot clearer now :) so when would we use the equation Change in K.E + Useful work done = Total k.E in overcoming friction + Change in P.E... like for what types of questions?
Post by: Deadly_king on November 13, 2010, 11:01:45 am
I see :) that makes it a lot clearer now :) so when would we use the equation Change in K.E + Useful work done = Total k.E in overcoming friction + Change in P.E... like for what types of questions?
Hmm...........this equation can be altered. You don't need to keep it fixed. It depends according to the question. I always develop my own equation in the exams though you must obey the principle of conservation of enrgy. ;)
It's only used when you need the principle of conservation of energy.
Post by: $!$RatJumper$!$ on November 13, 2010, 11:12:12 am
Post by: Deadly_king on November 13, 2010, 11:14:33 am
So isnt energy being conserved in this system of the car?
Not really since you're not told about frictional force. ;)
Post by: $!$RatJumper$!$ on November 13, 2010, 11:48:01 am
Thank you.. and sorry if its a lot of questions :)
Post by: Deadly_king on November 13, 2010, 12:22:37 pm
i see ok thank you :) Can u help with W09 P11 Q 14, 22, 23, 26, 27, 28
Thank you.. and sorry if its a lot of questions :)
Nov 09 p11
11. The centre of gravity of an object is defined as the point at which the weight can be considered to act.
Answer is C
14. Initial k.E is given to be E.
Since the angle of projection is 45o, the initial kinetic energy will divide equally to give the final P.E and the final K.E. ;)
So final k.E will be 0.5E
Answer is A.
22. Youngs Modulus is given as Y = TL/Ae where
T : Tension in the string
L : Length of the string
A : Area of cross-section of the string.
e : extension produced.
You're asked to find the % length the string contracts. Hence you need to find the percentage of e/L which is also equal to F/AY. (from equation above)
Hence replace the values given to you to get 20/(pie*(diameter/2)2*2.0 x 1011) x 100
It will come out to be 0.051%.
Hence answer is B
23. You should know this from the elctromagnetic spectrum. ;)
Max wavelength of U.V = 10-13 and c = 330ms-1
Hence c = f(lambda) -----> f = 330/10-13 = 1015 Hz
Answer is B
26. You should use the formula dsin(theta) = n(lambda) where d = 1/N. Since the beam of light is parallel, theta will be 90o. You should also convert the wavelength into millimetres since N is given as 300 lines/mm.
(1/300) x sin 90 = n(450 x 10-6) ----> n = 7.5
Hence there'll be 7.5 lambda above and 7.5 lambda below. This makes 15 lambda. Each lambda will form one maxima. ;)
Answer is D
27. Force = Electric field x charge
Electric field = Force / charge = F/q
Answer is D
28. Equilibrium => Electric force = Weight
Hence QE = mg ----> Q/m = g/E
Electric force is upwards(opposite to the weight). Since it is attracted towards the positive plate, it has to be negatively charged. ;)
Answer is B
Hope it helps :)
Post by: $!$RatJumper$!$ on November 13, 2010, 12:40:22 pm
Take your time :)
Post by: Deadly_king on November 13, 2010, 01:20:04 pm
Thank you so much buddy :) +rep
Take your time :)
Done. Check it out ;)
I modified my previous post. :D
Post by: $!$RatJumper$!$ on November 13, 2010, 01:42:44 pm
Post by: Deadly_king on November 13, 2010, 01:48:02 pm
Thank you so very much :) how is your preparation going for the exam
I guess am ready for p1. :)
But since am sitting for A2, i would be having p5 as well. And this paper is not as easy. I'll have to work out some more. ;)
How about you?
Post by: $!$RatJumper$!$ on November 13, 2010, 06:59:32 pm
Q 9, 10, 15, 21 (urgent), 25, 26, 27, 32
S08 Q26
Thank you :)
Post by: $!$RatJumper$!$ on November 13, 2010, 07:00:30 pm
I guess am ready for p1. :)
But since am sitting for A2, i would be having p5 as well. And this paper is not as easy. I'll have to work out some more. ;)
How about you?
Yeha i still need to clear up a few doubts .. but best of luck buddy! :)
Post by: TJ-56 on November 13, 2010, 07:02:51 pm
S03 Q20
S04 Q25
W05 Q34
Thankx for the help in advance!
Post by: TJ-56 on November 13, 2010, 07:26:48 pm
W06 P1
Q 9, 10, 15, 21 (urgent), 25, 26, 27, 32
Thank you :)
Q9
Horizontal component: using cosine, will give us u cos angle , since acceleration of free fall acts normal to the horiz. motion, u cos angle is the correct answer
Vertical: here we use sin to calculate the vertical component (opposite/adjacent), and so its u sin angle, and since we need to take the acceleration of free fall into consideration, we include -gt. I hope that was clear.
Q10
F=ma,
the resultant force MUST be in the same direction as the acceleration, so B. I think thats all to this question.
Q15
Using the parallelogram rule, or we could find the length of the dotted line (by length i mean force), then multiply it by 2. So to calculate this force: 10 cos(30) (the dotted line) * 2 = 17.32 N
17.32 - 10 = 7.3 (the resultant force).
21
Since the total pressure is 17.5 MPa:
17*10^6 = Pressure exerted by oil +Pressure exerted by water
= 830*9.81*x + 1000*9.81*(2000-x
Solving for x would give 1270m
And hence the answer is D
25
I have doubts about this question as well, sorry.
26
A smaller gap would give a greater diffraction,
http://www.knowledgerush.com/wiki_image/7/70/Diffraction.png
27
n lambda = d sin theta
n=1 theta=30
so lambda=d*0.5
and we have lambda*D=ax
D=1 and x=500d
so lambda=500d*a
solving simultaneously gives a=10^-3 so C
32
As R=V/I which is 1/gradient on the graph;
A filament lamps resistance increases as the temperature increases if the resistance increases then the gradient decreases and so D is the correct response.
Hope that was helpful
Post by: $!$RatJumper$!$ on November 13, 2010, 08:05:21 pm
Post by: $!$RatJumper$!$ on November 13, 2010, 08:07:06 pm
Can you help me with the following p1 MCQs;
S03 Q20
S04 Q25
W05 Q34
Thankx for the help in advance!
For S04 Q25
You need to remember this:
Node to Antinode = Downward (NAD)
Antinode to Node = Upwards (ANU)
:)
Post by: TJ-56 on November 13, 2010, 08:26:49 pm
WOW! :) Thank you so much! +rep
Sure ;)
thanx for ur answer, but just a question,
is Q always stationary at this point?
Post by: $!$RatJumper$!$ on November 13, 2010, 08:38:50 pm
Sure ;)
thanx for ur answer, but just a question,
is Q always stationary at this point?
That is the exact question i asked my teacher when i learnt it but i forgot what he said :/ Perhaps you could check out this question and make a meaning from it? S07 Q23
Let me know :)
Post by: $!$RatJumper$!$ on November 13, 2010, 08:56:27 pm
Post by: TJ-56 on November 13, 2010, 09:02:08 pm
Help with S09 Q29 and S08 Q26 please :)
So8 Q26
First you need to find the new intensity,
so I/I(new) = (1/r^2)/(1/(2r)^2)
Simplified, the new intensity is 0.25I
Now to find the new amplitude:
I/0.25I = (8^2)/(A[new]^2)
Simplified, the new amplitude would be root 16, which is 4 so D is the correct answer.
Hope that helped.
Post by: Dania on November 13, 2010, 09:21:29 pm
Post by: $!$RatJumper$!$ on November 13, 2010, 09:24:04 pm
and S09 Q29 if you may please? :)
Post by: TJ-56 on November 13, 2010, 09:25:34 pm
Why is the answer A?
Momentum before = momentum after so:
mv - mv = 2mV ( the negative sign is given due to the 2 trolleys moving in opposite directions ) so
60m - 40m = 2mV (V is the speed of the 2 together, and the masses are added because they are taken as 1 trolley now)
so simplified would give 10 cm/s
Post by: Hypernova on November 13, 2010, 09:29:08 pm
Help with S09 Q29 and S08 Q26 please :)
S09 P1 Q29
The electron has kinetic energy = 1/2 m v2
We know that the electron travels a distance x, then stops. This means that the it lost its kinetic energy since velocity changed from something to zero.
The loss in Ke should equal the work done on the force.
F x X = 1/2 x m x v2 E = F/e
F = Ee so...
EeX = 1/2 x m x v2
X = m x v2
2Ee
Post by: $!$RatJumper$!$ on November 13, 2010, 09:31:06 pm
S09 P1 Q29
The electron has kinetic energy = 1/2 m v2
We know that the electron travels a distance x, then stops. This means that the it lost its kinetic energy since velocity changed from something to zero.
The loss in Ke should equal the work done on the force.
F x X = 1/2 x m x v2 E = F/e
F = Ee so...
EeX = 1/2 x m x v2
X = m x v2
2Ee
+rep :)
Post by: Dania on November 13, 2010, 09:57:11 pm
Momentum before = momentum after so:
mv - mv = 2mV ( the negative sign is given due to the 2 trolleys moving in opposite directions ) so
60m - 40m = 2mV (V is the speed of the 2 together, and the masses are added because they are taken as 1 trolley now)
so simplified would give 10 cm/s
Thank You!
Post by: $!$RatJumper$!$ on November 13, 2010, 10:30:59 pm
Sorry for all my questions! :(
Post by: TJ-56 on November 13, 2010, 10:49:27 pm
W08 P1 Q 10, 27, 31, 36, 37 please :)
Sorry for all my questions! :(
Q10
Relative speed of approach = relative speed of separation
u1 + u2 = (-)v1 + v2
so:
u1+ u2 = v2 - v1 rearranged (v1 i given a negative sign due to its direction)
Q27
Between 2 maxima is 15 mm , so wavelength = 2 * 15 * 10-3 (remember its a stationary wave)
so speed c = 3 * 108
c=f * lambda
and f = c/wavelength
we have the 2 values required, so the frequency is calculated as 1 * 1010
Q31
Wire p has twice the diameter, not twice the area, its quadruple the area because the diameter is squared,
current I=V/R V is constant, so its a ratio of resistances, which is 4 to 1
Q36
Looked confusing to me at first, but turned out to be rly easy:
V=IR
V=7.5 (1.5 volts are lost to internal resistance)
R=15
Go for I to give 7.5/15
which is 0.5A
Q37
If we have the emf of cell E2, we could calculate pd per unit length along XY.
Im still not completely sure why i cant be response A, but B made much more sense.
Hope that helped.
Post by: $!$RatJumper$!$ on November 13, 2010, 11:00:14 pm
Q10
Relative speed of approach = relative speed of separation
u1 + u2 = (-)v1 + v2
so:
u1+ u2 = v2 - v1 rearranged (v1 i given a negative sign due to its direction)
Q27
Between 2 maxima is 15 mm , so wavelength = 2 * 15 * 10-3 (remember its a stationary wave)
so speed c = 3 * 108
c=f * lambda
and f = c/wavelength
we have the 2 values required, so the frequency is calculated as 1 * 1010
Q31
Wire p has twice the diameter, not twice the area, its quadruple the area because the diameter is squared,
current I=V/R V is constant, so its a ratio of resistances, which is 4 to 1
Q36
Looked confusing to me at first, but turned out to be rly easy:
V=IR
V=7.5 (1.5 volts are lost to internal resistance)
R=15
Go for I to give 7.5/15
which is 0.5A
Q37
If we have the emf of cell E2, we could calculate pd per unit length along XY.
Im still not completely sure why i cant be response A, but B made much more sense.
Hope that helped.
Thank you that did help a lot :)
I just have a few questions
For Q10, why cant we say u1 + -u2 = v1 + v2
Post by: TJ-56 on November 13, 2010, 11:10:08 pm
Thank you that did help a lot :)
I just have a few questions
For Q10, why cant we say u1 + -u2 = v1 + v2
Well the original equation is
u1 - u2 = - (v1 - v2) Which is derived by using both the principle of conservation of momentum and energy, where it is assumed all directions are the same (for example both to the left before and both with velocities to the left after)
since u1 and u2 are in different direction , its now u1 - -u2, which is u1 + u2
and on the right hand side, -(v1 -v2) which is v2 - v1
Thats all there is to it i guess, hope that cleared it up.
Post by: Hypernova on November 13, 2010, 11:14:25 pm
Q37
If we have the emf of cell E2, we could calculate pd per unit length along XY.
Im still not completely sure why i cant be response A, but B made much more sense.
Hope that helped.
Q37)
The potentiometer!!
In this circuit,
E2 = E1 x resistance of XT
resistance of XY + R
since the galvanometer is reading a null result, we know that the pd across XT is equal to E2.
The galvanometer is basically an ammeter. 0 amps means 0 p.d which means that the p.d across XT exactly nullifies the emf produced by E2
So we know the length of XT. All we need is E2 and divide it by the length to find its pd per length.
Post by: $!$RatJumper$!$ on November 13, 2010, 11:16:46 pm
Q37)
The potentiometer!!
In this circuit,
E2 = E1 x resistance of XT
resistance of XY + R
since the galvanometer is reading a null result, we know that the pd across XT is equal to E2.
The galvanometer is basically an ammeter. 0 amps means 0 p.d which means that the p.d across XT exactly nullifies the emf produced by E2
So we know the length of XT. All we need is E2 and divide it by the length to find its pd per length.
Thank you that helped :)
Well the original equation is
u1 - u2 = - (v1 - v2) Which is derived by using both the principle of conservation of momentum and energy, where it is assumed all directions are the same (for example both to the left before and both with velocities to the left after)
since u1 and u2 are in different direction , its now u1 - -u2, which is u1 + u2
and on the right hand side, -(v1 -v2) which is v2 - v1
Thats all there is to it i guess, hope that cleared it up.
Many thanks that helped a lot :)
+rep both
Post by: $!$RatJumper$!$ on November 13, 2010, 11:26:00 pm
:)
Post by: thecandydoll on November 14, 2010, 03:20:01 am
Q15
Post by: thecandydoll on November 14, 2010, 03:34:43 am
20/21
I dont get it :(
Post by: thecandydoll on November 14, 2010, 03:59:25 am
http://www.xtremepapers.net/CIE/International%20A%20And%20AS%20Level/9702%20-%20Physics/9702_w09_qp_11.pdf
Q15
Post by: ashish on November 14, 2010, 04:04:28 am
may/june 2009
20/21
I dont get it :(
no.20
young modulus will be constant in the same material
Y=young modulus=Stress/strain
Y=Fl/eA
Yp=Fpl/eA
YQ=FQ2l/e(A/2)
= 4FQl/eA
YQ=Yp
4FQl/eA=Fpl/eA
eliminating l,e and A
you will get
4FQ= Fp
Fp/Q= 4/1
Post by: ashish on November 14, 2010, 04:22:24 am
wrong questions.
http://www.xtremepapers.net/CIE/International%20A%20And%20AS%20Level/9702%20-%20Physics/9702_w09_qp_11.pdf
Q15
1st by law of conservation of momentum
m1v1=M2v2
2*2=1*v2
v2=4
kinetic energy of the two trolley = (0.5*2*22)+ 0.5*16
=4+8
=12
Post by: ashish on November 14, 2010, 04:31:32 am
may/june 2009
20/21
I dont get it :(
21) divide the graph into 2 trapeziums(0-20)and (20-30)
=((0.5*(20+10)*5)+ (0.5*(5+17)*10))/100
=1.85 + almost 6 squares out of the graph
=1.91
~ 2.0 A(approximation)
Post by: ashish on November 14, 2010, 04:55:43 am
S08 Q 24, 29, 32, 34 please
:)
24) directly C because for same materials young modulus is constant
32)
P=I2R
Px=I2R
R=Px/I2
Py= (I/2)22R
Py=I2R/2
R=2Py/I2
equating both you will get
Px=2Py
Py/Px= 1/2
34)rho will be constant in both wire
V=Ay*4L
V=Ax*L
equating both you will get
Ax =4Ay------(1)
R= (rho)L/A
rho=RA/L
for wire x
rho= Rx * Ax / L
for wire Y
rho = Ry * Ay /4L
equating both
Ry * Ay /4L = Rx * Ax / L
from 1 Ax =4Ay
now substitute in the equation
AyRy/4L = 4 AyRx/L
after simplifying
Ry=16Rx
Ry/Rx=16
Post by: $!$RatJumper$!$ on November 14, 2010, 07:42:37 am
24) directly C because for same materials young modulus is constant
32)
P=I2R
Px=I2R
R=Px/I2
Py= (I/2)22R
Py=I2R/2
R=2Py/I2
equating both you will get
Px=2Py
Py/Px= 1/2
34)rho will be constant in both wire
V=Ay*4L
V=Ax*L
equating both you will get
Ax =4Ay------(1)
R= (rho)L/A
rho=RA/L
for wire x
rho= Rx * Ax / L
for wire Y
rho = Ry * Ay /4L
equating both
Ry * Ay /4L = Rx * Ax / L
from 1 Ax =4Ay
now substitute in the equation
AyRy/4L = 4 AyRx/L
after simplifying
Ry=16Rx
Ry/Rx=16
Thank you :)
Though could you please explain Q24 in terms of E = pL/Ae ? Like for instance by doubling L wouldn't we double E because they are proportional in the equation?
I get that the modulus doesn't change, but in terms of doubling and stuff, could you please explain :)
Post by: birchy33 on November 14, 2010, 08:35:54 am
Which one completes the acceleration with the greatest momentum?
A ?-particle
B electron
C neutron
D proton
please explain. Its from ON07 QP1
Post by: Hypernova on November 14, 2010, 09:03:51 am
40 The following particles are each accelerated from rest through the same potential difference.
Which one completes the acceleration with the greatest momentum?
A ?-particle
B electron
C neutron
D proton
please explain. Its from ON07 QP1
All these particles will be accelarated in the same potential difference
the alpha particle has a charge of +2, the elecctron -1, the neutron 0 and the proton +1.
The Alpha particle will have the greatest force acting on it
Then the electron, then the proton.
Momentum = mass x velocity
Immediatelty ignore the neutron and electron. The neurton will not accelerate because it has no charge, so no velocity. The electron has negligble mass, so negligble momentum.
F=ma
Compare the proton and The Alpha particle
Proton has a force of 1 and The Alpha particle has a force of 2 on them, relative to each other
The mass of the proton is 1 and the mass of The Alpha particle is 4, relative to each other.
acceleration of the proton is 1 and the acc of The Alpha particle is 0.5
NOw we know that the velocity of the proton will increase faster than The Alpha particle
The both start from rest so using V=u+at, velocity = the acceleration x time
Momentum of the alpha particle = 4 x 0.5t = 2t
Momentum of the proton = 1 x 1t = t
Relative to each other, the momentum of the alpha particle increases by twice the momentum of the proton in 1 second.
This is a quantitative way of looking at it. But in the exam, I knew that the velocity of the alpha particle was only slightly smaller than the proton, and the alpha particle had 4 times the mass of the proton. I could confidently guess that the alpha particle would have the higher momentum.
Post by: birchy33 on November 14, 2010, 09:14:34 am
Post by: TJ-56 on November 14, 2010, 09:15:02 am
Can you help me with the following p1 MCQs;
S03 Q20
S04 Q25
W05 Q34
Thankx for the help in advance!
Can anyone help me with those pls ???
Post by: Hypernova on November 14, 2010, 09:24:13 am
W08 P1 Q 10, 27, 31, 36, 37 please :)S08 q29
Sorry for all my questions! :(
Use the formula lamda= fringe spacing X split spacing
Distance of slits from screen
7x10-9 = 3x10-3 x split
D
3.5x10-9 = ? x 2split
D
make things a bit neater
7x10-9 = 3x10-3 x K
3.5x10-9 = ? x 2k
sub in the 2 Ks
3.5x10-9 = ? x 2 (7x10-9 / 3x10-3)
solve to get ? = 0.00075 m
So ans is A
Post by: Deadly_king on November 14, 2010, 09:47:58 am
Can anyone help me with those pls ???
Nov 05 No 34
Use the formula V = IR. Make R the subject of formula ---> R = V/I
Hence smallest resistance will result when V is minimum while I is maximum.
Therefore answer is C
Jun 04 No 25
At the instant the point Q is at a maximum which means it is temporarily stationary and it is neither moving upwards nor downwards.
At the same instant the point P is a zero displacement but is moving downwards since it was previously at Q and is moving downwards upto the minimum. ;)
Answer is A
Jun 03 No 20
Decreasing pressure by 10% ----> New pressure 0.1Po
Hence 0.1Po = h
H becomes 0.1Po/
Answer is A
Hope it helps :)
Post by: TJ-56 on November 14, 2010, 10:04:12 am
Nov 05 No 34
Use the formula V = IR. Make R the subject of formula ---> R = V/I
Hence smallest resistance will result when V is minimum while I is maximum.
Therefore answer is C
Jun 04 No 25
At the instant the point Q is at a maximum which means it is temporarily stationary and it is neither moving upwards nor downwards.
At the same instant the point P is a zero displacement but is moving downwards since it was previously at Q and is moving downwards upto the minimum. ;)
Answer is A
Jun 03 No 20
Decreasing pressure by 10% ----> New pressure 0.1Po
Hence 0.1Po = h
H becomes 0.1Po/
Answer is A
Hope it helps :)
Thank you so much +rep
Just a question though
Jun 03 No 20
The child reduced p0 BY 10% and not TO 10%, isnt it supposed to be 0.9p0 rather than 0.1p0 , he reduced it BY 10%, so 90% is the new percentage, i understood this question once, but now it seems real confusing
and q34
Resistance = v/I
which is 1/gradient
So the greater the gradient, the lower the resistance, at B the gradient is greater than at C, so why cant B be the correct response?
Sorry for the questions, but i just need to get those clarified,
thank you
Post by: Dania on November 14, 2010, 10:14:04 am
Post by: $!$RatJumper$!$ on November 14, 2010, 10:16:09 am
Post by: $!$RatJumper$!$ on November 14, 2010, 10:19:19 am
Why is the answer B?
Momentum before = Momentum after
Momentum before: (1000 + 10) * 0 = 0 ---> There was no initial velocity thus * 0
Momentum after: (1000 * 5) + (-10 * v) = 0 ---> Cannon ball moving in opposite direction so since it is velecity we need to specify its direction
Thus v = 500 ms-1
Post by: Deadly_king on November 14, 2010, 10:20:32 am
Though could you please explain Q24 in terms of E = pL/Ae ? Like for instance by doubling L wouldn't we double E because they are proportional in the equation?
I get that the modulus doesn't change, but in terms of doubling and stuff, could you please explain :)
Since same material used the Youngs modulus is constant. Hence E remains the same which also implies pL/A
However we cannot use the formula to show that Youngs Modulus is constant as we have not been given anything about the extension produced or the force acting on the wire.
This formula would have been applicable only if we were told that the extension produced and the force were constant. ;)
Post by: Hypernova on November 14, 2010, 10:27:44 am
W07 Q 3, 6, 16, 22, 25, 37 :) thank you
w07
37)
a nuclear process the decay of the nucleus
the nucleon number is not conserved.
know these
in alpha decay, 2 protons and 2 neutrons leave the nucleus.
in beta decay, a neutron changes to an electron and a proton
in y decay, radiation is emitted
Post by: Hypernova on November 14, 2010, 10:31:43 am
W07 Q 3, 6, 16, 22, 25, 37 :) thank you
25)
nlamda=dsin@
maximum order= d / lamda
3lamda=dsin45
d / lamda = 3 / sin45
maximum order= 4.24
so 4
ans is B
Post by: Dania on November 14, 2010, 10:32:56 am
Momentum before = Momentum after
Momentum before: (1000 + 10) * 0 = 0 ---> There was no initial velocity thus * 0
Momentum after: (1000 * 5) + (-10 * v) = 0 ---> Cannon ball moving in opposite direction so since it is velecity we need to specify its direction
Thus v = 500 ms-1
Thanks!
Post by: Deadly_king on November 14, 2010, 10:37:04 am
Jun 03 No 20
The child reduced p0 BY 10% and not TO 10%, isnt it supposed to be 0.9p0 rather than 0.1p0 , he reduced it BY 10%, so 90% is the new percentage, i understood this question once, but now it seems real confusing
Nov 05 No 34
Resistance = v/I
which is 1/gradient
So the greater the gradient, the lower the resistance, at B the gradient is greater than at C, so why cant B be the correct response?
No 20. You're actually right. Am sorry to have got you more confused as I misplaced my words. :-[
The new pressure is indeed 90% but to be able to drink the liquid, its not the new pressure which is required but rather the difference in pressure which is 10%(100 - 90).
No 34
You have a point there. :-\
But i was told by my teacher that for a graph involving I and V, gradient of tangent is never used. Rather we should use gradient of chord from the origin. From the gradient of chords, we can easily note that C has a greater value than B. ;)
Post by: TJ-56 on November 14, 2010, 10:43:37 am
W07 Q 3, 6, 16, 22, 25, 37 :) thank you
3
X remains at the same direction, but since they asked for X-Y, Y is now at the opposite direction,
The resultant is Z. X goes diagonally right-upwards, and Y now goes horizontally to the left, so Z would be upwards, and hence b is the correct answer.
6
When u look at the meter, as the current increases, the deflection decreases, and so the gradient decrease with increasing current, so its either A or C
When the current is zero, there is a deflection X as stated on the question, so when the current is 0, there must be an intersection at the y axis and so A is the answer.
16
At position 3, the person is not moving (or moving very slowly) , unlike position 2, where he has kinetic energy moving downwards, stretching the rope. Moreover, at 3, the rope stores maximum elastic energy, since its stretched the most. so C
22
Radio waves have frequencies 10^3 to 10^9, i have uploaded previously a document comprising these, and if u cant find it i have no problem uploading it again.
Hope that helped,
Post by: $!$RatJumper$!$ on November 14, 2010, 10:53:13 am
3
X remains at the same direction, but since they asked for X-Y, Y is now at the opposite direction,
The resultant is Z. X goes diagonally right-upwards, and Y now goes horizontally to the left, so Z would be upwards, and hence b is the correct answer.
6
When u look at the meter, as the current increases, the deflection decreases, and so the gradient decrease with increasing current, so its either A or C
When the current is zero, there is a deflection X as stated on the question, so when the current is 0, there must be an intersection at the y axis and so C is the answer.
16
At position 3, the person is not moving (or moving very slowly) , unlike position 2, where he has kinetic energy moving downwards, stretching the rope. Moreover, at 3, the rope stores maximum elastic energy, since its stretched the most. so C
22
Radio waves have frequencies 10^3 to 10^9, i have uploaded previously a document comprising these, and if u cant find it i have no problem uploading it again.
Hope that helped,
Thank you it would be gr8 if you could upload it again :)
And for question 6 the answer is actually A :(
Post by: Dania on November 14, 2010, 10:58:01 am
Post by: TJ-56 on November 14, 2010, 11:02:27 am
Why is the answer D?
21
Since the total pressure is 17.5 MPa:
17*10^6 = Pressure exerted by oil +Pressure exerted by water
= 830*9.81*x + 1000*9.81*(2000-x)
Solving for x would give 1270m
And hence the answer is D
Post by: TJ-56 on November 14, 2010, 11:04:34 am
Thank you it would be gr8 if you could upload it again :)
And for question 6 the answer is actually A :(
Post by: Dania on November 14, 2010, 11:07:15 am
21
Since the total pressure is 17.5 MPa:
17*10^6 = Pressure exerted by oil +Pressure exerted by water
= 830*9.81*x + 1000*9.81*(2000-x)
Solving for x would give 1270m
And hence the answer is D
Thanks! :)
Post by: TJ-56 on November 14, 2010, 11:07:54 am
Yeah the answer is A, where there is a y-intersection
Post by: Dania on November 14, 2010, 11:10:58 am
Can someone please explain to me why the answer is C?
Post by: $!$RatJumper$!$ on November 14, 2010, 11:19:42 am
25)
nlamda=dsin@
maximum order= d / lamda
3lamda=dsin45
d / lamda = 3 / sin45
maximum order= 4.24
so 4
ans is B
Thank you! :)
Oh sorry, i mistyped it, was typing in a hurry,
Yeah the answer is A, where there is a y-intersection
Makes sense. thankx! :)
Post by: Dania on November 14, 2010, 11:26:28 am
Post by: TJ-56 on November 14, 2010, 11:33:15 am
Why is the answer A?
Q=It (charge=current*time)
There are for charges so 4Q
I=Q\T
=4Q\T
Frequency=1/T
so replacing f=1/T into I= 4Q \ T
will give I=4Qf
Post by: Dania on November 14, 2010, 11:40:27 am
Q=It (charge=current*time)
There are for charges so 4Q
I=Q\T
=4Q\T
Frequency=1/T
so replacing f=1/T into I= 4Q \ T
will give I=4Qf
Thanks :)
Post by: thecandydoll on November 14, 2010, 11:55:50 am
Post by: thecandydoll on November 14, 2010, 12:12:08 pm
When there is force/extension curve not a straight line,how to find strain energy
Post by: Deadly_king on November 14, 2010, 12:13:52 pm
S08 P1 Q34!
Resistance of a wire is given by the formula R =
L : length of wire
A : Area of cross-section of wire.
Since both wires are made of copper, they have the same resistivity and they will cancel out.
It is said that both wires have same volume which implies that Vx = Vy
Vx = AxL and Vy = 4AyL
Now we can find Ax in terms of Ay and L ----> Ax = 4Ay
Now we replace in the formula.
Rx = L/Ax or L/4Ay and Ry = 4L/Ay
Now Ry/Rx = (4L/Ay) / (L/4Ay)
Simplify and you'll get it to be 16.
Answer is C
Hope it helps :)
Post by: cs on November 14, 2010, 12:17:22 pm
edited: whoops, its paper 11..
Post by: Deadly_king on November 14, 2010, 12:32:01 pm
Deadly king to the rescue.. glad that you are here, i don't dare to see June 2010 paper 1, i can't do most of them cos my teacher didn't go through, can you help me Q5 and 6 ( any variant cos they are the same) before i go to sleep? =)
Hehe.........indeed they are all the same though the numbers are not in the same order. :-[
Anyway i'll do No 5 and 6 from variance 1 but if they are not the numbers you need, let me know. ;)
Jun 10 p1
5. let theta be
This is a right-angled triangle. If you resolve the resultant V in the X direction, it should be in terms of cos
As the angle is increased, we can note the sin x will rise as well. But cos
Hence X is decreasing and Y is increasing. :)
Answer is C.
6. The equation is Density = Mass / Volume ---->
Therefore % uncertainty in
^
Now you need to find ^
NOTE : The uncertainty in mass is multiplied by two since two values are measured.
(initial : empty beaker and final : beaker + liquid).
Answer is B
Post by: $!$RatJumper$!$ on November 14, 2010, 01:20:53 pm
Thank you
Post by: Deadly_king on November 14, 2010, 01:59:35 pm
2009 JUNE -Q10/21
When there is force/extension curve not a straight line,how to find strain energy
You need to find the area under graph. Most of the time you need to estimate since it's not easy to calculate the area under graph when a curve is involved.
Jun 09 p01
21.
You need to find are over the curve since it is rubber and the question ask for the energy stored.
Try to divide it into small triangles and trapeziums.
Answer is A
Post by: Deadly_king on November 14, 2010, 02:13:51 pm
S07 Q 3, 7, 8, 23, 31, 40
Thank you
Jun 07 p1
3. Do it in terms of units. The first one itself is the answer.
V has unit ms-1.
g has unit ms-2 while lambda has unit m. Hence g(lamnda) will have unit m2s-2. Its square root will come out to be ms-1.
Answer is A.
7. You should use the equation v = u + at to describe the motion of this object.
From vector diagram we can note that V = u + X
Hence X comes out to be at. ;)
Answer is C
I'll complete the rest asap
Post by: TJ-56 on November 14, 2010, 02:14:53 pm
S07 Q 3, 7, 8, 23, 31, 40
Thank you
3
Its not about deriving a formula here, its about using SI units to get the result:
Balancing SI units on each side of the 4 choices, only A is balanced,
m/s = root(m s-2 * m ) sqr both sides
m2 s-2 = m2 s-2
8
I found the time taken to fall 40m - the time taken to fall 30 m , which is equivalent to the required time
s=40 t=? g=a=9.81 u=0
s=ut + 0.5at2 is used to find t, with using 40m once as s and 30m the other, and finding the difference, which is the answer A
31
Ohm’s Law: if a conductor obeys Ohm’s law, then the current I through it is directly proportional to the potential difference V across its ends provided that its temperature remains constant. That is current I ? V if temperature is constant.
40
You need to calculate the ratios proton/nucleons of each to find the speed, with the lowest being 0.428 of lithium
Post by: Dania on November 14, 2010, 02:16:10 pm
Post by: Alpha on November 14, 2010, 02:17:55 pm
3
Its not about deriving a formula here, its about using SI units to get the result:
Balancing SI units on each side of the 4 choices, only A is balanced,
m/s = root(m s-2 * m ) sqr both sides
m2 s-2 = m2 s-2
8
I found the time taken to fall 40m - the time taken to fall 30 m , which is equivalent to the required time
s=40 t=? g=a=9.81 u=0
s=ut + 0.5at2 is used to find t, with using 40m once as s and 30m the other, and finding the difference, which is the answer A
31
Ohm’s Law: if a conductor obeys Ohm’s law, then the current I through it is directly proportional to the potential difference V across its ends provided that its temperature remains constant. That is current I ? V if temperature is constant.
40
You need to calculate the ratios proton/nucleons of each to find the speed, with the lowest being 0.428 of lithium
Thank you so much. +rep
Now Dead King can rest a lil bit.
Post by: $!$RatJumper$!$ on November 14, 2010, 02:28:11 pm
3
Its not about deriving a formula here, its about using SI units to get the result:
Balancing SI units on each side of the 4 choices, only A is balanced,
m/s = root(m s-2 * m ) sqr both sides
m2 s-2 = m2 s-2
8
I found the time taken to fall 40m - the time taken to fall 30 m , which is equivalent to the required time
s=40 t=? g=a=9.81 u=0
s=ut + 0.5at2 is used to find t, with using 40m once as s and 30m the other, and finding the difference, which is the answer A
31
Ohm’s Law: if a conductor obeys Ohm’s law, then the current I through it is directly proportional to the potential difference V across its ends provided that its temperature remains constant. That is current I ? V if temperature is constant.
40
You need to calculate the ratios proton/nucleons of each to find the speed, with the lowest being 0.428 of lithium
Thank you :)
so why cant we do this: 10 = 0.5 * 9.81 * t2?
and so the ratio of the two gives the speed of them?
Post by: TJ-56 on November 14, 2010, 02:51:17 pm
Thank you :)
so why cant we do this: 10 = 0.5 * 9.81 * t2?
and so the ratio of the two gives the speed of them?
Because at 30m, it has an initial velocity, whereas ur equation assumes initial velocity is 0 (ut=0)
i remember reading it off a book, so i think yeah thats true.
Post by: $!$RatJumper$!$ on November 14, 2010, 03:07:13 pm
Because at 30m, it has an initial velocity, whereas ur equation assumes initial velocity is 0 (ut=0)
i remember reading it off a book, so i think yeah thats true.
But then by doing the method you mentioned, arent we also assuming that initial velocity for the 30m is zero?
Post by: Deadly_king on November 14, 2010, 03:14:46 pm
Thank you :)
so why cant we do this: 10 = 0.5 * 9.81 * t2?
and so the ratio of the two gives the speed of them?
You cannot use this equation since you'll be finding the time taken to cover the first 10m of the fall, which is not the required answer ;)
TJ-56 described the appropriate method. :)
Ok Alpha.........am gonna go rest a bit now.
Good luck for the exams tomorrow guys. :D
Post by: $!$RatJumper$!$ on November 14, 2010, 03:19:31 pm
You cannot use this equation since you'll be finding the time taken to cover the first 10m of the fall, which is not the required answer ;)
TJ-56 described the appropriate method. :)
Ok Alpha.........am gonna go rest a bit now.
Good luck for the exams tomorrow guys. :D
OK yeah i see how he did it :) it makes sense thankx
Post by: $!$RatJumper$!$ on November 14, 2010, 03:20:06 pm
W06 Q25
S07 Q23
Post by: TJ-56 on November 14, 2010, 03:45:29 pm
W06 Q25
S06 Q9 & Q33
S07 Q23
W07 Q35
S08 Q6 & Q12
W09 P11 Q17 Why is it C and not D
Sorry for all the questions but these are all i have, thanx for any help, rly appreciated
Post by: TJ-56 on November 14, 2010, 03:55:26 pm
W05 Q 24 (URGENT please)
W06 Q25
S07 Q23
W05 Q 24
Frequency P = 1/T = 1/ t20
So I is proportional to A2/T2
I0 / I(new) = ( x20/t20 ) / ( (2x)20/(2t)20 )
Simplifying would give I0, since both the time T and amplitudes are doubled , and since they are squared, they cancel each other out, and its back to the original so B
W06 Q25
I dont know it either, sorry
S07 Q23
Have doubts about that too
Post by: $!$RatJumper$!$ on November 14, 2010, 06:34:49 pm
I need help with the following pls
W06 Q25
S06 Q9 & Q33
S07 Q23
W07 Q35
S08 Q6 & Q12
W09 P11 Q17 Why is it C and not D
Sorry for all the questions but these are all i have, thanx for any help, rly appreciated
I need help on these too :( Someone?
Post by: TJ-56 on November 14, 2010, 06:44:47 pm
I need help on these too :( Someone?
Those are rly important, some1 help
and did u understand q24 RatJumper?
Post by: Dania on November 14, 2010, 07:15:19 pm
Post by: $!$RatJumper$!$ on November 14, 2010, 07:30:09 pm
Those are rly important, some1 help
and did u understand q24 RatJumper?
i did thank you buddy :)
Post by: $!$RatJumper$!$ on November 14, 2010, 07:31:18 pm
Post by: TJ-56 on November 14, 2010, 07:38:07 pm
S03 Q22 :S
Think of it this way: if it were just 1 spring at the beginning, the extension is 3x, since they are 3, the extension is x
1 is removed, so now its 3x / 2 , which is 1.5x
The load is doubled, so 2 * 1.5x
Which is 3x, response D.
Hope that was clear, i understood it that way
Post by: $!$RatJumper$!$ on November 14, 2010, 07:47:28 pm
Post by: Dania on November 14, 2010, 09:00:47 pm
The answer is B.
Post by: TJ-56 on November 14, 2010, 09:47:42 pm
Anyone?Energy= mgh
The answer is B.
The distance is halved
And the mass is also halved, because half move into the other vessel
so mgh/4
Post by: Dania on November 14, 2010, 09:48:40 pm
Energy= mgh
The distance is halved
And the mass is also halved, because half move into the other vessel
so mgh/4
So silly of me. Thanks!
Post by: Deadly_king on November 15, 2010, 04:54:38 am
I need help with the following pls
W06 Q25
S06 Q9 & Q33
S07 Q23
W07 Q35
S08 Q6 & Q12
W09 P11 Q17 Why is it C and not D
Sorry for all the questions but these are all i have, thanx for any help, rly appreciated
Nov 06 No 25
From X to Y, there is 1.5 cycle. This implies that X is 1.5T ahead of Y.
T is taken to be 2pie.
Phase difference = 1.5 x 2pie = 3pie. Hence n = 3.
Answer is C
Jun 06 p1
9. It says that the object moves up and down. So initially it is found at the bottom from where it is released. It's velocity increases until it passes the equilibrium position and then decreases to zero when it reaches its maximum at the top. This is point B.
The question asks the velocity at the lowest point of the motion.
Lowest point => V = 0ms-1
Since B shows the highest point, D is bound to be the answer. As the ball comes back to its initial lowest position, its velocity comes back to zero. ;)
Answer is D
33. A p.d of 12V is applied across P and Q. This implies p.d across both PXQ and PYQ is 12V since they are connected in parallel.
We need to find p.d in each resistor.
Along PXQ, p.d across 500 Ohm resistor is given by R500/Rtotal x p.d
p.d acorss 500 Ohm resistor will be 500/1500 x 12 = 4V. Therefore p.d across 1000 Ohm resistor will be (12 - 4) = 8V
Use the same formula(in bold) and you'll find p.d across 2000 Ohm resistor to be 8V while that across 1000 Ohm resistor to be 4V.
Hence p.d across X and Y = 8 - 4 = 4V (using either 500 and 2000 Ohms resistor or the two 1000 Ohms resistor) ;)
Answer is B
Jun 07 No 23
Am not sure about the explanation. So i'll proceed by elimination.
A : The speed is maximum at P -----> False since at maximum positions v= 0 ms-1.
B : The displacement at Q is always zero -----> this is not true since it is zero only at that instant. It will increase or decrease in other cases.
C : The energy at R is entirely kinetic. ----> At maximum or minimum positions, energy is rather entirely potential and zero kinetic. ;)
D : The acceleration at S is maximum. -----> Magnitude of acceleration should have been constant throughout the wave. Constant but maximum.
I find D as the most appropriate answer but am not sure if the reason I provided for D is valid. :-\
Answer is D
Nov 07 No 35
From diagram 1, we can note that both the voltmeter and the battery are connected in parallel to the system of resistors.
This is also the case in diagram 2. Both the voltmeter and the battery are still connected in parallel to the system of resistors. So both diagrams are the same and will have same readings. ;)
Voltmeter reading is zero since all resistors being identical will draw same amount of voltage. Hence p.d will come out to be zero.
Answer is A
Jun 08 p1
6. Acceleration is always perpendicular to its motion => Here we are talking about centripetal acceleration.
Hence speed remains constant but the velocity is said to be continuously changing since its direction changes. However its magnitude is constant.
Answer is C.
12. Using Newton's 2nd law of motion.
Resultant force = ma
Driving force - Resistive force = ma
2000 - Fr = 750(2) -----> Fr = 500 N or 0.5 kN
Answer is A
Nov 09 p11 No 17
Nope. It can't be D since ice is a solid and has a compact shape with strong intermolecular forces of attraction between its molecules. Therefore they won't be having same potential energies. So total energies won't be the same either. ;)
The answer is C because K.E is proportionl to temperature and both the water and the ice are at 0oC.
Hope it helps :)
Post by: cs on November 15, 2010, 05:11:13 am
Post by: Deadly_king on November 15, 2010, 05:19:08 am
Thanks Deadly King, done with the exam. =)
Glad that you're over. :)
No discussions please because i've not yet sit for it :P
Post by: TJ-56 on November 15, 2010, 06:17:04 am
Nov 06 No 25
From X to Y, there is 1.5 cycle. This implies that X is 1.5T ahead of Y.
T is taken to be 2pie.
Phase difference = 1.5 x 2pie = 3pie. Hence n = 3.
Answer is C
Jun 06 p1
9. It says that the object moves up and down. So initially it is found at the bottom from where it is released. It's velocity increases until it passes the equilibrium position and then decreases to zero when it reaches its maximum at the top. This is point B.
The question asks the velocity at the lowest point of the motion.
Lowest point => V = 0ms-1
Since B shows the highest point, D is bound to be the answer. As the ball comes back to its initial lowest position, its velocity comes back to zero. ;)
Answer is D
33. A p.d of 12V is applied across P and Q. This implies p.d across both PXQ and PYQ is 12V since they are connected in parallel.
We need to find p.d in each resistor.
Along PXQ, p.d across 500 Ohm resistor is given by R500/Rtotal x p.d
p.d acorss 500 Ohm resistor will be 500/1500 x 12 = 4V. Therefore p.d across 1000 Ohm resistor will be (12 - 4) = 8V
Use the same formula(in bold) and you'll find p.d across 2000 Ohm resistor to be 8V while that across 1000 Ohm resistor to be 4V.
Hence p.d across X and Y = 8 - 4 = 4V (using either 500 and 2000 Ohms resistor or the two 1000 Ohms resistor) ;)
Answer is B
Jun 07 No 23
Am not sure about the explanation. So i'll proceed by elimination.
A : The speed is maximum at P -----> False since at maximum positions v= 0 ms-1.
B : The displacement at Q is always zero -----> this is not true since it is zero only at that instant. It will increase or decrease in other cases.
C : The energy at R is entirely kinetic. ----> At maximum or minimum positions, energy is rather entirely potential and zero kinetic. ;)
D : The acceleration at S is maximum. -----> Magnitude of acceleration should have been constant throughout the wave. Constant but maximum.
I find D as the most appropriate answer but am not sure if the reason I provided for D is valid. :-\
Answer is D
Nov 07 No 35
From diagram 1, we can note that both the voltmeter and the battery are connected in parallel to the system of resistors.
This is also the case in diagram 2. Both the voltmeter and the battery are still connected in parallel to the system of resistors. So both diagrams are the same and will have same readings. ;)
Voltmeter reading is zero since all resistors being identical will draw same amount of voltage. Hence p.d will come out to be zero.
Answer is A
Jun 08 p1
6. Acceleration is always perpendicular to its motion => Here we are talking about centripetal acceleration.
Hence speed remains constant but the velocity is said to be continuously changing since its direction changes. However its magnitude is constant.
Answer is C.
12. Using Newton's 2nd law of motion.
Resultant force = ma
Driving force - Resistive force = ma
2000 - Fr = 750(2) -----> Fr = 500 N or 0.5 kN
Answer is A
Nov 09 p11 No 17
Nope. It can't be D since ice is a solid and has a compact shape with strong intermolecular forces of attraction between its molecules. Therefore they won't be having same potential energies. So total energies won't be the same either. ;)
The answer is C because K.E is proportionl to temperature and both the water and the ice are at 0oC.
Hope it helps :)
Thank you so much Deadly King! +rep (need to spread the love but i owe u one)
Thank u all for ur help
>GOOD LUCK TO EVERYONE<
Post by: Deadly_king on November 15, 2010, 02:03:42 pm
Thank you so much Deadly King! +rep (need to spread the love but i owe u one)
Thank u all for ur help
>GOOD LUCK TO EVERYONE<
Anytime dude ;)
Hope you manage well in the exams. :D
Post by: $!$RatJumper$!$ on November 15, 2010, 04:10:32 pm
Thank guys and best of luck in your other subjects :)
Post by: Deadly_king on November 15, 2010, 05:12:14 pm
Well i guess im done with this topic in the forums here :) I have learnt so much here its unbelievable! Thank you all so much for all the help... every question i asked was answered by someone and im really grateful for that :)
Thank guys and best of luck in your other subjects :)
Hehe..............it has been a pleasure to clear your doubts and help you in your studies. :D
I just hope that now that your exams are over, you're not going to leave SF. :-[
I don't know if you've noticed but we're going to have kinda festival on the occasion of Halloween. It would be really nice if you all participated.
This is an event where most if not all members of SF are invited. :D
Here (https://studentforums.biz/sf-halloween-festival/) is the link if you need any details. ;)
Post by: $!$RatJumper$!$ on November 15, 2010, 05:39:16 pm
Festival sounds gr8!
Post by: Ivo on November 17, 2010, 12:22:27 am
1) A golf ball is hit at 60ms-1. At what angle should it leave the club in order to travel 250m horizontally? You will need to use the double angle formula
2) Prove that, in the absence of air resistance, the maximum range of any projectile is achieved when it is launched at 45o to the horizontal.
3) How can two projectiles launched with the same speed but at different angles have he same range?
Post by: elemis on November 17, 2010, 07:47:47 am
Post by: Hypernova on November 17, 2010, 08:03:23 am
Small question : when a projectile is thrown upwards.... at the top/apex of its motion does it have only potential energy OR KE and PE ?
When a projectile is throw directly upwards it comes to an instantaneous rest at the apex. Since KE=1/2mv2, and v=0, it has no KE at the apex, just PE.
Its different if it was thrown at an angle
Post by: elemis on November 17, 2010, 08:05:12 am
When a projectile is throw directly upwards it comes to an instantaneous rest at the apex. Since KE=1/2mv2, and v=0, it has no KE at the apex, just PE.
Its different if it was thrown at an angle
Thanks. Just wanted to confirm.
How would it be different ?
Post by: Deadly_king on November 17, 2010, 08:15:23 am
Thanks. Just wanted to confirm.
How would it be different ?
This is because it will still have a horizontal component of velocity. Only the vertical component will be zero. ;)
So it's going to have P.E as well as K.E ;D
Post by: elemis on November 17, 2010, 08:21:30 am
This is because it will still have a horizontal component of velocity. Only the vertical component will be zero. ;)
So it's going to have P.E as well as K.E ;D
Yeah, I knew that. Thanks guys. ;)
+rep to both of you.
Post by: Deadly_king on November 17, 2010, 08:40:38 am
Yeah, I knew that. Thanks guys. ;)
+rep to both of you.
Anytime buddy :D
Post by: $H00t!N& $t@r on November 20, 2010, 01:04:58 pm
Find the diameter of a copper wire which has the same resistance as an aluminium wire of equal length and diameter 1.20 mm. The resistivities of copper and aluminium at room temperature are 1.7 x 10^-8 ohm metres and 2.6 x 10^-8 ohm metres respectively.
The answer is 0.97 mm
Post by: Katatonia on November 20, 2010, 01:40:30 pm
Find R for aluminium. l is constant, so don't bother. rho is resistivity.
R= 2.6x10^-8/ pi X 0.6^2
R= 2.3X10^-8
So diameter of copper wire for same Resistance is given by
2.3X10^-8= 1.7x10^-8/ pi X r^2
r=0.485mm
d=0.97 mm (2r)
Post by: elemis on November 20, 2010, 01:48:05 pm
Glad to see you finally posting around :D
Post by: elemis on November 26, 2010, 02:29:15 pm
How do we determine the answer ?
I've tried arranging them all in vector triangles, yet all of the lines are apparently in sequence and hence all systems are in equilibrium.
How do you do questions like these ?
Post by: ashish on November 27, 2010, 05:02:22 am
Question 15
How do we determine the answer ?
I've tried arranging them all in vector triangles, yet all of the lines are apparently in sequence and hence all systems are in equilibrium.
How do you do questions like these ?
according to me C and D cancels out because the length of one force in one direction is too big
and the answer is for sure A
i try to resolved the two perpendicular forces and the length is equal to the diagonal force
hope it helps :D
Post by: elemis on November 27, 2010, 08:50:43 am
What is the answer and why ?
Post by: blackberry on November 27, 2010, 09:23:35 am
Post by: elemis on November 27, 2010, 10:00:56 am
How is it that the acceleration at T = Q
Post by: ashish on November 27, 2010, 10:05:58 am
No, I dont understand.
How is it that the acceleration at T = Q
there was something wrong in my last post(i removed it )
here is the way to get the answer
at T there is only gravitational force which acts on the body as the body has stop decelerating (v=0)
after point T only gravitational force act on the body thus Accn at T should be equal to accn at Q
Post by: elemis on November 27, 2010, 10:34:54 am
there was something wrong in my last post(i removed it )
here is the way to get the answer
at T there is only gravitational force which acts on the body as the body has stop decelerating (v=0)
after point T only gravitational force act on the body thus Accn at T should be equal to accn at Q
Yeah, I think I've got it.
Thanks man :) +rep
Post by: elemis on December 02, 2010, 06:24:39 am
Post by: Deadly_king on December 02, 2010, 07:39:51 am
Part (a) whats the direction ?
It's got to be vertically upwards since it is a torque.
The torques of a couple only results when the two forces are parallel to each other. ;)
Post by: elemis on December 02, 2010, 08:35:37 am
3 (c)(iii)
Whats the direction ? I'd appreciate a diagram, as that would make things clearer.
Post by: Deadly_king on December 02, 2010, 11:31:52 am
http://www.xtremepapers.net/CIE/International%20A%20And%20AS%20Level/9702%20-%20Physics/9702_w01_qp_2.pdf (http://www.xtremepapers.net/CIE/International%20A%20And%20AS%20Level/9702%20-%20Physics/9702_w01_qp_2.pdf)
3 (c)(iii)
Whats the direction ? I'd appreciate a diagram, as that would make things clearer.
According to me, the direction should be from the hinge along the lid.
Try here (http://www.engin.brown.edu/courses/en3/notes/Statics/Constraints/Constraints.htm), it might clear your doubts. ;)
AM sorry but I can't provide for diagrams. :-[
Post by: elemis on December 05, 2010, 10:46:04 am
The vector triangles for both options are in sequence so whats the deciding factor that makes B the right answer ?
Post by: ashish on December 05, 2010, 12:10:59 pm
Why is the answer B and not A ?
The vector triangles for both options are in sequence so whats the deciding factor that makes B the right answer ?
the answer is B , hmm extend all the line of forces and you will notice that the lines meet at a single place.
Post by: elemis on December 05, 2010, 12:49:29 pm
the answer is B , hmm extend all the line of forces and you will notice that the lines meet at a single place.
Fine, I agree they do meet, but what does that indicate ?
Post by: ashish on December 05, 2010, 12:58:03 pm
Fine, I agree they do meet, but what does that indicate ?
oops i forgot to mention that in my last post ,, when all these lines of forces meet at a single point the means that the body is in equilibrium
Post by: elemis on December 05, 2010, 01:14:09 pm
oops i forgot to mention that in my last post ,, when all these lines of forces meet at a single point the means that the body is in equilibrium
Splendid. Thanks :)
Post by: ashish on December 05, 2010, 01:20:28 pm
Splendid. Thanks :)
you are welcome ..
Post by: Monopoly on December 08, 2010, 04:44:36 pm
the question:
a sinusoidal carrier wave has a frequency of 750kHz and an unmodulated amplitude of 4.0V. the carrier wave is to be amplitude-modulated by a sinusoidal signal of frequency 3kHz and the amplitude 0.5V. describe the modulated carrier wave.
now the answer at the end of the book says the amplitude varies from 3.5V to 4.0V. Can someone explain why this is so.
Post by: elemis on December 12, 2010, 02:25:05 pm
Question 7 (b) (ii)
To be honest I dont understand what the voltmeter is measuring i.e. I dont know across what the voltmeter is measuring a voltage.
Can someone please explain both parts in (ii) ?
Post by: Deadly_king on December 13, 2010, 08:40:16 am
http://www.xtremepapers.net/CIE/International%20A%20And%20AS%20Level/9702%20-%20Physics/9702_w05_qp_2.pdf (http://www.xtremepapers.net/CIE/International%20A%20And%20AS%20Level/9702%20-%20Physics/9702_w05_qp_2.pdf)
Question 7 (b) (ii)
To be honest I dont understand what the voltmeter is measuring i.e. I dont know across what the voltmeter is measuring a voltage.
Can someone please explain both parts in (ii) ?
The voltmeter is measuring the potential difference across the thermistor. When the voltmeter shows zero deflection, it means that the p.d across the thermistor is equal to the p.d across wire MQ. This is why the potential difference is zero as shown by the voltmeter.
7.(b) (ii)
1. Potential difference = 2.70 V -----> Part (a) describes it.
2. Resistance of wire is proportional to both the p.d(V) across wire and the length(L) of the wire.
V = IR -----> R is proportional to V
R =
Hence we can deduce that V is also proportional to L.
This implies that V1/L1 = V2L2 -----> 4.5/100 = 2.7/L2 ----> L2 = 60.0 cm
Hope it helps :)
Post by: elemis on December 13, 2010, 08:49:22 am
Post by: Deadly_king on December 13, 2010, 09:56:11 am
I understood everything you explained and thank you for that, but I still dont get how the voltmeter is measuring the p.d. across the thermistor.
Just a mere precision. ;)
It's not only measuring the p.d across the thermistor but rather the p.d across the circuit BAQM.
When it shows zero deflection ----> p.d across BA = p.d across QM
Post by: $H00t!N& $t@r on December 14, 2010, 07:41:52 pm
Post by: Freaked12 on December 14, 2010, 08:48:41 pm
as 2m has a greater mass,it will exert more force on the first
sphere
hence the spheres velocity will increse due to collision with a heavier object
and it will move in an opposite direction.
put the values of initial mass and velocity into vf1=vf1= ((m1-m2)/(m1+m2))(v1i)+ (2m2/(mi+m2)vi2
u will get the answer
Post by: elemis on December 15, 2010, 03:28:30 am
can some1 please help me with q9 and 10 - O/N 2009 P1 variant 11
You dont need to do what requiem said.
A simpler method is to consider the velocities :
U1 - U2 = V2 - V1
Where U = initial velocity before collision
V = final velocity after collision.
For an elastic collision the left hand side of the above equation should be equal to the right hand side.
Plug in the necessary values and remember velocity is a VECTOR.... you should get the right answer ;)
Post by: elemis on December 15, 2010, 03:34:47 am
Consider the picture below.
Using Impulse = Force * time
Impulse = 60*0.5
Impulse = 30Ns
Therefore, the change in momentum of the trolley should be equal to 30Ns.
Hence, using m(v-u) = change in momentum
We get : 30(-v+3) = 30
-v = -2
Hence, v = 2 ms-1
Post by: Sue T on December 19, 2010, 09:27:29 am
(l) recall and solve problems using the realtion hf = E1 - E2
ny idea wth ths s bout? - all i no is :
hf = phi + 1/2mv2
is it related 2 tht or wat?
Post by: 6394 on December 20, 2010, 11:02:59 am
and "work done on a system" ??
thnq :)
Post by: Deadly_king on December 24, 2010, 05:22:50 am
plz explain the difference btwn "work dne by a system"
and "work done on a system" ??
thnq :)
For the explanation, i'll take the system to be a gas. ;)
1. Consider a gas which expands at constant pressure.
(http://exploration.grc.nasa.gov/education/rocket/Images/work2.gif)
This will cause the final volume(Vf) to be larger than the initial volume.(Vi)
Hence this will be work done by gas = F x L = P x A x L = P ( Vf - Vi)
2. Consider a gas being compressed at constant pressure.
Work done on gas = P (Vi - Vf)
Work done on gas is just the opposite of work done by gas.
EXPANSION.
Work done by gas = P ( Vf - Vi) is positive while Work done on gas = P (Vi - Vf) is negative.
COMPRESSION.
Work done by gas = P ( Vf - Vi) is negative while Work done on gas = P (Vi - Vf) is positive.
Hope it helps :D
Post by: aloha32 on January 03, 2011, 12:07:56 pm
1.Two vessels one having 3 times the volume of the other are connected by a narrow tube of negligible volume. Initially the whole system is filled with a gas at a temperature of 290 K. The smaller vessel is cooled to 250 k and the larger vessel is heated to 400 K. Find the final pressure of the system.
2. The molar mass of nitrogen is 28g. A sample of gas contains 6.02X10 ^23 molecules Calculate the number of moles of the gas , the mass of the gas and the volume of the gas at .110 Pa and 290 K
3. An oxygen cyclinder contains .50 kg of gas at a pressure of .50 PA and 7 degree centigrade. What mass of oxygen must be pumped into the cylinder to raise its pressure to 3 mega pascals at 27 degrees. If the molar mass of oxygen is 32g calculate the volume of the cylinder
I often get confused b/w the whole number of moles and number of molecules thing :S
Thanks!
To Monopoly : Are you sure it isn't 4.5 and 3.5 because then it makes sense..because the amplitude would vary between the voltages
Post by: $H00t!N& $t@r on January 04, 2011, 03:55:02 pm
O/N 2009 PAPER 1 VAR 11 QUESTION 9, 10, 13, 22
here is the link: http://www.freeexampapers.com/get_past_papers.php?l=Past_Papers%2FA+Level%2FPhysics%2FCIE%2F2009+Nov%2F9702_w09_qp_11.pdf (http://www.freeexampapers.com/get_past_papers.php?l=Past_Papers%2FA+Level%2FPhysics%2FCIE%2F2009+Nov%2F9702_w09_qp_11.pdf)
Post by: The Golden Girl =D on January 04, 2011, 05:58:10 pm
Post by: elemis on January 05, 2011, 05:18:53 am
hi i have a question..
O/N 2009 PAPER 1 VAR 11 QUESTION 9, 10, 13, 22
here is the link: http://www.freeexampapers.com/get_past_papers.php?l=Past_Papers%2FA+Level%2FPhysics%2FCIE%2F2009+Nov%2F9702_w09_qp_11.pdf (http://www.freeexampapers.com/get_past_papers.php?l=Past_Papers%2FA+Level%2FPhysics%2FCIE%2F2009+Nov%2F9702_w09_qp_11.pdf)
Question 9
Use the formula : U1 - U2 = V2 - V1
That is, for an elastic collision the relative velocities of approach are equal to the relative velocities of separation.
NOTE : You are considering Velocity which is a vector quantity and who's sign changes with direction.
Question 10
Using impulse = force * time it
60 * 0.5 = 30 Ns
Hence, it is given that the retarding force caused a 30 Ns change in momentum.
That is : 30(-v+3) = 30Ns where v = final velocity
Hence, v = 2
Question 22
Stress/strain = Young's Modulus
Stress = 20 / [pi*(2.5*10-4)2] <---- let this be equation 1
Divide 1 by 2*1011 = 5.1*10^-4
However, this is equal to strain which is basically change in length divided by original length. It is ratio so we multiply by 100 to give :
5.1*10^-2
Post by: Sue T on January 08, 2011, 09:16:03 am
ppr 4 mj 07 ques4
The output from the transformer is to be full-wave rectified. Fig. 4.1 shows part of the
rectifier circuit.
(attached - we had 2 put in th diodes on our own)
ii) The resistance of the resistor R is doubled. On Fig. 4.2, sketch the variation with
time t of the potential difference V across the resistor.
fig 4.2 is a graph also attached
my question is (ii)
the ans was:
(ii) sketch: same peak values
ripple reduced and reasonable shape
reduced ripple as in theres less 'curviness' or like it goes down lower?
and if its less 'curviness' then why?
Post by: Sue T on January 08, 2011, 01:38:06 pm
Q)A cobalt-60 source having a half-life of 5.27 years is calibrated and found to have an
activity of 3.50 × 105 Bq. The uncertainty in the calibration is ±2%.
Calculate the length of time, in days, after the calibration has been made, for the stated
activity of 3.50 × 105 Bq to have a maximum possible error of 10%.
ANS)source must decay by 8%
A = A0 exp(–ln2 t / T½) or A/ A0 = 1 / (2t/T)
0.92 = exp(–ln2 × t / 5.27) or 0.92 = 1 / (2t/5.27)
t = 0.634 years
= 230 days
why must th source decay by 8%? how did they get tht?
my mock xam s 2moro - i need urgent help :(
Post by: $H00t!N& $t@r on January 22, 2011, 01:55:10 pm
You dont need to do what requiem said.
A simpler method is to consider the velocities :
U1 - U2 = V2 - V1
Where U = initial velocity before collision
V = final velocity after collision.
For an elastic collision the left hand side of the above equation should be equal to the right hand side.
Plug in the necessary values and remember velocity is a VECTOR.... you should get the right answer ;)
I'm not getting the right answer :-[ can you solve it for me :-\
Post by: $H00t!N& $t@r on January 22, 2011, 02:22:09 pm
Question 9
Use the formula : U1 - U2 = V2 - V1
That is, for an elastic collision the relative velocities of approach are equal to the relative velocities of separation.
NOTE : You are considering Velocity which is a vector quantity and who's sign changes with direction.
Question 10
Using impulse = force * time it
60 * 0.5 = 30 Ns
Hence, it is given that the retarding force caused a 30 Ns change in momentum.
That is : 30(-v+3) = 30Ns where v = final velocity
Hence, v = 2
Question 22
Stress/strain = Young's Modulus
Stress = 20 / [pi*(2.5*10-4)2] <---- let this be equation 1
Divide 1 by 2*1011 = 5.1*10^-4
However, this is equal to strain which is basically change in length divided by original length. It is ratio so we multiply by 100 to give :
5.1*10^-2
Thanks for the answer. :) However, question 9 is not quite clear... :-\
Post by: donhassan on February 03, 2011, 08:41:26 pm
i have set of doubts in may june 2003 cie paper question number 5 ,15,18,20,22
please please please try ur best to explain them in a very lucid way
Post by: elemis on February 09, 2011, 03:06:53 pm
Question 30
Post by: astarmathsandphysics on February 09, 2011, 09:11:58 pm
Post by: elemis on February 10, 2011, 06:33:14 am
I said the path difference must be a odd number of non integer half wavelengths
and
the vector sum of the amplitudes must be zero.
The mark scheme says :
either same amplitude / intensity at M
or ratio of amplitudes is 1.28 / ratio of intensities is 1.28^2
Can someone please explain the statements in bold ? How can same amplitudes lead to an intensity of zero ?
Post by: Deadly_king on February 10, 2011, 06:54:02 am
Question 5 (a)
I said the path difference must be a odd number of non integer half wavelengths and the vector sum of the amplitudes must be zero.
The mark scheme says :
either same amplitude / intensity at M
or ratio of amplitudes is 1.28 / ratio of intensities is 1.28^2
Can someone please explain the statements in bold ? How can same amplitudes lead to an intensity of zero ?
My answers would have been :
1. The waves should meet in antiphase at M.(Phase difference = pie)
2. The sources must emit waves having the same amplitudes.
Same amplitudes meaning that their value should be the same but in antiphase.
Example : Amplitude of S1 = 10mm, amplitude of S2 = -10mm.
What you said about vector sum should be zero is correct but for amplitude we do not use the term vector. We just say they are equal in amplitude but in antiphase ;)
By ratio of amplitudes, the MS used it in terms of scale drawing. Use pythagoras theorem to find the direct distance from S2 to M. You'll see that it comes out to be 128cm.
Length is directly proportional to the amplitude and also to the square of intensity. ;)
(b) Distance that waves from S1 have to covered is 100cm while those from S2 have to cover 128cm. The difference therefore is 28cm.
Path difference = nV/F
Using the speed of sound and n=1 ----> F1 = (1 x 330)/0.28 = 1179
Hence F2 will be (2 x 1179) = 2358 and F3 = 3(1179) = 3535.
It can be easily seen that F4 will exceed the limit of 4.0 KHz.
So we'll be having 3 maxima at 1179, 2358 and 3535. This implies that we'll be having only 2 minima.
Hence answer is 2.
Hope it helps :)
Post by: elemis on February 10, 2011, 07:02:10 am
Path difference = nV/F
Woah ! Where's this formula from ?
I've never heard of it. What do all the variables stand for ?
Post by: Deadly_king on February 10, 2011, 07:27:05 am
Woah ! Where's this formula from ?
I've never heard of it. What do all the variables stand for ?
Hehe......this is derived form the formula V = f(lambda)
Lambda is replaced by n
Post by: elemis on February 10, 2011, 07:30:44 am
Hehe......this is derived form the formula V = f(lambda)
Lambda is replaced by nwhere n represents the number of maximum or minimum and
So can I also use this formula to determine the frequency at which a particular minima/maxima occurs at ?
Like if I want to find the frequency at which the 5th minima occurs at can this formula be used ?
One other thing, supposing the question asked me to find the no. of MAXIMA would the answer still be THREE ?
Post by: Deadly_king on February 10, 2011, 07:32:57 am
So can I also use this formula to determine the frequency at which a particular minima/maxima occurs at ?
Like if I want to find the frequency at which the 5th minima occurs at can this formula be used ?
One other thing, supposing the question asked me to find the no. of MAXIMA would the answer still be THREE ?
Actually Path difference = n(lamda) ---> little mistake I made above.
Yupz.......you can. :)
Three it will be. :D
Post by: elemis on February 10, 2011, 07:35:58 am
Actually Path difference = n(lamda) ---> little mistake I made above.
Hmm ? I dont understand ?
Post by: Deadly_king on February 10, 2011, 07:42:14 am
Hmm ? I dont understand ?
Earlier I said Lamda is replaced by n
The true equation is
Get me, now?
Post by: elemis on February 10, 2011, 08:05:58 am
Earlier I said Lamda is replaced by n
The true equation is
Get me, now?
Yeah, yeah. I understand.
Post by: HUSH1994 on February 11, 2011, 05:06:48 pm
Post by: SauD~ on February 11, 2011, 05:45:13 pm
paper 11 ON 2010 Q 25,why is it A not B :S and isnt there the editor's reports?is this IGCSE or A'Level doubt?
it seems to me that it is IGCSE:
so the answer is A because the word echo means that the sound has traveled twice times, so half the time 0.4 -> 0.2
now apply the formula... distance = Speed x time
1200 x 0.2 = 240 ;)
Post by: The Golden Girl =D on February 12, 2011, 11:11:55 am
is this IGCSE or A'Level doubt?
it seems to me that it is IGCSE:
so the answer is A because the word echo means that the sound has traveled twice times, so half the time 0.4 -> 0.2
now apply the formula... distance = Speed x time
1200 x 0.2 = 240 ;)
Saud this is in the A level board so it's A Level not IGCSE ;)
Post by: elemis on February 12, 2011, 11:16:24 am
paper 11 ON 2010 Q 25,why is it A not B :S and isnt there the editor's reports?
I = k A2
Hence, we square the amplitude to find the intensity.
Thus, 0.52 = 0.250
Post by: SauD~ on February 12, 2011, 04:21:29 pm
Saud this is in the A level board so it's A Level not IGCSE ;)i am never able to help anyone :'( :'( :(
Post by: aqws0987 on February 13, 2011, 04:12:42 am
Question:
The manufacturer of a digital ammeter quotes its uncertainty as +/-1.5% +/-2 digits
The meter is used to measure the current from a a.c. power supply. The currernt is found to fluctuate randomly between 1.58A and 2.04A. Determine the most likely value of the current, with its uncertainty
Answer given is (2.01 +/- 0.09) A
How is the answer obtained?
Post by: Sue T on February 14, 2011, 12:59:05 pm
pg 8 at the bottom ''the input impedance of the op-amp....'' and fig. 1.13
i dont get it :-\
Post by: astarmathsandphysics on February 14, 2011, 08:30:27 pm
What does +/- 2 digits mean
Post by: aqws0987 on February 16, 2011, 12:57:28 pm
Don't know what it means tho :-\
Post by: tmisterr on February 20, 2011, 05:21:51 pm
i have an A2 doubt from the app booklet:
pg 8 at the bottom ''the input impedance of the op-amp....'' and fig. 1.13
i dont get it :-\
Electrical Impedance, is the total opposition that a circuit presents to alternating current. It is a combination of resistance and capacitive reactance I think, it's quite complex but for A2 physics all you need to is they use impedance to mean resistance. since I=V/R, if an op-amp has a very large input impedance(i.e resistance), the value of current will tend toward zero so essentially the op-amp is not drawing any current from the input
Here read these short notes, they will help with your understanding of Op-amps:
http://www.williamson-labs.com/480_opam.htm
I recommend starting with the paragraph that reads "Anyway, because of the aforementioned...." up to "This high impedance input can be used to great advantage: in sample & hold circuits, peak detectors--you name it. "
the rest is un-important literature in the scope of A2 physics
Post by: aloha32 on February 21, 2011, 05:54:55 am
V (Volts) : 5.0 +/- .2
I (10^-3 Amps) : 15 +/- .2
X (Ohms) : 330 (as i calculated to 2 sf)
I found the percentage error for V as 4%. But I don't know what to do for I. Do I keep it as (.2/15 X 100) or (.2/.015 X 100)?
Also, for the value of 1/f why do we have to give 3 sf when all the values given to us are in 2 sf? ???
Here is the link of the mark scheme :
http://www.freeexampapers.com/get_past_papers.php?l=Past_Papers%2FA+Level%2FPhysics%2FCIE%2F2010+Jun%2F9702_s10_ms_51.pdf
Thank you!
Post by: cs on March 06, 2011, 07:19:02 am
Nov 2004 paper 4 Q5
I couldn't upload the attachment sorry
Post by: Deadly_king on March 07, 2011, 10:47:28 am
Can someone please help me with P5 June 2010 Q2b? I need to know what would be the correct absolute error for this
V (Volts) : 5.0 +/- .2
I (10^-3 Amps) : 15 +/- .2
X (Ohms) : 330 (as i calculated to 2 sf)
I found the percentage error for V as 4%. But I don't know what to do for I. Do I keep it as (.2/15 X 100) or (.2/.015 X 100)?
Also, for the value of 1/f why do we have to give 3 sf when all the values given to us are in 2 sf? ???
Here is the link of the mark scheme :
http://www.freeexampapers.com/get_past_papers.php?l=Past_Papers%2FA+Level%2FPhysics%2FCIE%2F2010+Jun%2F9702_s10_ms_51.pdf
Thank you!
It should be of same units. So in this case where current has been given as I (10^-3 Amps) : 15 +/- .2 the calculations for absolute errors are :
% error for I = (0.2 x 10-3)/(15 x 10-3) x 100 = 13%
Usually we are to give answers to the same number of sf or one better. I think that answers with 2 sf should also be accepted.
I think from now on, you can proceed with the rest of the question yourself. :)
Post by: sarsam on March 12, 2011, 06:25:05 pm
So I came back with a question in my hand ;D
Time taken for 10 oscillation is given 2.83s with uncertainity 0.01
How to calculate uncertainity of time period in this case?
Do we have to divide the uncertainity of time i.e 0.01 with 10?
Post by: xlane on March 12, 2011, 06:43:57 pm
Post by: tmisterr on March 13, 2011, 11:10:22 am
may june 2007.....paper 5 ,code 9702......question no. 2.....part b.....hiw do u find the uncertainity of l square......L=6.0 (+_) 0.4.....so wat will be l square ??
62=36
fractional uncertainty in l is 0.4/6. so fractional uncertainty for l2 which is l*l it will be 2*(0.4/6)=0.8/6
so the absolute error in l2= (0.8/6)*36=4.8
since absolute error is written on one significant figure, it becomes 5
so L2 is 36+-5 cm2
Post by: xlane on March 21, 2011, 02:33:42 pm
Post by: sabrina on March 25, 2011, 09:32:26 am
i dont know anything. plz help
Post by: elemis on March 26, 2011, 07:59:27 am
Post by: basoom16 on March 28, 2011, 10:18:32 am
two +30 micro columb charges are placed on a straight line 0.4 m apart. A +0.5 micro columb charge is to be moved a distance of 0.10 m along the line from a point midway between the charges. how much work must be done.
can anyone please show me how its done.
thankyou
Post by: elemis on March 29, 2011, 09:27:06 am
question
two +30 micro columb charges are placed on a straight line 0.4 m apart. A +0.5 micro columb charge is to be moved a distance of 0.10 m along the line from a point midway between the charges. how much work must be done.
plz answer thnk u
I havent done A2 Physics yet.
Post by: HUSH1994 on April 01, 2011, 02:34:25 pm
Post by: ashish on April 01, 2011, 02:50:04 pm
MJ 2002 Paper 1 Q26,how do u solve this type of questions?
well 1 complete wave is formed in 4 squares therefore T= (4*2ms)
=8ms
f= 1/(8*10^-3)
=125
Post by: HUSH1994 on April 01, 2011, 02:57:23 pm
Post by: ashish on April 01, 2011, 03:08:35 pm
thanks man,can u plz help me with Q28
At x there is destructive interference, therefore the path difference should be and odd multiple of of lambda by 2
ans :c
Post by: HUSH1994 on April 01, 2011, 04:16:13 pm
Post by: elemis on April 01, 2011, 04:48:11 pm
sorry ashish im not getting what you say,sorry for the trouble but please someone help with Q31
Can you attach the paper and mark scheme ?
Post by: ashish on April 03, 2011, 04:05:08 am
sorry ashish im not getting what you say,sorry for the trouble but please someone help with Q31
Have you understand WAVE completely?
there is a chapter called interference! it states that when two wave completely out of phase meet there is destructive interference! no sound/ light will be heard , the experiment is just like YOUNG's double slits experiment. you have just to apply the formula (2m+1)(wavelength/2)
ques 31 EMF is defined as the work done/ energy dissipated per unit charged moved round the circuit
I=Q/t
Ans:C
Post by: ashish on April 03, 2011, 04:30:37 am
Post by: ashish on April 03, 2011, 04:40:07 am
Have you understand WAVE completely?
there is a chapter called interference! it states that when two wave completely out of phase meet there is destructive interference! no sound will be heard , the experiment is just like YOUNG's double slits experiment. you have just to apply the formula (2m+1)(wavelength/2)
ques 31 EMF is defined as the work done/ energy dissipated per unit charged moved round the circuit
I=Q/t
Ans:C
Post by: iluvme on April 04, 2011, 04:20:38 pm
Post by: elemis on April 04, 2011, 04:56:54 pm
November 2009 Paper 2 Varient 2, Q5
Its a rather daft question on their part; they're trying to confuse you.
The informaion in the question states that the frequency is f.
Also, the amplitude is simply A since it has been stated in the question as such.
Post by: aloha32 on April 06, 2011, 08:09:38 am
June 2004 - Medical Physics- Question 8 (b)..I can't figure out how they got the final length.I got to the part where from the graph, the time b/w pulse in 20 microseconds but can't go further.
Also, another question (not included in the paper though) is :
Give 3 reason why MR scanning would not be used to diagnose a simple bone fracture.
Post by: muzxx on April 08, 2011, 01:31:31 pm
[N90/III/1]
Please help mii.. :$
Post by: elemis on April 08, 2011, 01:35:53 pm
Explain, why in practice trains do not have a constant acceleration.
[N90/III/1]
Please help mii.. :$
Can you attach the paper ? I dont have that question paper.
Post by: muzxx on April 08, 2011, 04:07:54 pm
Post by: ashish on April 10, 2011, 06:21:59 am
i got this question from a past exam book (topic by topic), in the chapter kinematics...the question is like this.. sorry :(
hmm one factor is air resistance, as the speed increases (acceleration) the net force acting on the body is affected therefore there is a change in acceleration.
Post by: Master_Key on April 11, 2011, 12:36:52 pm
hmm one factor is air resistance, as the speed increases (acceleration) the net force acting on the body is affected therefore there is a change in acceleration.
Net force = Force by engine - Retarding Force(Air resistance, Friction, etc.)
As the speed increase the Retarding Force(Air resistance, Friction, etc.) increases. You have learn't this. Falling bodies have a maximum terminal velocity. Their velocity will not increase after this. So at this time NF = 0.
F=MA
0=MA
A=0.
As F decrease F is directly proportional to A.
Acceleration decreases. The acceleration is not constant.
Hope it helps.
Post by: mdwael on April 11, 2011, 10:53:43 pm
(a) Express "M" interms of "m".
(b) Determine the fraction of the original kinetic energy retained by the ball of mass "m" after the collision.
Pls urgently answer this question from my text book its so confusing and I dont get the correct answer in the back of the text book :(
Post by: EMO123 on April 12, 2011, 09:53:39 am
Is this question complete? I am trying to answer it, i think it must have something else too.something is missing in this question
(a)If both the balls move with half the velocity of ball "m" then
By using momentum
m1v1 = m1v1+M2v2
Take v = 1
m = 0.5m + 0.5M
m = M
If "m" stays stationary and only ball "M" moves with half the velocity then
m1v1 = M2v2
v2 = 2v1
M = 2m.
(b) Fraction of Ek
Ek = 1/2mv2.
v=0.5u
Ek before collision
Ek = 1/2m(u)2.
Ek after collision
Ek = 1/2m(.5u)2.
(1/2m(.5u)2)/(1/2m(u)2)
=.25u/u
=1/4th Ek
Post by: HUSH1994 on April 12, 2011, 02:46:56 pm
Post by: mdwael on April 12, 2011, 03:19:37 pm
Is this question complete? I am trying to answer it, i think it must have something else too.
(a)If both the balls move with half the velocity of ball "m" then
By using momentum
m1v1 = m1v1+M2v2
Take v = 1
m = 0.5m + 0.5M
m = M
If "m" stays stationary and only ball "M" moves with half the velocity then
m1v1 = M2v2
v2 = 2v1
M = 2m.
(b) Fraction of Ek
Ek = 1/2mv2.
v=0.5u
Ek before collision
Ek = 1/2m(u)2.
Ek after collision
Ek = 1/2m(.5u)2.
(1/2m(.5u)2)/(1/2m(u)2)
=.25u/u
=1/4th Ek
That is actually wrong because you didnt use the law (velocities of approach= velocities of seperation) M=3m
Post by: mdwael on April 12, 2011, 03:23:30 pm
" In ordere to strengthen her legs, an athlete steps on a box and then down again 30 times per minute. The girl has mass 50kg and the box is 35cm high. The exercise lasts 4.0 minutes and as a result of the exercise, her leg muscles generate 120kJ of heat energy. Calcualate the efficiency of the leg muscles (g= 10m/s^2) "
Post by: tmisterr on April 12, 2011, 08:07:58 pm
work done for one step up is (50*10)*0.35=175J. while stepping down, her weight does the work so don't worry about that. ok so she does this 30 times in one minute for four minutes so the total work done (energy from legs) is equal to 175*30*4=21,000J. Her legs dissipate 120,000 joules in the form of heat energy. so the total energy from the legs is 21,000+120,000=141,000J
efficiency is equal to (work out/work in) * 100. Work input by the legs was 141,000J. but 120,000J was dissipated as heat energy so the useful work (work out, which is the energy used during the exercise) is 21,000J. replacing these values into the equation above
efficiency=(21,000/141,000)*100=14.8936......% which is approximately 15% efficiency
Post by: Master_Key on April 13, 2011, 02:54:05 pm
That is actually wrong because you didnt use the law (velocities of approach= velocities of seperation) M=3m
Does the question specify that the BALLS MOVE IN OPPOSITE DIRECTION?
Sorry for that but in this case :-
m1v1 + M1v1 = M2v2 - m1v1
Ball "M" is stationary.
m1v1 = M2v2 - m1v2
m1v1 + m1v2 = M2v2
m1+.5m1 = .5M2
1.5m1 = .5M2
M2 = 2*1.5m1
M2 or M = 3m
So did you get your answer. I thought something was missing and it was the diagram. I forgot to add case 3 in my first reply.
Case 3 :- ball "m" moves in opposite direction with .5 velocity
ball "M" moves in direction of impact with .5 velocity.
Post by: Master_Key on April 13, 2011, 02:58:28 pm
Ek = 1/2mv2.
v=0.5u
Ek before collision
Ek = 1/2m(u)2.
Ek after collision
Ek = 1/2m(.5u)2.
(1/2m(.5u)2)/(1/2m(u)2)
= .25u
------------
u
= 1
-- th Ek
4
Post by: Master_Key on April 13, 2011, 03:07:34 pm
MJ 2006 P1 Q15, I get the answer 40N anticlockwise as the moment at the 300N force is 120 while at the 200N force 160,so the difference is 40N and should be anticlockwise,can anyone explain it to me?and question 31 on the same paper please?
Q15 - Should be clockwise.
as 160N acting anti-clockwise
to get in equilibruim
160 - 160 = 0
-160N is 160 N in clockwise direction.
And you correctly got 40 N so i am not doing it.
Q31
charge = 1.6 * 10-19
4.8A = 4.8C
Rate of Flow = 4.8
------------
1.6 * 10-19
Rate of Flow = 3 * 1019 s-1
Post by: mdwael on April 13, 2011, 04:04:41 pm
Does the question specify that the BALLS MOVE IN OPPOSITE DIRECTION?
Sorry for that but in this case :-
m1v1 + M1v1 = M2v2 - m1v1
Ball "M" is stationary.
m1v1 = M2v2 - m1v2
m1v1 + m1v2 = M2v2
m1+.5m1 = .5M2
1.5m1 = .5M2
M2 = 2*1.5m1
M2 or M = 3m
So did you get your answer. I thought something was missing and it was the diagram. I forgot to add case 3 in my first reply.
Case 3 :- ball "m" moves in opposite direction with .5 velocity
ball "M" moves in direction of impact with .5 velocity.
Yes now its correct, the question did not specify anything other than what I wrote..its fully compete and nothing is missing thats why it took me a lot to know that the ball moves in opposite direction..
Post by: HUSH1994 on April 13, 2011, 04:49:37 pm
Q15 - Should be clockwise.
as 160N acting anti-clockwise
to get in equilibruim
160 - 160 = 0
-160N is 160 N in clockwise direction.
And you correctly got 40 N so i am not doing it.
Q31
charge = 1.6 * 10-19
4.8A = 4.8C
Rate of Flow = 4.8
------------
1.6 * 10-19
Rate of Flow = 3 * 1019 s-1
Thanks dude +rep ;)
Post by: Master_Key on April 14, 2011, 08:50:55 am
Yes now its correct, the question did not specify anything other than what I wrote..its fully compete and nothing is missing thats why it took me a lot to know that the ball moves in opposite direction..
I think the same happened with me.
Post by: joel on April 19, 2011, 04:31:14 am
- May/June 2002
Q6 (b) (iii) - where should capacitor be placed? Mark Scheme states "capacitor connected across SQ". Should two capictors be conected one on each side of the load after the bridge rectifier.
- October/November 2002
Q3 (b) (i) and (ii) - Can someone explain the concept and solve the questions.
Q6 (c) (i) Why is E upwards?. Mark Scheme states "arrow pointing up page"
- May/June 2003
Q1 (c) (i) and (ii)
Q2 (b). Also please explain the concept.
Post by: HUSH1994 on April 19, 2011, 04:52:29 pm
Post by: $H00t!N& $t@r on April 19, 2011, 07:58:46 pm
Can someone please help me with Q 17, 18, 21, 25 and 27 from o/n 2006 p1
Here is the link ---> http://www.freeexampapers.com/past_papers.php?l=Past_Papers%2FA+Level%2FPhysics%2FCIE%2F2006+Nov/ (http://www.freeexampapers.com/past_papers.php?l=Past_Papers%2FA+Level%2FPhysics%2FCIE%2F2006+Nov/)
Post by: srk on April 19, 2011, 10:10:30 pm
May/June 2002, Q4(b) and Q7(c)
Oct/Nov 2002, Q2 (b)
and Oct/Nov 2009 Q11(c) i)
Awaiting your response..
Post by: EMO123 on April 20, 2011, 03:00:16 pm
Hey guys,
Can someone please help me with Q 17, 18, 21, 25 and 27 from o/n 2006 p1
Q=17
ans= as 93 Joule of Energy is emitted as Thermal Energy and left 7 joule is emitted as light energy :. 7% is the efficiency of electrical energy is converted to light energy. :. A
Post by: EMO123 on April 20, 2011, 03:05:54 pm
I have doubts in Physics/9702/Paper- IV:you are only awaiting for the completing the diagrams
May/June 2002, Q4(b) and Q7(c)
Oct/Nov 2002, Q2 (b)
and Oct/Nov 2009 Q11(c) i)
Awaiting your response..
Post by: Dania on April 21, 2011, 06:27:30 am
Post by: EMO123 on April 21, 2011, 06:46:12 am
Hey guys,
Can someone please help me with Q 17, 18, 21, 25 and 27 from o/n 2006 p1
Here is the link ---> http://www.freeexampapers.com/past_papers.php?l=Past_Papers%2FA+Level%2FPhysics%2FCIE%2F2006+Nov/ (http://www.freeexampapers.com/past_papers.php?l=Past_Papers%2FA+Level%2FPhysics%2FCIE%2F2006+Nov/)
Q 18
4 MW is maximum is 80%.
Then 100% is 5 MW.
P = IV
5 X 106 = I X 25 X 103
I = 200 A
Q 25
Count the phase difference from X. 3 (Crest to trough) = 720 = 3 pie. n = 3.
Q 27
n X Lamda = d sin x
n = 1 (first order)
N = 500
d = 1/N
Lamda = d X sin x
------------
n
= (1/500) X Sin 30
= 1 X 10-3 m
Post by: EMO123 on April 21, 2011, 06:47:50 am
Post by: HUSH1994 on April 21, 2011, 08:27:47 am
Pressure of oil = pgh = 830*9.81*X
Total Pressure = 17.5*10^6 Pa(From Mega *one million)
17.5*10^6 = (19620000 - 9810X) + 8142.3X
(17.5*10^6) - (1.962*10^7) = -9810X + 8142.3X
-2.12*10^6 = -1667.7X
X = -2.12*10^6
------------ = 1270 to 3 sf
-1667.7
Hope it helped :)
Post by: EMO123 on April 21, 2011, 04:22:34 pm
Q21.Pressure of water = pgh = 1000*9.81*(2000-X)thanx but after sometime i just solved from my friend but i could not post that time
Pressure of oil = pgh = 830*9.81*X
Total Pressure = 17.5*10^6 Pa(From Mega *one million)
17.5*10^6 = (19620000 - 9810X) + 8142.3X
(17.5*10^6) - (1.962*10^7) = -9810X + 8142.3X
-2.12*10^6 = -1667.7X
X = -2.12*10^6
------------ = 1270 to 3 sf
-1667.7
Hope it helped :)
then also thanxx for the try
Post by: HUSH1994 on April 21, 2011, 04:59:36 pm
Post by: HUSH1994 on April 23, 2011, 07:13:33 am
A wave of amplitude a has intensity of 3.0 Wm^-2
What is the intensity of a wave of the same frequency that has an amplitude 2a?
how come the answer is 12
Post by: Vin on April 23, 2011, 08:04:45 am
Post by: HUSH1994 on April 23, 2011, 10:03:49 am
Post by: red_911 on April 23, 2011, 04:16:05 pm
The amount of electrical energy transferred when a charge of 8 mC moves through a
potential difference of 12 V ??
Post by: EMO123 on April 23, 2011, 04:26:43 pm
Post by: Vin on April 23, 2011, 04:57:02 pm
hey can anyone answer this ???
The amount of electrical energy transferred when a charge of 8 mC moves through a
potential difference of 12 V ??
Product of the two? convert mC to C
Post by: Vin on April 23, 2011, 04:59:58 pm
can anybody tell me a perfect definition of Stress and Strain
Strain's just when an object is extended by, for eg, delta L , right? Wonder how'd you define it.
Stress is F/A , so tensile force F acting normally to an area A, i suppose?
Post by: EMO123 on April 23, 2011, 05:08:18 pm
Strain's just when an object is extended by, for eg, delta L , right? Wonder how'd you define it.thanx man
Stress is F/A , so tensile force F acting normally to an area A, i suppose?
Post by: $H00t!N& $t@r on April 24, 2011, 11:12:59 pm
Q 18
4 MW is maximum is 80%.
Then 100% is 5 MW.
P = IV
5 X 106 = I X 25 X 103
I = 200 A
Q 25
Count the phase difference from X. 3 (Crest to trough) = 720 = 3 pie. n = 3.
Q 27
n X Lamda = d sin x
n = 1 (first order)
N = 500
d = 1/N
Lamda = d X sin x
------------
n
= (1/500) X Sin 30
= 1 X 10-3 m
Thanks a lot! +rep :)
Post by: $H00t!N& $t@r on April 25, 2011, 02:48:45 pm
O/N 2004 p1 Q 9, 13 and 37
Here is the link:http://www.freeexampapers.com/past_papers.php?l=Past_Papers%2FA+Level%2FPhysics%2FCIE%2F2004+Nov/ (http://www.freeexampapers.com/past_papers.php?l=Past_Papers%2FA+Level%2FPhysics%2FCIE%2F2004+Nov/)
Post by: astarmathsandphysics on April 25, 2011, 03:16:56 pm
10=1/2*10t^2 so t=sqrt(2)=1.43
v=u+at=0+9.8*1.43=14m/s
D
13.moments about pivot 1.5W=2.5*300=750
W=750/1.5=500
37)A current though 3 ohm resistor is twice the current though the 6 ohm resisitor but the current thought the 2 ohm resistors are the same.
total resistance of 3 &6 ohm branch is 2 ohms (1/3+1/6)^-1 =2 but total resistance of 2& 2 ohm branch os 1 (2+2)^-1 =1 so more volts need to drive cuurent through 3 & 6 ohm brach.
Post by: $H00t!N& $t@r on April 25, 2011, 04:09:51 pm
9)s=1/2at^2
10=1/2*10t^2 so t=sqrt(2)=1.43
v=u+at=0+9.8*1.43=14m/s
D
13.moments about pivot 1.5W=2.5*300=750
W=750/1.5=500
37)A current though 3 ohm resistor is twice the current though the 6 ohm resisitor but the current thought the 2 ohm resistors are the same.
total resistance of 3 &6 ohm branch is 2 ohms (1/3+1/6)^-1 =2 but total resistance of 2& 2 ohm branch os 1 (2+2)^-1 =1 so more volts need to drive cuurent through 3 & 6 ohm brach.
Thanks sir, but I don't understand your working for q13... :-\ How is the answer 480N ?
Post by: HUSH1994 on April 25, 2011, 04:19:14 pm
300*2 = 1.25X
X= 600/1.25 = 480N
1.25 is the middle of the beam,X is the weight and as the distance between the 300N force and the pivot is 2 m.
Hope you get it :)
Post by: $H00t!N& $t@r on April 25, 2011, 04:33:07 pm
As the weight acts at the middle,and the beam is in equilibrium,you can make the following equation:
300*2 = 1.25X
X= 600/1.25 = 480N
1.25 is the middle of the beam,X is the weight and as the distance between the 300N force and the pivot is 2 m.
Hope you get it :)
oh! thanks i get it :)
Post by: HUSH1994 on April 25, 2011, 04:33:46 pm
Post by: Vin on April 25, 2011, 05:02:57 pm
P1 Q27 ON 2002.
A wave of amplitude a has intensity of 3.0 Wm^-2
What is the intensity of a wave of the same frequency that has an amplitude 2a?
how come the answer is 12
Did you get Q36 from the same paper? I dont seem to get it :/
Post by: HUSH1994 on April 25, 2011, 05:39:19 pm
Im not really good at explaining but i hope you get it
Post by: EMO123 on April 25, 2011, 06:24:36 pm
each of the 3 resistors has 5 K(ohms),and the potential differnce across the circuit is 2 volts,so each resistor will get 2/3V(the six of them as they are connected in parallel)and the V at X is 4/3(as you have 2 resistors before)and 2/3 at Y,so 4/3 - 2/3 = 2/3.but still you explained the whole doubt
Im not really good at explaining but i hope you get it
Post by: Aadeez || Zafar on April 25, 2011, 06:41:12 pm
Post by: EMO123 on April 26, 2011, 02:56:42 pm
hehehewhat hehehehehe
Post by: da_rockstar on April 27, 2011, 01:57:40 pm
please explain with reasoning the answers to part (a) and (b).
Post by: astarmathsandphysics on April 27, 2011, 02:15:24 pm
force max when x=90
b)i)torqu=force*distance
2.1*10^-3 =F*0.028 so F=0.075N
ii)0 since parallel to field
c)F=BIlN
0.075=B*0.170*4.5*140
B=0.07T
Post by: HUSH1994 on April 27, 2011, 02:36:32 pm
Post by: da_rockstar on April 27, 2011, 03:53:30 pm
Post by: Vin on April 27, 2011, 05:16:08 pm
ON 2003 P1 Q11
Initial momentum = mu
Final momentum = - mu
( why u, because the speed remains unchanged, however direction changes, here it is perpendicularly opposite therefore velocity -ve. Agreed?)
Change in momentum = Final m. - Initial m.
= (- mu) - (mu)
= -2mu
Post by: HUSH1994 on April 27, 2011, 06:08:01 pm
Post by: Vin on April 27, 2011, 07:51:48 pm
each of the 3 resistors has 5 K(ohms),and the potential differnce across the circuit is 2 volts,so each resistor will get 2/3V(the six of them as they are connected in parallel)and the V at X is 4/3(as you have 2 resistors before)and 2/3 at Y,so 4/3 - 2/3 = 2/3.
Im not really good at explaining but i hope you get it
Oh yes, gave me a great idea. Thanks a lot! ;)
Post by: HUSH1994 on April 28, 2011, 03:52:55 am
Post by: HUSH1994 on April 29, 2011, 05:07:51 am
Post by: astarmathsandphysics on April 29, 2011, 08:03:17 am
Junn 04 p1 q9
Use suvat vertically to find t
s=1.25
u=0
v=
a=9.8
t=
s=ut+1/2at^2
1.25=0+1/2*4.9t^2 so t=0.505s
horizontal speed constant so use distance =speed*time
10=s*0.505 so speed=19.8 D
Post by: HUSH1994 on April 29, 2011, 01:37:38 pm
Post by: physichemaths on April 30, 2011, 06:22:59 pm
Post by: SkyPilotage on April 30, 2011, 06:53:34 pm
help me with november02 question 14 paper 1 mcqFor the rule to be in equilibrium, clockwisew moments = anticlockwise moments
Taking moments about the pivot at 40 cm mark:
- Clockwise moments = (0.1kg x 10) x 0.1 m = 0.10 Nm *P.s: since the rule is uniform, the centre of gravity is on the 50 cm mark
- Anticlockwise moments = (0.02kg x 10) x 0.6 m = 0.12 Nm
Therefore we know that the 50 g mass must be hanged to the right of the pivot, hence :
CLockwise moments = 0.1 + 0.5x(distance from pivot)=0.12 ---> 0.5 x (d)=0.12 Nm --> d=0.04 m= 4 cm to the right of the pivot!
Therefore the 50 g mass must be suspended at the (40 + 4) cm mark= 44cm which is C!
I hope you got it! tried to make the explanation as simple as possible :)
Post by: SkyPilotage on April 30, 2011, 06:58:57 pm
A simple question on this was on June 2003 Question 5...
-And on paper 3 practicals, can we use a FO and calculate c from the eqn by substitution?
WHen do use s.f and d.p in our table of results?
Thanks ...
Post by: SkyPilotage on May 03, 2011, 10:17:03 am
Post by: elemis on May 03, 2011, 10:22:49 am
-Can someone tell me how the percentage uncertainity determines the number of significant figures???
A simple question on this was on June 2003 Question 5...
-And on paper 3 practicals, can we use a FO and calculate c from the eqn by substitution?
WHen do use s.f and d.p in our table of results?
Thanks ...
What's an FO ?
Post by: SkyPilotage on May 03, 2011, 10:24:56 am
What's an FO ?A False Origin meaning not to start the x-axis from 0 to find the y-intercept... I find it in the markscheme "If not 0, Mark as FO" My question is tht sometimes my scale wont fit if i start from zero!?
Post by: $H00t!N& $t@r on May 05, 2011, 08:23:21 am
Here is the link: http://www.xtremepapers.me/CIE/International%20A%20And%20AS%20Level/9702%20-%20Physics/9702_w07_qp_2.pdf (http://www.xtremepapers.me/CIE/International%20A%20And%20AS%20Level/9702%20-%20Physics/9702_w07_qp_2.pdf)
Post by: Master_Key on May 05, 2011, 12:31:02 pm
question! o/n 2007 q5a
Here is the link: http://www.xtremepapers.me/CIE/International%20A%20And%20AS%20Level/9702%20-%20Physics/9702_w07_qp_2.pdf (http://www.xtremepapers.me/CIE/International%20A%20And%20AS%20Level/9702%20-%20Physics/9702_w07_qp_2.pdf)
The Graph has amplitude of 10 boxes.
Intensity is directly proportional to (amplitude)2.
I = 102 = 100.
0.5 I = (102)/2 = 50
I = sq. root of 50.
= 7.07
therefore the amplitude will be 7 boxes.
Phase difference is 60 degrees.
So 5 boxes of difference between waves. Got this by calculation.
HOPE IT HELPS.
Post by: Vin on May 05, 2011, 01:07:34 pm
The Graph has amplitude of 10 boxes.
Intensity is directly proportional to (amplitude)2.
I = 102 = 100.
0.5 I = (102)/2 = 50
I = sq. root of 50.
= 7.07
therefore the amplitude will be 7 boxes.
Phase difference is 60 degrees.
So 5 boxes of difference between waves. Got this by calculation.
HOPE IT HELPS.
Can you please explain how 5 boxes? I get 10, actually :/ Which's wrong, I know.
Post by: Master_Key on May 05, 2011, 01:09:52 pm
Can you please explain how 5 boxes? I get 10, actually :/ Which's wrong, I know.
Calculate the distance between 2 Crest
X boxes = 360 degrees.
Calculate for 60 degrees.
SHORTEST, SIMPLEST AND EASIEST WAY.
Post by: Arthur Bon Zavi on May 05, 2011, 03:10:10 pm
Can you please explain how 5 boxes? I get 10, actually :/ Which's wrong, I know.
Take the first crest.
15 boxes gives you a phase difference of 180o;
So a phase difference of 60o will be given by :
( 60 X 15 ) / 180 = 5
Post by: $H00t!N& $t@r on May 05, 2011, 07:40:57 pm
The Graph has amplitude of 10 boxes.
Intensity is directly proportional to (amplitude)2.
I = 102 = 100.
0.5 I = (102)/2 = 50
I = sq. root of 50.
= 7.07
therefore the amplitude will be 7 boxes.
Phase difference is 60 degrees.
So 5 boxes of difference between waves. Got this by calculation.
HOPE IT HELPS.
Thanks ;D
Post by: muskaan on May 07, 2011, 09:03:39 am
Q6/b/part 3..how does capacitor does smoothing? (explain de concept plz)
Q2/b..isnt workdone takn positive whn vol increases?
Post by: Master_Key on May 07, 2011, 03:25:00 pm
june02/ppr 4/
Q6/b/part 3..how does capacitor does smoothing? (explain de concept plz)
Q2/b..isnt workdone takn positive whn vol increases?
Q2/b.. U = W + Q. Energy is needed for boiling so Q is positive. Take all values for "change of/delta".
Q6/b/part 3..
Vrms = 0.707 * Vmax
Vrms / 0.707 = Vmax
Or equivalent
Vrms * Sq.rt. 2.
Post by: mmanikandan on May 08, 2011, 06:25:41 am
just been two days i started physics degree and so i have some doubts
so my first doubt,
Q1.When you whistle out air, on to your palm held close to your mouth the air fills cold but, when you blow out air from your mouth keeping it wide open the air feels hot why?
Post by: EMO123 on May 08, 2011, 08:15:36 am
cauz of hollowness
Post by: HUSH1994 on May 08, 2011, 10:21:56 am
Post by: aloha32 on May 08, 2011, 10:26:48 am
I have major doubts regarding CIE Physics Paper 4 Nov 10 Variant 3 : question 3 part b (ii) and (iii) how do you know how high it is above the piston with the angle?
and also question 2 part ii ; why have they taken 1.2E-12 instead of the total kinetic 2.4E-12 Joule?
Post by: Sue T on May 08, 2011, 03:31:17 pm
the second thing u asked - i think bcuz that eqn applies for a single molecule u have to take only half th kinetic energy since a mixture of those two contains two molecules and assuming both have the same kinetic energy, u divide by 2 as to put in just a single molecule in ur eqn (i thought bout it alot n ths s th best i cod come up wth - sometimes th information is soo less...)
Post by: AN10 on May 08, 2011, 06:17:19 pm
O/N 10 P41 q2 (b) iii
anyone plz help ???
Post by: Sue T on May 08, 2011, 08:08:30 pm
Post by: cashem'up on May 09, 2011, 09:44:47 am
1.A hot air balloon its basket and passengers have a total mass of 450kg. The inflated balloon holds 3000m^3 of air.
(a) minimum forced needed to lift balloon of the ground? - 4500 N
(b) Atm pressure is 1.03 x 10^5 abd density of air is 1.23kg/m^3
(i) calculate mass of 3000m^3 air? - 3690kg
(ii) Calculate Number of Moles in the balloon, Consider air relative Mr as 29 ? 1.27 x 10^5 mol
THIS IS PRIOR INFORMATION AND CALCULATIONS DONE BY ME..the next two parts i dont
know
c. As the air is heated some of it is expelled through a vent at the top of the balloon. Calculate max mass of air that can remain in balloon to give sufficient upthrust for the balloon to just the leave the ground???? DONT KNOW
d. Calculate the minimum temperature of the air inside the balloon for the balloon to just lift off the ground??? DONT KNOW
Post by: physichemaths on May 09, 2011, 09:56:16 am
Post by: aldehyde1612 on May 09, 2011, 12:25:01 pm
from ppr 43 ques10b you have 2 situations - open switch and closed switch - ive attached an expalanation
When V- is more than V+ ... isnt the output negative??
Vout= G x [V+ - V-] .. so if V- is 3V and V+ is 2V, then 2-3 = -1 .. So Vout is G x -1, which is negative?
And vice versa when V+ is more than V- ??
Thats the logic we were taught, but then with the marking scheme that logics wrong.. so can you please explain?? ???
Post by: Sue T on May 09, 2011, 12:34:29 pm
Post by: Sue T on May 09, 2011, 12:37:35 pm
Help me with nov 08 question 10 part b?
so you have : Vout/Vin = Rfeedback/R1
so at 100kohms calculate the Rfeedback using the fact that the LDR is in parallel to the 50kohm resistor
so 1/50k + 1/100k = Rfeedback
once you get Rfeedback just substitute the values and obtain Vout
the second part uses the same method
Post by: aldehyde1612 on May 09, 2011, 12:52:39 pm
humans make marking schemes - it can be wrong you no - cuz ryt in front of me in the booklet the rule is set and an xample is done - so yes wat i said is the ryt one - i asked my teacher also ;)
I was saying you're explanation agrees with the marking scheme... but me and all my physics teachers got it the other way around (the way I said). :-\
But true. humans do make mistakes.
Thanks for the help.
Post by: Dthdfyng on May 09, 2011, 01:19:02 pm
O/N 05 (Q5)(ii) and (Q6)(c)(ii)
M/J 06 (Q3)(a) and (Q6)(c) and (Q7)(c)
O/N 06 (Q7)(b)
M/J 10 (Q3)(b) and (Q9)(a)(ii)
O/N 10 (Q8)(c) and (Q11)(b)
will appreciate any help!
Post by: Master_Key on May 09, 2011, 01:25:53 pm
Hey everyone! i really need some help with Physics P4 past paper questions
O/N 05 (Q5)(ii) and (Q6)(c)(ii)
M/J 06 (Q3)(a) and (Q6)(c) and (Q7)(c)
O/N 06 (Q7)(b)
M/J 10 (Q3)(b) and (Q9)(a)(ii)
O/N 10 (Q8)(c) and (Q11)(b)
will appreciate any help!
O/N 05 (Q5)(b)(ii)
1. v is small so deflection is larger.
2. More force = More deflection.
Post by: Master_Key on May 09, 2011, 01:29:09 pm
Hey everyone! i really need some help with Physics P4 past paper questions
O/N 05 (Q5)(ii) and (Q6)(c)(ii)
M/J 06 (Q3)(a) and (Q6)(c) and (Q7)(c)
O/N 06 (Q7)(b)
M/J 10 (Q3)(b) and (Q9)(a)(ii)
O/N 10 (Q8)(c) and (Q11)(b)
will appreciate any help!
M/J 06 (Q3)(a)
The gradient is sensitivity. It is not linear and varies with temperature.
Post by: Sue T on May 09, 2011, 01:31:00 pm
I was saying you're explanation agrees with the marking scheme... but me and all my physics teachers got it the other way around (the way I said). :-\
But true. humans do make mistakes.
Thanks for the help.
hey lsn - i meant the markin scheme is rong - i typed it out wen i ws sleepy - i mixed it up - jus switch tht part round - so bottom line if v- is more it becomes negative - sorry 4 th confusion - i ws bout 2 go 2 bed :p
Post by: da_rockstar on May 09, 2011, 01:50:00 pm
if you have 2 capacitors in series together. you obviously say that the charge on each is the same right?
so when you take their energy individually and on the other hand take their combined energy...won't the two values differ. i mean the combined energy will fall according to the formula of Q^2/2C. that will give a reduced overall energy compared to what you have from the individuals because your capacitance decreases in series.
and in one question i saw that while calculating individual energies of the capacitances they reduced the charge on them by 2. that is divided the charge that you get through their combined charge to give charge on one of the capacitance. but that confused me greatly since in my opinion the charges across them both should remain the same. please do let me know where i'm wrong in my assumptions. reply asap please :) Thanks in advance
Post by: Sue T on May 09, 2011, 03:31:55 pm
Guys i had a question i wanted to ask.
if you have 2 capacitors in series together. you obviously say that the charge on each is the same right?
so when you take their energy individually and on the other hand take their combined energy...won't the two values differ. i mean the combined energy will fall according to the formula of Q^2/2C. that will give a reduced overall energy compared to what you have from the individuals because your capacitance decreases in series.
and in one question i saw that while calculating individual energies of the capacitances they reduced the charge on them by 2. that is divided the charge that you get through their combined charge to give charge on one of the capacitance. but that confused me greatly since in my opinion the charges across them both should remain the same. please do let me know where i'm wrong in my assumptions. reply asap please :) Thanks in advance
it'll be hard 2 really answer ur ques without an example - so if u could find that ques i can c wat i can do -- but anyways in terms of energy the thing is when capacitors are in series, the charge is Q for both thats ryt - but the potential is shared. When in parallel the charge is shared, but the voltage is same. E=Q2/2C can only be used in certain situations... like amm...ugh u have 2 gimme an example... :-\ :s
Post by: your sister on May 10, 2011, 05:12:16 am
2. graph of gravitational force and radius : gradient gives gravitational field strength
which one is correct 1 or 2 ??
Post by: HUSH1994 on May 10, 2011, 06:23:10 am
Post by: your sister on May 10, 2011, 06:32:33 am
is this A2?yes
Post by: aldehyde1612 on May 10, 2011, 07:15:43 am
1. graph of gravitational potential and radius : gradient gives gravitational field strength
2. graph of gravitational force and radius : gradient gives gravitational field strength
which one is correct 1 or 2 ??
1. is right...
gradient would be Grav.Potential/ radius. Potential=GM/r .. so that divided by 'r' give GM/r^2 ... which is the Grav.Field Strength per unit mass. right??
Post by: your sister on May 10, 2011, 07:30:37 am
1. is right...
gradient would be Grav.Potential/ radius. Potential=GM/r .. so that divided by 'r' give GM/r^2 ... which is the Grav.Field Strength per unit mass. right??
ya you gottaay point :)
Post by: your sister on May 10, 2011, 07:56:40 am
thankyou :D
Post by: EMO123 on May 10, 2011, 08:12:06 am
hope it helps you
Post by: your sister on May 10, 2011, 01:19:08 pm
http://www.thestudentroom.co.uk/showthread.php?t=398675
hope it helps you
oh Thanks :D
Post by: TJ-56 on May 10, 2011, 01:23:33 pm
May June 2002 Q4 (b) Posted, just want to make sure it is correct.
May June 2002 Q6 (a) Posted as well, though I have no idea how to draw it.
October November 2002 Q1 (b) is it an under or over estimation? pls explain
October November 2002 Q3 (b) (ii) pls explain as well
I would greatly appreciate help for any of the questions
thanx!
Post by: your sister on May 10, 2011, 01:23:41 pm
ques 3, b, i
ms says = 3050 J
but i am getting 3047
is it fine if i use my answer in the iii part ?
Post by: TJ-56 on May 10, 2011, 01:26:42 pm
P4 oct/nov 05
ques 3, b, i
ms says = 3050 J
but i am getting 3047
is it fine if i use my answer in the iii part ?
The required answer should probably be rounded to 3 significant figures, so your answer should be correct, because 3047 becomes 3050.
hope that was helpful
Post by: your sister on May 10, 2011, 01:47:32 pm
The required answer should probably be rounded to 3 significant figures, so your answer should be correct, because 3047 becomes 3050.
hope that was helpful
is it because the values given in the question are to 3sf ??
Post by: TJ-56 on May 10, 2011, 01:50:26 pm
Yup, thats exactly why. All of them are to 3sf if you have noticed.
is it because the values given in the question are to 3sf ??
Post by: Sue T on May 10, 2011, 02:08:55 pm
Can someone help with the following paper 4 questions:mj 2002 Q4b) yes its ryt
May June 2002 Q4 (b) Posted, just want to make sure it is correct.
May June 2002 Q6 (a) Posted as well, though I have no idea how to draw it.
October November 2002 Q1 (b) is it an under or over estimation? pls explain
October November 2002 Q3 (b) (ii) pls explain as well
I would greatly appreciate help for any of the questions
thanx!
mj 2002 Q6a) https://studentforums.biz/sciences-149/physics-doubts-12323/ (last post last attachment):
nov 2002 Q 1b) see you used: Pt=mL ryt? now if u include energy losses (lets call them S)
Pt - S = mL
so ur L was an overestimation, since th ryt equation involves subtracting energy loss to surroundings.
nov 2002 Q 3bii) imagine a child on a swing, you push it at the right time to give it more amplitude - so if its resonant frequency is fo you provide a frequency of f0
but if you push it once yes and once no (as in alternate cycles) u are providing a frequency of 1/2f0 - you still build up amplitude but its not the max
hope i helped
Post by: TJ-56 on May 10, 2011, 02:26:36 pm
mj 2002 Q4b) yes its ryt
mj 2002 Q6a) https://studentforums.biz/sciences-149/physics-doubts-12323/ (last post last attachment):
nov 2002 Q 1b) see you used: Pt=mL ryt? now if u include energy losses (lets call them S)
Pt - S = mL
so ur L was an overestimation, since th ryt equation involves subtracting energy loss to surroundings.
nov 2002 Q 3bii) imagine a child on a swing, you push it at the right time to give it more amplitude - so if its resonant frequency is fo you provide a frequency of f0
but if you push it once yes and once no (as in alternate cycles) u are providing a frequency of 1/2f0 - you still build up amplitude but its not the max
hope i helped
Wow, thanks, that was REALLY REALLY helpful, I really appreciate the help
+rep
Post by: TJ-56 on May 10, 2011, 02:58:00 pm
thanx
Post by: Sue T on May 10, 2011, 04:39:34 pm
May June 2005 Q6 (b) Posted, again i just want to make sure its the correct answeryes - the differential of sine is cosine so yes
thanx
Post by: TJ-56 on May 10, 2011, 04:48:46 pm
yes - the differential of sine is cosine so yesur awesome, thanx again!
+rep
Post by: TJ-56 on May 10, 2011, 04:49:28 pm
May June 2006 6 (a) (i) what exactly do we have to draw?
thanx in advance
Post by: physichemaths on May 10, 2011, 10:26:56 pm
Post by: Sue T on May 11, 2011, 02:11:21 am
May June 2006 2 (b) (ii) The volume in the ‘baloon’ conditions become 5.405 x10^6, so why cant we just divide that value be 7.24 x 10^3 and round the answer from 746.54 to 746. the answer on the ms is 741, and what is meant by: (answer 740 or ““fails to allow for gas in cylinder””, max 2/3) ?i codnt log in - some network errror - u might not c th answer till after the xam but - wateve:
May June 2006 6 (a) (i) what exactly do we have to draw?
thanx in advance
for bii) the original cylinder's volume must be included cuz some might be left over in it and it dsnt fill any balloons - so you have 2 include that in the calculation of ur new volume-thts as far as i can help you
6ai) it is a vertical downward arrow for direction of B (think of th flux lines atound y) and an arrow to the right for th force - im talkin bout those 2 lines being xactly on point Q - 2 find the direction of th force, use flemings left hand
Post by: MistahNobodeh on May 11, 2011, 09:03:54 am
i codnt log in - some network errror - u might not c th answer till after the xam but - wateve:In the balloon question u cannot just divide the Volumes cuz they are both at different pressures and according to PV=nRT Vis directly propotional to n only and only if all other quantities are constant.
for bii) the original cylinder's volume must be included cuz some might be left over in it and it dsnt fill any balloons - so you have 2 include that in the calculation of ur new volume-thts as far as i can help you
6ai) it is a vertical downward arrow for direction of B (think of th flux lines atound y) and an arrow to the right for th force - im talkin bout those 2 lines being xactly on point Q - 2 find the direction of th force, use flemings left hand
In the wire question....The arrow B should be downwards since according to the right hand grip rule that will be the component of wire XY's magnetic field onto the wire PQ. The force should be towards the right hand side(towards XY) in accordance with the Fleming's left hand rule.
Post by: NidZ- Hero on May 12, 2011, 08:38:36 am
what is the magnitude of the resultant force?
A. 0N
B. 1.3N
C. 7.3 N
D. 10 N
can yu plz show the working
tahnx ;)
Post by: elemis on May 12, 2011, 08:43:47 am
three coplanar forces, each of magnitude of 10N ACT, through the same point of a body in the direction shown
what is the magnitude of the resultant force?
A. 0N
B. 1.3N
C. 7.3 N
D. 10 N
can yu plz show the working
tahnx ;)
Can you attach the diagram or link me to the original question ?
Post by: cashem'up on May 12, 2011, 11:01:08 am
three coplanar forces, each of magnitude of 10N ACT, through the same point of a body in the direction shown
what is the magnitude of the resultant force?
A. 0N
B. 1.3N
C. 7.3 N
D. 10 N
can yu plz show the working
tahnx ;)
if they complete a vector triangle its zero... otherwise plz attach the diagram..
Post by: mdwael on May 12, 2011, 11:01:44 pm
I attached the picture of the question.
The mark scheme says exactly..
(ii) moment = force × perpendicular distance
= 0.051 × 9.8 × 0.61 × sin18
= 0.094 N
I dont get this explanation :/
This question is from october november 2007
Post by: NidZ- Hero on May 13, 2011, 07:42:27 am
can u aid the explanation with a graPh plz
Post by: NidZ- Hero on May 14, 2011, 05:33:02 am
Can you attach the diagram or link me to the original question ?
Post by: NidZ- Hero on May 14, 2011, 07:21:41 am
that move apart with speeds v1 and v2 respectively
what is the ratio of v1/v2?
A. M1/M2
B. M2/M1
C. (M1/M2)^1/2
D. (M2/M1)^1/2
Post by: elemis on May 14, 2011, 07:33:26 am
three coplanar forces, each of magnitude of 10N ACT, through the same point of a body in the direction shown
what is the magnitude of the resultant force?
A. 0N
B. 1.3N
C. 7.3 N
D. 10 N
can yu plz show the working
tahnx ;)
Resolving vertically.
10cos30 + 10cos30 - 10 = 7.32 N
Post by: EMO123 on May 14, 2011, 07:35:30 am
Resolving vertically.thanx ari
10cos30 + 10cos30 - 10 = 7.32 N
Post by: NidZ- Hero on May 14, 2011, 01:01:38 pm
1. A rocket of mass m is travelling at a constant speed v away form the stationary observer when it instantaneously ejects one-third of its mass at a speed of 6v toward the observer. what is the new speed of the rocket?
A. 1.5v
B. 2.0v
C. 4.5v
D. 5.0v
2. A body of mass 10kg is being pulled up a smooth incline at uniform acceleration by a force of 200N which acts parallel to the plane. If the incline is at an angle of 30. to the horizontal, whta is the resultant force acting on the body parallel to the plane?
Assume the g=10Nkg-1
A. 100N
B. 10N
C. 150N
D. 200N
plz show the working
ThankzZz
Post by: Ghost Of Highbury on May 16, 2011, 07:13:56 am
If the question asks, calculate the %uncertainty in this value.
The absolute uncertainty is 0.1cm or 1mm, therefore, the percentage uncertainty will be 0.1/33.7*100 = 0.296735...%
My doubt - i) To how many s.f do i round off the percentage uncertainty?
ii) I'm not sure if an uncertainty of just 0.3% sounds right. I mean, what I generally hear is % uncertainty of >1% . Have I done
the calculations correctly?
Post by: elemis on May 16, 2011, 09:59:34 am
Say, I measure the length of a pendulum using a metre rule and it is 33.7cm
If the question asks, calculate the %uncertainty in this value.
The absolute uncertainty is 0.1cm or 1mm, therefore, the percentage uncertainty will be 0.1/33.7*100 = 0.296735...%
My doubt - i) To how many s.f do i round off the percentage uncertainty?
ii) I'm not sure if an uncertainty of just 0.3% sounds right. I mean, what I generally hear is % uncertainty of >1% . Have I done
the calculations correctly?
Your measurements are to a minimum of 2 s.f. (0.10 cm), so I would round it up to 2 s.f.
Yes, your calculations seem correct BUT you must factor other features. Was it difficult to hold ruler vertically ? Was a clamp available ? Was the distance very small ?
All of those may increase your uncertainty. As you yourself said, when in doubt use the average of the range of measurements ;)
Your not doing chemistry this time round, right ?
Post by: Ghost Of Highbury on May 16, 2011, 10:25:41 am
Your measurements are to a minimum of 2 s.f. (0.10 cm), so I would round it up to 2 s.f.
Yes, your calculations seem correct BUT you must factor other features. Was it difficult to hold ruler vertically ? Was a clamp available ? Was the distance very small ?
All of those may increase your uncertainty. As you yourself said, when in doubt use the average of the range of measurements ;)
Your not doing chemistry this time round, right ?
Thank you.
Nah, not this session, later.
Post by: aneeta on May 16, 2011, 12:49:08 pm
Post by: elemis on May 16, 2011, 12:53:39 pm
P3 TOMORROWWW!! any Suggestions =\
Whatever you do... Don't fail :P
Post by: elemis on May 16, 2011, 12:54:37 pm
Thank you.
Nah, not this session, later.
Good luck then.
Post by: aneeta on May 16, 2011, 12:56:07 pm
Whatever you do... Don't fail :P
Thankyou for your oh so awesome suggestion =P
Post by: elemis on May 16, 2011, 12:59:10 pm
Thankyou for your oh so awesome suggestion =P
My advice doesnt come free... Would you prefer to pay by cheque or credit card ?
Post by: username on May 16, 2011, 03:09:26 pm
like i think its (+/-) .1 cm for ruler measurments
wat about screw gauge, volts, current, vernier calipers etc. etc???
Post by: aneeta on May 16, 2011, 03:37:22 pm
My advice doesnt come free... Would you prefer to pay by cheque or credit card ?
Lol cash...
Post by: aneeta on May 16, 2011, 03:38:39 pm
Post by: elemis on May 16, 2011, 03:54:46 pm
Has anyone any idea of what's coming in p3?!
There's definitely going to be addition, though I'm not sure subtraction is in the syllabus.
On a serious note, be prepared to draw a graph, know how to derive the units for the gradient and go through some papers to determine the improvements that can be made to experiments.
Post by: your sister on May 16, 2011, 08:44:48 pm
Post by: $H00t!N& $t@r on May 19, 2011, 04:57:52 pm
Post by: EMO123 on May 19, 2011, 05:08:25 pm
Post by: EMO123 on May 19, 2011, 05:10:29 pm
use this formula
Post by: EMO123 on May 19, 2011, 05:14:18 pm
Post by: $H00t!N& $t@r on May 19, 2011, 05:27:28 pm
5 (b) (lambda*D) / a
use this formula
no, your not meant to use that for this question.. its quite confusing actually...
what I have done so far is as follows:
path difference between S1 and S2 = 28 cm
v = f x (lambda)
S1 330/1000 = 33 cm
S2 330/4000 = 8.25 cm
after this I tried using (n + 1/2)(lambda) but I'm getting a wrong answer :-\
Post by: EMO123 on May 19, 2011, 05:58:28 pm
suppose take 650nm so change it to meters so that will be= 650*10-9
D is = 2.4
a is 0.86 mm change it to meters so that will be = 8.6*10-4
so substitute the values
((650*10-9)*2.4)/(8.6*10-4)
change the answer to mm
Post by: $H00t!N& $t@r on May 19, 2011, 05:59:35 pm
wavelength of red light is between 620 to 750 so take any from it
suppose take 650nm so change it to meters so that will be= 650*10-9
D is = 2.4
a is 0.86 mm change it to meters so that will be = 8.6*10-4
so substitute the values
((650*10-9)*2.4)/(8.6*10-4)
change the answer to mm
hey I asked about variant 1 :D
Post by: SkyPilotage on May 19, 2011, 06:00:45 pm
no, your not meant to use that for this question.. its quite confusing actually...Ok what you did using v=f*lambda is that you found the range of possibilities of lambda...
what I have done so far is as follows:
path difference between S1 and S2 = 28 cm
v = f x (lambda)
S1 330/1000 = 33 cm
S2 330/4000 = 8.25 cm
after this I tried using (n + 1/2)(lambda) but I'm getting a wrong answer :-\
To find out about all the wavelenghts remember tht for destructive interference the phase difference or path difference should be equal to 1/2 lamda or 3/2 lambda or 5/2 lamda and soo onn. So you just find the lambda by 28x2 or 28x2/3 or 28x2/5 etc... and then count ho wmany wavelelgnths you have in the range.....
Post by: EMO123 on May 19, 2011, 06:02:11 pm
hey I asked about variant 1 :Dit's k i thought it is variant 2
Post by: EMO123 on May 19, 2011, 06:03:20 pm
need help with m/j 2009 variant 2 q 5bi found something written here variant 2 (it's k this also helped me in practicing this question)
Post by: $H00t!N& $t@r on May 19, 2011, 06:06:19 pm
i found something written here variant 2 (it's k this also helped me in practicing this question)
oops! my bad.. sorry :-[
Post by: $H00t!N& $t@r on May 19, 2011, 06:07:10 pm
Ok what you did using v=f*lambda is that you found the range of possibilities of lambda...
To find out about all the wavelenghts remember tht for destructive interference the phase difference or path difference should be equal to 1/2 lamda or 3/2 lambda or 5/2 lamda and soo onn. So you just find the lambda by 28x2 or 28x2/3 or 28x2/5 etc... and then count ho wmany wavelelgnths you have in the range.....
ohh! so thats how you do it! thanks ;D
Post by: EMO123 on May 19, 2011, 06:09:13 pm
oops! my bad.. sorry :-[it's all right
Post by: SkyPilotage on May 19, 2011, 06:13:39 pm
ohh! so thats how you do it! thanks ;Dglad to hellp :)
Post by: polltery on May 20, 2011, 10:33:19 am
Post by: Ghost Of Highbury on May 20, 2011, 11:04:18 am
plz help me out wid this! cant understand the method used here ? ???
Draw tangent to the curve at 0.4s, calculate the gradient of the tangent.
Post by: Ghost Of Highbury on May 20, 2011, 11:05:41 am
(CIE AS level Physics)
Please explain the steps to determine the no. of s.f in 'g' and the uncertainty clearly.
If any rules are to be followed to determine them, please state them too. Thanks a lot
Post by: polltery on May 20, 2011, 11:17:52 am
Draw tangent to the curve at 0.4s, calculate the gradient of the tangent.THANKS ALOT MATE! :D
Post by: elemis on May 20, 2011, 11:19:14 am
Winter 2009, paper 22, Q1)c)ii)
(CIE AS level Physics)
Please explain the steps to determine the no. of s.f in 'g' and the uncertainty clearly.
If any rules are to be followed to determine them, please state them too. Thanks a lot
Hang on.
Post by: Ghost Of Highbury on May 20, 2011, 11:21:29 am
Hang on.
Sure thing
Post by: elemis on May 20, 2011, 11:28:53 am
Since we are calculating the absolute uncertainty of g we let it be delta x :
Looking at the initial data we can see that the absolute uncertainties we used were accurate to only 1 s.f. Hence, we express our calculated absolute uncertainty (delta x) to a maximum of 1 s.f.
Look at our value of g and comparing it to the original data we can see each value was accurate to a minimum of 2 s.f. Hence we express g to a max of 2 s.f.
Post by: $H00t!N& $t@r on May 20, 2011, 11:34:53 am
Post by: Ghost Of Highbury on May 20, 2011, 11:37:54 am
First calculate the value of g using the equation, which is 9.75064.... The equate the fractional uncertainties of the left hand side of the equation to the right hand side of the equation.
Since we are calculating the absolute uncertainty of g we let it be delta x :
Looking at the initial data we can see that the absolute uncertainties we used were accurate to only 1 s.f. Hence, we express our calculated absolute uncertainty (delta x) to a maximum of 1 s.f.
Look at our value of g and comparing it to the original data we can see each value was accurate to a minimum of 2 s.f. Hence we express g to a max of 2 s.f.
Raw data, ofcourse! DARN!
Thanks Ari. Been a long time man, mind catching up? yahoo?
Post by: elemis on May 20, 2011, 11:39:32 am
Raw data, ofcourse! DARN!
Thanks Ari. Been a long time man, mind catching up? yahoo?
No worries.
Hahaha !! It has been very long. Unfortunately, I'm neck deep in revision, maybe after the exams ?
Post by: Ghost Of Highbury on May 20, 2011, 11:42:11 am
need help with O/N 2010 variant 23 q6b (ii) :-\
Dark fringe, destructive interference, out of phase, 180o
2. Superposition can cause the amplitudes to add up,( Crest Crest), or get subtracted.
Max amplitude = 2+1.4 , min = 2-1.4
3.4, 0.6
I = kA^2
Ratio = k*3.4^2 / k*0.6^2 = 3.4^2/0.6^2 = 32
Post by: Ghost Of Highbury on May 20, 2011, 11:43:09 am
No worries.
Hahaha !! It has been very long. Unfortunately, I'm neck deep in revision, maybe after the exams ?
Ah, same here man. Past papers -_-
You're giving physics paper 2 on Monday too right?
After exams, done. Good luck
Post by: elemis on May 20, 2011, 11:43:18 am
need help with O/N 2010 variant 23 q6b (ii) :-\
To be out of phase the phase diff. has to be 180 degrees.... ALWAYS.
A bright fringe is formed when two crests meet.
Amplitude of crest A = 2
Amplitude of Crest B = 1.4
Sum = 3.4
Intensity = k3.42
Can you figure out the rest ?
Post by: $H00t!N& $t@r on May 20, 2011, 11:47:21 am
Post by: elemis on May 20, 2011, 11:48:41 am
Ah, same here man. Past papers -_-
You're giving physics paper 2 on Monday too right?
After exams, done. Good luck
Yup, though I've finished all of the papers.
What variant are you doing ?
Same to you.
Post by: Ghost Of Highbury on May 20, 2011, 12:08:53 pm
Yup, though I've finished all of the papers.
What variant are you doing ?
Same to you.
Paper 22
Post by: $H00t!N& $t@r on May 20, 2011, 12:13:13 pm
but i thought you have to find the phase difference using 360(x/lambda) ... when do we use this?
hello?
Post by: Ghost Of Highbury on May 20, 2011, 12:16:59 pm
but i thought you have to find the phase difference using 360(x/lambda) ... when do we use this?
First, you can't use x/lambda here as the x-axis is time. What you mean is 360(t/T)
And second, this formula is easier to use when 2 curves are plotted on the same graph with the values of t given.
It's pretty obvious here that the phase difference is 180o
Post by: $H00t!N& $t@r on May 20, 2011, 12:25:44 pm
First, you can't use x/lambda here as the x-axis is time. What you mean is 360(t/T)
And second, this formula is easier to use when 2 curves are plotted on the same graph with the values of t given.
It's pretty obvious here that the phase difference is 180o
oh alrite.. yeah i see my mistake.. thanks
Post by: mdwael on May 20, 2011, 03:35:34 pm
Post by: EMO123 on May 20, 2011, 03:45:58 pm
Hello people! Please advice me, I have an exam in 3 days and I cant solve a pastpaper without losing atleast 30 marks, even with the text book infront of me open!! Yes, my textbook is that bad.. so can you suggest me reference material/ Livebooks/ websites that will help me in the coming paper 2 physics?http://www.astarmathsandphysics.com/a_level_physics_notes_menu.html
short out the notes of as soon and start reading precise and easy
https://studentforums.biz/revison-notes/physics-notes-by-me/
this is also helpful download it
hope i helped
Post by: mdwael on May 20, 2011, 03:47:58 pm
http://www.astarmathsandphysics.com/a_level_physics_notes_menu.html
short out the notes of as soon and start reading precise and easy
https://studentforums.biz/revison-notes/physics-notes-by-me/
this is also helpful download it
hope i helped
Thank you + :)
Post by: EMO123 on May 20, 2011, 03:50:51 pm
Thank you + :)glad to help
Post by: aneeta on May 20, 2011, 03:59:45 pm
Can anyone help me with o/n paper 2 question 4C :s I don't get it!!
Post by: EMO123 on May 20, 2011, 04:00:59 pm
Gosh I hope paper 2 is as easy as paper 3 was...which year
Can anyone help me with o/n paper 2 question 4C :s I don't get it!!
Post by: SkyPilotage on May 21, 2011, 08:02:04 am
which yearwhy dont you guess? :P
Post by: EMO123 on May 21, 2011, 08:03:57 am
why dont you guess? :Pas i it is not given :P
Post by: SkyPilotage on May 21, 2011, 08:08:12 am
as i it is not given :PMan your gonna FAIL Physics...How come you dont know question 4c????
AHAH Hopefully your gonna Ace it IsA :)
Post by: EMO123 on May 21, 2011, 08:10:11 am
Man your gonna FAIL Physics...How come you dont know question 4c????actually i had asked which year :P
AHAH Hopefully your gonna Ace it IsA :)
after this you are goin to fail in your diving licensee test
Post by: aneeta on May 21, 2011, 08:50:06 am
which yearLoool haha my bad:$ I totally forgot abt the year... It's 2007
Post by: elemis on May 21, 2011, 08:53:21 am
Loool haha my bad:$ I totally forgot abt the year... It's 2007
Summer or winter ?
Post by: EMO123 on May 21, 2011, 08:55:04 am
Loool haha my bad:$ I totally forgot abt the year... It's 2007it is written ari oct-nov
so winter afcos :D
will get to it soon
Post by: aneeta on May 21, 2011, 08:58:04 am
it is written ari oct-novExactly what he said:p OCTOBER/NOVEMBER 2007 QUESTION 4C
so winter afcos :D
will get to it soon
Post by: EMO123 on May 21, 2011, 09:07:38 am
stress = force / area
make area subject of the formula
so that will be area=force/stress
area(mini) = (1.9*103) / (9.5*108)
= 2.0*10–6 m2
as it is stretched so the max-mini
area (max)= (3.2 – 2.0)*10–6
= 1.2*10–6 m2
so mini area=2.0*10-6m2
and max area=1.2*10-6m2
Post by: yasser37 on May 21, 2011, 09:48:18 am
Paper 5
June 07
Question 2
The error bars are horizontal yet too small to be drawn
can someone attach a drawing? Thanks
Post by: aneeta on May 21, 2011, 10:41:22 am
Post by: NidZ- Hero on May 21, 2011, 01:46:47 pm
Post by: EMO123 on May 21, 2011, 02:13:40 pm
June 08 paper 2 question 6(b)6 (b) 1st is open so power supply so that is zero
2nd from the first the power is supplied so it will move to s2 but s3 is closed so it will not move from s3
so only of s2 which is 1.5kW so for second it is 1.5kW
3rd the power is supplied from s2 and s3 so both are added 1.5kW+1.5kW=3kW
4th - Current flows through A and B (not the part of the junction with S2 as it has infinite resistance)
R = 38.4ohms, V one heater receives is 120V , P = V2/R = 1202/38.4 = 375W
Thus, for two heaters = 375*2 = 750W or 0.75kW
5th - Current flows through all heaters. Power in A and B as calculated previously = 0.75kW
Power in C = V2/R = 2402/38.4 = 1.5kW Total - 1.5 + 0.75 = 2.25kW
Post by: EMO123 on May 21, 2011, 02:19:00 pm
can someone explain the difference between stationary waves and progressive wavesThe notes we hear from a cello are created by the vibrations of strings. The wave patterns on the vibrating strings are called stationary waves.
The waves in the air which carry the sound to our ears transfer energy, and so are called progressive waves.
Hope it helps.
Post by: NidZ- Hero on May 21, 2011, 02:36:21 pm
The notes we hear from a cello are created by the vibrations of strings. The wave patterns on the vibrating strings are called stationary waves.
The waves in the air which carry the sound to our ears transfer energy, and so are called progressive waves.
Hope it helps.
:)
But in my book there is this part which i dont understand the amplitude of the vibration varies with position along the string: its zero at the nodes and maximum at an antinode .in progressive waves all the points have the same amplitude
Post by: Ghost Of Highbury on May 21, 2011, 04:11:20 pm
:)
But in my book there is this part which i dont understand the amplitude of the vibration varies with position along the string: its zero at the nodes and maximum at an antinode .in progressive waves all the points have the same amplitude
Yes, u see, the stationary waves are made out of 2 waves traveling in opposite directions. The phase difference is 180o, and because of the superposition of the 2 waves, the resultant displacement at the nodes is 0 and is maximum at the antinodes.
Post by: aneeta on May 21, 2011, 07:17:05 pm
Post by: Ghost Of Highbury on May 21, 2011, 07:21:03 pm
A hard ball and a soft ball, with equal masses and volumes are thrown at a glass window. The balls hit the window at the same speed. Suggest why the hard ball is more is more likely than the soft ball to break the window.
Newton's second law of motion. Force is the rate of change of momentum. F = Change in momentum / t
The ball which exerts more force on the window is, obviously, more likely to break the window.
The change in momentum is the same as the mass and velocity is the same for both the balls.
ALthough, the time the balls are in contact with the window is greater in the soft one than the hard one (deformation)
An increase in time, means decrease in Force. Hence, hard ball is more likely to break the window.
PS : Does anybody know how to put the 'delta' sign here? (Yes, the triangle symbol, Ari? )
Post by: TJ-56 on May 21, 2011, 07:21:37 pm
A hard ball and a soft ball, with equal masses and volumes are thrown at a glass window. The balls hit the window at the same speed. Suggest why the hard ball is more is more likely than the soft ball to break the window.
Force = rate of change of momentum = mv\t
So the masses and the velocities are equal. So the different variable here is the time t.
Soft ball: the ball gets compressed more and hence is more in contact with the window than the hard ball. Thus, momentum is much less and so it doesn't break the window. The hard ball has negligable compression and hence has a high momentum, breaking the window.
Post by: TJ-56 on May 21, 2011, 07:43:44 pm
Thanx in advance.
Post by: aneeta on May 21, 2011, 07:58:52 pm
I don't get why it's sin and not cos, or is it cos I didn't do part (i) correctly? :s
Post by: Arthur Bon Zavi on May 21, 2011, 11:12:24 pm
Newton's second law of motion. Force is the rate of change of momentum. F = Change in momentum / t
The ball which exerts more force on the window is, obviously, more likely to break the window.
The change in momentum is the same as the mass and velocity is the same for both the balls.
ALthough, the time the balls are in contact with the window is greater in the soft one than the hard one (deformation)
An increase in time, means decrease in Force. Hence, hard ball is more likely to break the window.
PS : Does anybody know how to put the 'delta' sign here? (Yes, the triangle symbol, Ari? )
See the code.
Post by: elemis on May 22, 2011, 03:56:06 am
PS : Does anybody know how to put the 'delta' sign here? (Yes, the triangle symbol, Ari? )
Type : \Delta
The select it and select the pi symbol in the editing bar above the message box.
Post by: Ghost Of Highbury on May 22, 2011, 07:00:04 am
Post by: elemis on May 22, 2011, 07:12:15 am
THNK YOU
Wilkommen.
Post by: NidZ- Hero on May 22, 2011, 07:57:47 am
Post by: elemis on May 22, 2011, 08:04:04 am
Ok why does a standing wave is produced only at a particular frequency and i guess this frequency is called resonance
Unless you are doing A2 Physics, I dont think you need to know that.
Post by: EMO123 on May 22, 2011, 08:05:08 am
Unless you are doing A2 Physics, I dont think you need to know that.still you can tell this to us
Post by: aneeta on May 22, 2011, 08:49:32 am
May/June 2005 paper 2 question 6b(iii)
I don't get why it's sin and not cos, or is it because I didn't do part (i) correctly? :s
Help!!
Post by: elemis on May 22, 2011, 08:54:29 am
May/June 2005 paper 2 question 6b(iii)
I don't get why it's sin and not cos, or is it because I didn't do part (i) correctly? :s
Help!!
The torque of a couple is the perpendicular distance from the line of action of one force multiplied by the magnitude of one of the forces.
Post by: TJ-56 on May 22, 2011, 08:55:59 am
Can someone explain October/November 2010 Paper 54 question 2 (e)? The mark scheme is useless on this one.
Thanx in advance.
Anyone? Please the exam is tomorrow.
Post by: elemis on May 22, 2011, 08:56:44 am
Anyone? Please the exam is tomorrow.
Examiners report, maybe ?
Post by: aneeta on May 22, 2011, 09:03:21 am
The torque of a couple is the perpendicular distance from the line of action of one force multiplied by the magnitude of one of the forces.
Yes I know that but in what direction is the force acting? Is it perpendicular to the direction of the electric field? Or parallel to the electric field?
Ps. Thanks a lot x
Post by: elemis on May 22, 2011, 09:12:48 am
Yes I know that but in what direction is the force acting? Is it perpendicular to the direction of the electric field? Or parallel to the electric field?
Ps. Thanks a lot x
Parallel. Always parallel.
Post by: TJ-56 on May 22, 2011, 09:41:01 am
Examiners report, maybe ?Thanx.. I solved it, but I cant believe all that is for only 2 marks.
Anyways: May June 2010 Paper 5, question 1, what should we draw exactly for the diagram?
Can someone post their drawing if possible?
Really appreciate the help.
Post by: aneeta on May 22, 2011, 10:14:29 am
Parallel. Always parallel.
Life saver :)
Post by: TJ-56 on May 22, 2011, 03:18:32 pm
Thanks for any help in advance.
Post by: °o.O-hash94-O.o° on May 22, 2011, 03:56:32 pm
http://www.xtremepapers.me/CIE/International%20A%20And%20AS%20Level/9702%20-%20Physics/9702_w10_qp_21.pdf (http://www.xtremepapers.me/CIE/International%20A%20And%20AS%20Level/9702%20-%20Physics/9702_w10_qp_21.pdf)
Post by: EMO123 on May 22, 2011, 04:10:53 pm
pls can anyone answer the question no 2 from following paper...i also need help
http://www.xtremepapers.me/CIE/International%20A%20And%20AS%20Level/9702%20-%20Physics/9702_w10_qp_21.pdf (http://www.xtremepapers.me/CIE/International%20A%20And%20AS%20Level/9702%20-%20Physics/9702_w10_qp_21.pdf)
Post by: aneeta on May 22, 2011, 04:37:00 pm
Post by: SkyPilotage on May 22, 2011, 06:20:44 pm
Can someone please explain to me how to draw a wave with a certain phase difference according to the wave already given in the question? =/ please.I guess you got that question from June 2010 P23 :D
Well you gotta know that one wave phase difference is 360..half is 180...quarter is 90 soo its rlly just approximations in the draawing...And they werent very picky in the marking scheme either...
Post by: leebux101 on May 22, 2011, 06:28:04 pm
October November 2010 Paper 51 question 1: the question says that the coil y is tightly winded around coil x, but the diagram required by the mark scheme says 'two independent coils labeled X and Y'. So what is the correct and acceptable drawing here? One tube with both coils and just labeling the 2?yes as long as u label them both, i suggest to draw with different coloured inks. :)
Thanks for any help in advance.
Post by: SkyPilotage on May 22, 2011, 06:30:36 pm
pls can anyone answer the question no 2 from following paper...Part I)
http://www.xtremepapers.me/CIE/International%20A%20And%20AS%20Level/9702%20-%20Physics/9702_w10_qp_21.pdf (http://www.xtremepapers.me/CIE/International%20A%20And%20AS%20Level/9702%20-%20Physics/9702_w10_qp_21.pdf)
Resolve the velocity into its vertical adn horizontal components...Meaning draw an arrow down adn to the right and you have an angle of 60 degress and you will have to use tan(v/8.2) Man it would bee easier if I had a scanner :-\
Part II)
since its thrown horizontally the initaila speed is 0 and the accelerration is 9.81 and final velocity is 14.2..So use the formula s=ut + 0.5(a)t^2
Part III)
horizontal speed is constant when there is no air resistance soo x= speed x time.... To find the time we can use the vertical component of velocity using v=u+at where v= 14.2 and u=0 and a=9.81 soo t=14.2/9.81.......Multiplay the answer by the horizontal component(8.2) to get the distance.....
Post by: TJ-56 on May 22, 2011, 06:33:05 pm
yes as long as u label them both, i suggest to draw with different coloured inks. :)
Thanks for the reply. I don't think they would accept any diagrams drawn with anything other than a soft pencil, so I though about it and I think its best that we draw them and label them (using the same pencil), and on the side of the diagram we should state that the 2 circuits are not connected together.
Post by: Vin on May 22, 2011, 07:12:22 pm
Q6 b)
suchan easy question i dont get it
coming up with few more in 2.
Post by: Vin on May 22, 2011, 07:17:12 pm
http://www.freeexampapers.com/get_past_papers.php?l=Past_Papers%2FA+Level%2FPhysics%2FCIE%2F2009+Jun%2F9702_s09_qp_2.pdf
Q1 a . why not vernier calliper? It can measure 1 mm alright. (its not mentioned in the mark scheme)
Q 5 b. how'd you get the path difference at 28?
Q 7 b iii)
Post by: SkyPilotage on May 22, 2011, 08:04:59 pm
http://www.freeexampapers.com/get_past_papers.php?l=Past_Papers%2FA+Level%2FPhysics%2FCIE%2F2008+Jun%2F9702_s08_qp_2.pdfUse the formula P=v^2/R
Q6 b)
suchan easy question i dont get it
coming up with few more in 2.
Am sure you will be able to do it...Emo explained it a couple of pages back :D
Post by: SkyPilotage on May 22, 2011, 08:07:24 pm
couple of stupid questions, but i idc-Just stick to micrometer :)
http://www.freeexampapers.com/get_past_papers.php?l=Past_Papers%2FA+Level%2FPhysics%2FCIE%2F2009+Jun%2F9702_s09_qp_2.pdf
Q1 a . why not vernier calliper? It can measure 1 mm alright. (its not mentioned in the mark scheme)
Q 5 b. how'd you get the path difference at 28?
Q 7 b iii)
-Distancew travelled by S1 is 100 cm and distance travelled by s2 is hypotuneus which is 128 cm... Path difference is 28 cm
-between A and Z you have two R resistors and 2 more R is parallel..Treat the 2 R as one resistor or 0.5 R sooo Total Resistance is R+R+0.5R=5/2R
Post by: SkyPilotage on May 22, 2011, 08:21:23 pm
Post by: aneeta on May 22, 2011, 08:53:08 pm
Good luck everyone that's sitting an exam tomorrw :) x
Post by: Arthur Bon Zavi on May 22, 2011, 09:06:36 pm
Any tips for tomorrows exam? =/
Good luck everyone that's sitting an exam tomorrw :) x
Don't leave your paper blank. That's all. :P ::)
Post by: leebux101 on May 22, 2011, 10:38:57 pm
Thanks for the reply. I don't think they would accept any diagrams drawn with anything other than a soft pencil, so I though about it and I think its best that we draw them and label them (using the same pencil), and on the side of the diagram we should state that the 2 circuits are not connected together.i dont understand. y wudnt they accept diagrams drawn in stuff other than a pencil , the isnstructions state that >> u may use a pencil for any diagrams.
that means that its ok to draw in stuff other than a pencil. what do u think?
Post by: astarmathsandphysics on May 22, 2011, 11:16:56 pm
Post by: MKL on May 23, 2011, 11:03:47 am
http://www.xtremepapers.me/CIE/International%20A%20And%20AS%20Level/9702%20-%20Physics/9702_s05_qp_1.pdf
Post by: Vin on May 23, 2011, 04:05:08 pm
I was talking about the other subpart for q7. But its all cool. You're a life saver :)
Post by: NidZ- Hero on May 28, 2011, 09:47:48 am
Post by: EMO123 on May 28, 2011, 09:50:14 am
What m i suppose to study for practicals >:(you need to learn what to do while your practicals and be accurate
don't make mistake in the uncertainties
and
enjoy your practicals
Post by: NidZ- Hero on May 28, 2011, 06:02:40 pm
Question 9 and 12
Pls Show the working
Thanks
Post by: elemis on May 28, 2011, 06:13:59 pm
Use the formula :
Final velocity of second Ball - final velocity of first ball = initial velocity of FIRST ball - initial velocity of second ball.
Make sure to use the VELOCITIES and not the SPEEDS.
Post by: NidZ- Hero on May 30, 2011, 12:46:07 pm
It has practical stuff on it
Post by: EMO123 on May 30, 2011, 06:02:41 pm
Why wont this http://www.freeexampapers.com/past_papers.php?l=Past_Papers%2FA+Level%2FPhysics%2FCoursework+and+Practicals/ wont openbecause it shows invalid file
It has practical stuff on it
Post by: Aadeez || Zafar on May 31, 2011, 08:53:35 pm
Why wont this http://www.freeexampapers.com/past_papers.php?l=Past_Papers%2FA+Level%2FPhysics%2FCoursework+and+Practicals/ wont openi have it
It has practical stuff on it
for p3
http://www.mediafire.com/?rwdbs8a9tds6nsj
for p5
http://www.mediafire.com/?vwanw9r4c2z848z
Post by: EMO123 on June 01, 2011, 02:26:14 am
i have itthanxx but a bit late
for p3
http://www.mediafire.com/?rwdbs8a9tds6nsj
for p5
http://www.mediafire.com/?vwanw9r4c2z848z
Post by: 7amwakingupinthemorning on June 02, 2011, 09:24:19 am
And tell me how the the answer to Question 5 in http://www.xtremepapers.me/Edexcel/Advanced%20Level/Physics/2010%20Jan/Edexcel%20Physics%20Unit%202%206PH02%20January%202010%20QP.pdf (January 2010) is 50 ?? :-\
Thanks xx
Post by: Ghost Of Highbury on June 02, 2011, 11:18:47 am
Can someone please tell me how we could find the frequency of a wave using a cathode ray oscilloscope??????
And tell me how the the answer to Question 5 in http://www.xtremepapers.me/Edexcel/Advanced%20Level/Physics/2010%20Jan/Edexcel%20Physics%20Unit%202%206PH02%20January%202010%20QP.pdf (January 2010) is 50 ?? :-\
Thanks xx
The time period of the wave is 20ms (4 blocks)
20ms = 0.02s
Frequency = 1/0.02 = 100/2 = 50Hz
Post by: Ghost Of Highbury on June 05, 2011, 10:50:30 am
http://www.xtremepapers.me/CIE/index.php?dir=International%20A%20And%20AS%20Level/9702%20-%20Physics/&file=9702_w02_qp_1.pdf
Guys...could you help me out with Qns 11, 19 and 36...It would be awesome if someone could give me a detailed explanation... cus im just not getting it
Oct/Nov 2002 By the way...
11. Gimme some time
19. KE (original) = 0.5mv^2
v^2 = 2as = 3920
therefore KE = 1960m
40% loss = 40/100 * 1960m loss = 784m loss
Thus, remaining = 1176m
1176m = 0.5mv^2
v = sqrt(2*1176) ~ 49
36) Total resistance = 1/(1/15 + 1/15) = 7.5ohms
Resistance between X and Y = 1/(1/10 + 1/5) = 10/3
VOltage = (10/3)/7.5 V = 2/3 V
Post by: NidZ- Hero on June 05, 2011, 11:12:00 am
11. Gimme some time
19. KE (original) = 0.5mv^2
v^2 = 2as = 3920
therefore KE = 1960m
40% loss = 40/100 * 1960m loss = 784m loss
Thus, remaining = 1176m
1176m = 0.5mv^2
v = sqrt(2*1176) ~ 49
36) Total resistance = 1/(1/15 + 1/15) = 7.5ohms
Resistance between X and Y = 1/(1/10 + 1/5) = 10/3
VOltage = (10/3)/7.5 V = 2/3 V
VOltage = (10/3)/7.5 V = 2/3 V
Which formula did yu used
Post by: JACKRABBIT on June 05, 2011, 11:36:08 am
Quote
11. Gimme some time
19. KE (original) = 0.5mv^2
v^2 = 2as = 3920
therefore KE = 1960m
40% loss = 40/100 * 1960m loss = 784m loss
Thus, remaining = 1176m
1176m = 0.5mv^2
v = sqrt(2*1176) ~ 49
36) Total resistance = 1/(1/15 + 1/15) = 7.5ohms
Resistance between X and Y = 1/(1/10 + 1/5) = 10/3
VOltage = (10/3)/7.5 V = 2/3 V
VOltage = (10/3)/7.5 V = 2/3 V
Which formula did yu used
Awesome..thx for number 19...but for 36 I kinda need a bit more detail on that please...and yeah which formula did you use for the last part of qn 36
Post by: EMO123 on June 05, 2011, 11:38:54 am
11. Gimme some timefor Kinetic Energy = 1/2mv2
19. KE (original) = 0.5mv^2
v^2 = 2as = 3920
therefore KE = 1960m
40% loss = 40/100 * 1960m loss = 784m loss
Thus, remaining = 1176m
1176m = 0.5mv^2
v = sqrt(2*1176) ~ 49
36) Total resistance = 1/(1/15 + 1/15) = 7.5ohms
Resistance between X and Y = 1/(1/10 + 1/5) = 10/3
VOltage = (10/3)/7.5 V = 2/3 V
then v2=2as
for resistance
if in parallel then = 1/R1+1/R2=1/R
if in series then = R1+R2=R
Voltage as such = IR
but i dint get it here HIghburry
Post by: Ghost Of Highbury on June 05, 2011, 03:25:05 pm
Potential divider formula.
VOltage = (10/3)/7.5 V = 2/3 V
Which formula did yu used
V(out) = R1/R1+R2 *V(in)
Post by: NidZ- Hero on June 05, 2011, 03:43:14 pm
V0ut=V*R^1/(R^1 + R^2)
Post by: Malak on June 05, 2011, 04:22:32 pm
I in my book the formula they say isIts the same, Multiply by V first or in the end, wont make a difference :)
V0ut=V*R^1/(R^1 + R^2)
Post by: Ghost Of Highbury on June 05, 2011, 04:45:11 pm
Awesome..thx for number 19...but for 36 I kinda need a bit more detail on that please...and yeah which formula did you use for the last part of qn 36
I'm sry, i can't figure out Q11. Would be great if someone could help out. FIDATO? Ari? Astar?
Post by: EMO123 on June 05, 2011, 05:15:52 pm
I'm sry, i can't figure out Q11. Would be great if someone could help out. FIDATO? Ari? Astar?ask this to
master_key
Post by: elemis on June 05, 2011, 05:22:02 pm
I'm sry, i can't figure out Q11. Would be great if someone could help out. FIDATO? Ari? Astar?
Take east as the positive direction and west as the negative direction.
Use the idea :
Vi of Body A - Vi of Body B = Vf of Body B - Vf of Body A
Where Vi = initial VELOCITY and Vf = final velocity.
Post by: Ghost Of Highbury on June 05, 2011, 05:26:42 pm
Post by: EMO123 on June 05, 2011, 05:27:05 pm
Take east as the positive direction and west as the negative direction.thanxx
Use the idea :
Vi of Body A - Vi of Body B = Vf of Body B - Vf of Body A
Where Vi = initial VELOCITY and Vf = final velocity.
ari
Post by: bulono on June 06, 2011, 01:13:48 am
Thx :)
Post by: Ghost Of Highbury on June 06, 2011, 06:33:13 am
can somebody explain the answers for Q9,13,15 & 28 O/N 2009/11
Thx :)
9. otal Ke b4 collision => 0.5*2m*u^2 + 0.5*m*(-u)^2 = mu^2 + mu^2/2 = 3/2 mu^2
Total momentum b4 => 2mu - mu = mu
A. KE after collision -> 0.5*2m*(u/3)^2 + 0.5*m*(5u/3)^2 = mu^2 /9 + 25mu^2/18 = 3/2 mu^2
Momentum = 2m*(-u/3) + m*(5u/3) = 5mu/3 - 2mu/3 = 3mu/3 = mu
13.Im not sure about variant 1. The torque is obviously 4.5Nm
{Tension = Torque/diameter = 3/0.1 = 30
Torque = tension*diameter = 30*0.15=4.5}
Although, i'm not sure why the tension is 60. Maybe, because the lower belt has no tension, and for a torque two forces acting in opposite directions
are required, I guess, the tension in the lower belt is transferred to the upper belt which adds up to 60 (30+30).
15. First find out the velocity of the 1kg trolley
(2*-2)+(1*x) = 0
-4+x = 0
x = 4m/s
KE = 0.5mv^2 = 0.5*2*(-2)^2 + 0.5*1*(4)^2 = 12
D
28. Charge to mass
We know, that Force acting on it = Weight
E = F/q
F = Eq
Eq = mg
m = Eq/g
Charge to mass = q/(Eq/g) = g/E
Polarity should be negative because the positive plate will pull it upwards and its weight will pull it downwards. Hence they balance o ut
Post by: JACKRABBIT on June 06, 2011, 06:43:27 am
Quote
VOltage = (10/3)/7.5 V = 2/3 V
isnt 10/3/7.5 equal to 4/9?? and its root is 2/3...so...i get that its the potential divider formula..and its resistance you want to find the Vout of divided by total resistance into the Vin..which is 2 in this case right?? so is it 10/3 * 2 /7.5 ...which is 8/9
Post by: Ghost Of Highbury on June 06, 2011, 07:06:24 am
isnt 10/3/7.5 equal to 4/9?? and its root is 2/3...so...i get that its the potential divider formula..and its resistance you want to find the Vout of divided by total resistance into the Vin..which is 2 in this case right?? so is it 10/3 * 2 /7.5 ...which is 8/9
you're right, my bad, should be (10/3)/7.5 * 2 = 8/9
Hmm, y is the answer 2/3 then? Gimme some time..
Post by: Ghost Of Highbury on June 06, 2011, 08:40:25 am
the p.d across the 5 ohm resistor in the 1st part of the circuit = 5/15 * 2V = 2/3
THe p.d across the 2 5 ohms resistors in the second part of the circuit = 10/15 * 2 = 4/3
Hence the total potential "DIFFERENCE" = 4/3 - 2/3 = 2/3 V
------
If u apply this formula to find the p.d across, say the all three resistors in the 1st branch
The voltage across the point at the bottom and the negative terminal = 0V
the voltage across the point at the top and the negative terminal = 2V
Difference = 2-0 = 2V -> which is the P.d across the 3 resistors in the first branch..
Post by: physichemaths on June 06, 2011, 09:04:54 am
how to do question 40?? how we know the ans is C?
Post by: EMO123 on June 06, 2011, 09:46:46 am
http://www.xtremepapers.me/CIE/International%20A%20And%20AS%20Level/9702%20-%20Physics/9702_s07_qp_1.pdfon taking ratio you an find that
how to do question 40?? how we know the ans is C?
take the ratio of proton:Neutron
For H= 1:1 i.e 1
for He= 2:2 i.e 1
for Li= 3:4 i.e 0.75
for Be= 4:5 i.e 0.8
so the least ratio has the lowest speed that is C
Post by: JACKRABBIT on June 06, 2011, 04:03:07 pm
Quote
k, im not sure about this method but i got the answer somewhow
the p.d across the 5 ohm resistor in the 1st part of the circuit = 5/15 * 2V = 2/3
THe p.d across the 2 5 ohms resistors in the second part of the circuit = 10/15 * 2 = 4/3
Hence the total potential "DIFFERENCE" = 4/3 - 2/3 = 2/3 V
------
If u apply this formula to find the p.d across, say the all three resistors in the 1st branch
The voltage across the point at the bottom and the negative terminal = 0V
the voltage across the point at the top and the negative terminal = 2V
Difference = 2-0 = 2V -> which is the P.d across the 3 resistors in the first branch..
Aiight thx man much appreciated...Now i have a few more...and these are like really the hard ones.....anyway...again a full detailed explanation would be awesome cos i dont take maths and that kinda makes it hard to understand all these equations and stuff
http://www.xtremepapers.me/CIE/index.php?dir=International%20A%20And%20AS%20Level/9702%20-%20Physics/&file=9702_w10_qp_12.pdf
Qns. 7, 8(my teacher couldnt solve it) ,9,15,34
AAAAND finally..
http://www.xtremepapers.me/CIE/index.php?dir=International%20A%20And%20AS%20Level/9702%20-%20Physics/&file=9702_w07_qp_1.pdf
Qns, 23 and 25
Post by: username on June 06, 2011, 05:47:53 pm
exam on wednesday
and loads of doubts
http://www.xtremepapers.me/CIE/International%20A%20And%20AS%20Level/9702%20-%20Physics/9702_w10_qp_11.pdf
questions: 7,8,9,11,15,22,29,40
http://www.xtremepapers.me/CIE/International%20A%20And%20AS%20Level/9702%20-%20Physics/9702_w10_qp_12.pdf
questions: 9,13,34,37
Post by: Ghost Of Highbury on June 06, 2011, 06:19:30 pm
Aiight thx man much appreciated...Now i have a few more...and these are like really the hard ones.....anyway...again a full detailed explanation would be awesome cos i dont take maths and that kinda makes it hard to understand all these equations and stuff
http://www.xtremepapers.me/CIE/index.php?dir=International%20A%20And%20AS%20Level/9702%20-%20Physics/&file=9702_w10_qp_12.pdf
Qns. 7, 8(my teacher couldnt solve it) ,9,15,34
AAAAND finally..
http://www.xtremepapers.me/CIE/index.php?dir=International%20A%20And%20AS%20Level/9702%20-%20Physics/&file=9702_w07_qp_1.pdf
Qns, 23 and 25
8. XY - 40m - 12s
XZ - 80m - 18s
80 = 18u + 0.5a*18^2
40 = 12u + 0.5a*12^2
Solve the simultaneous equations, by eliminating u and calculating a.
9. There's a little bit of guess work involved here.
Given it's inelastic, the rebound velocity won't be v. Say, 40% of energy is lost, say around 0.6v
Change = (-0.6v*2m) - 2mv = -3.2mv
SO the change is almost 3mv.
(Even if u take 0.7v, u won't get 4mv+)
15. Steady speed v, therefore weight is balanced by retarding force
mg = kv
v = mg/k
KE = 0.5mv^2 = 0.5m*(mg/k)^2 = m/2 * m^2g^2 /k^2 = m^3g^2 /2k^2
34. When x = 0, the voltage measured is across the whole of the wire, so it has to be V.
when x decreases, and the pointer is moved at the right end of the part with resistivity p, the voltage is divided into 2 parts (p and 2p+3p)
So the voltage across the length x is now, not V, but a lil less than V because some VOltage is lost in the remaining length of the wire.
Answer: B (note the graph is not a straight line because as x decreases, the resistance doesn't decrease proportionately)
---
23. frequency = 1/(time taken for 1 complete oscillation = 50m (check x-axis))
It covers 8m in 1 second (speed)
so 50m in 50/8 s = 6.25
That's its time period
Hence, frequency = 1/t = 1/6.25 = 0.16Hz
Speed = 2pi*a*f = 2pi*2*0.16 = 2.01m/s
KE = 0.5mv^2 = 4mJ
25. dsin45 = 3*lambda
d = 3lambda/sin 45 = 3sqrt(2)*lambda
maximum order of diffraction = dsin90 = n*lambda
n = d/lambda
n = (3sqrt(2)*lambda)/lambda = 3sqrt(2) = 4.24264.... ~ 4
Post by: Ghost Of Highbury on June 06, 2011, 06:52:40 pm
hey guys
exam on wednesday
and loads of doubts
http://www.xtremepapers.me/CIE/International%20A%20And%20AS%20Level/9702%20-%20Physics/9702_w10_qp_11.pdf
questions: 7,8,9,11,15,22,29,40
http://www.xtremepapers.me/CIE/International%20A%20And%20AS%20Level/9702%20-%20Physics/9702_w10_qp_12.pdf
questions: 9,13,34,37
qp11
7. I have trouble in this too. How does horizontal component of velocity decrease to zero? Isn't it constant in a projectile?
8. It goes up 5m, comes down 3m. Displacement = 5-3 = 2m
9. v^2 = u^2 +2as
s = v^2-u^2 / 2a
v = 0, so 0 = u^2+2as -> 2a = -u^2/s or -u^2/x
Now, the new u = U+0.2u = 1.2u
V = 0
so s = 0-1.44u^2 /2a
s = -1.44u^2/(-u^2/x) = 1.44x
11. M1V1 -M2V2 = 0
M1V1 = M2V2
V1/V2 = M2/M1
B
15. It can be either A or D (because they are closed vector triangles)
It's D, because the direction of the arrow representing the force by the cable is correct.
22. The length of the wire doesnt affect the extension. The load and cross-sectional area do. Here both are the same. So extensions of
both the wires are same (1.5 and 2), that's y total e = 1.5 + 2 = 3.5mm
29. It's upward because electrons are attracted towards +ve plate. And ofcourse, it's not a part of the circle because the path is not exactly an arc.
40. Count th4e no. of protons in Argon, it's 18. Only option with 18 protons for argon is C. So C it is
qp12
9. There's a little bit of guess work involved here.
Given it's inelastic, the rebound velocity won't be v. Say, 40% of energy is lost, say around 0.6v
Change = (-0.6v*2m) - 2mv = -3.2mv
SO the change is almost 3mv.
(Even if u take 0.7v, u won't get 4mv+)
34. When x = 0, the voltage measured is across the whole of the wire, so it has to be V.
when x decreases, and the pointer is moved at the right end of the part with resistivity p, the voltage is divided into 2 parts (p and 2p+3p)
So the voltage across the length x is now, not V, but a lil less than V because some VOltage is lost in the remaining length of the wire.
Answer: B (note the graph is not a straight line because as x decreases, the resistance doesn't decrease proportionately)
13. Ofcourse it's D, gravitational pull is constant (mg). Mass is constant and a is a constant.
37. Check 1/R = 1/R1 + 1/R2 + 1/R3
And see when u get value of R as 50.
Post by: SkyPilotage on June 06, 2011, 07:19:16 pm
Post by: Ghost Of Highbury on June 06, 2011, 07:24:28 pm
How is it 15?
(1/300000)/(450*10^-9) = 7.4 orders of diffraction
Maxima on both sides = 7.4*2 = 14.8
Meaning, myou would see 7 on both sides, so it should add up to 14 right
Why is the answer 15?
And Can someone also explain question 27_w09_qp11
@Sky-pilotage - Ofcourse, it is a projectile? SO what if there's air resistance?
Post by: Ghost Of Highbury on June 06, 2011, 07:32:23 pm
Post by: SkyPilotage on June 06, 2011, 07:34:32 pm
DoubtIt Is projected..Correct..But It will NOT show uniform Projectile motion....
How is it 15?
(1/300000)/(450*10^-9) = 7.4 orders of diffraction
Maxima on both sides = 7.4*2 = 14.8
Meaning, myou would see 7 on both sides, so it should add up to 14 right
Why is the answer 15?
And Can someone also explain question 27_w09_qp11
@Sky-pilotage - Ofcourse, it is a projectile? SO what if there's air resistance?
Therefore Its not a regular projjectille So there is no constand vertical acceleration nor constant horizontal velocity...
A Projectile is an object upon which ONLY the force of Gravity acts on it...
Post by: Ghost Of Highbury on June 06, 2011, 07:37:50 pm
It Is projected..Correct..But It will NOT show uniform Projectile motion....
Therefore Its not a regular projjectille So there is no constand vertical acceleration nor constant horizontal velocity...
A Projectile is an object upon which ONLY the force of Gravity acts on it...
Agreed. Could you help me with question 27_w09_qp11 and q26.
Post by: SkyPilotage on June 06, 2011, 07:57:32 pm
Agreed. Could you help me with question 27_w09_qp11 and q26.Question 26:
You will find the number of orders on each side...So to get the maximium Possible of orders on one side You will consider sinX to be sin90....
THerefore n(4.5 x 10^-7)=(1 / (300/0.001) ) ( sin90) *P.S:-we need to get d which is the seperation of 2 slits.
You will get n as 7.4.. Which means that there are 7 orders on each side including the zero ordder..Hence 15 orders..
Question 27:
The rule is F/charge..They want the field strength acting on charge q only..SO its F/q...
Post by: Ghost Of Highbury on June 06, 2011, 08:35:35 pm
Question 26:
You will find the number of orders on each side...So to get the maximium Possible of orders on one side You will consider sinX to be sin90....
THerefore n(4.5 x 10^-7)=(1 / (300/0.001) ) ( sin90) *P.S:-we need to get d which is the seperation of 2 slits.
You will get n as 7.4.. Which means that there are 7 orders on each side including the zero ordder..Hence 15 orders..
Question 27:
The rule is F/charge..They want the field strength acting on charge q only..SO its F/q...
Ah! The zero order! Damn, thanks a lot man.
Post by: wstrawberries on June 06, 2011, 09:18:34 pm
http://www.xtremepapers.me/CIE/International%20A%20And%20AS%20Level/9702%20-%20Physics/9702_s06_qp_1.pdf
Post by: SkyPilotage on June 06, 2011, 10:07:42 pm
I need help with question 9 and 33. If anyone could spare the time?Question 9:-( I just need a confirmation on this )
http://www.xtremepapers.me/CIE/International%20A%20And%20AS%20Level/9702%20-%20Physics/9702_s06_qp_1.pdf
Since the mass is going up and down..
At A its halfway going up..
At B its at its highest point...
At C its halfway going down..
And finally At D its at the bottom which is the lowest point of its motion..
Question 33:
Find the Pd of the resistors accross XQY or XPY..
The voltage across the 2 parallel wires is the same i.e 12 V...
To find the voltage of the 500 ohm resistor its 500/(500+1000) x12 = 4 V
ANd the voltage across the 2000 ohm resistor is 2000/(2000+1000) = 8 V
-->Therefore The potential DIFFERENCE is 8-4 volts = 4volts..
* you could also go using XQY and find the voltages accross the 1000 oh resistors and you will find the top one to be 4 and the bottom on eot be 8 so THe P.d between XandY is 4 volts...
@Ghost I forgot it too ;D
Glad to help :)
Post by: wstrawberries on June 06, 2011, 10:37:43 pm
Question 9:-( I just need a confirmation on this )
Since the mass is going up and down..
At A its halfway going up..
At B its at its highest point...
At C its halfway going down..
And finally At D its at the bottom which is the lowest point of its motion..
Question 33:
Find the Pd of the resistors accross XQY or XPY..
The voltage across the 2 parallel wires is the same i.e 12 V...
To find the voltage of the 500 ohm resistor its 500/(500+1000) x12 = 4 V
ANd the voltage across the 2000 ohm resistor is 2000/(2000+1000) = 8 V
-->Therefore The potential DIFFERENCE is 8-4 volts = 4volts..
* you could also go using XQY and find the voltages accross the 1000 oh resistors and you will find the top one to be 4 and the bottom on eot be 8 so THe P.d between XandY is 4 volts...
@Ghost I forgot it too ;D
Glad to help :)
Thanks a lot :D uhhmm for q 33 what formula are you using to calculate 4V and 8V?
Post by: SkyPilotage on June 06, 2011, 11:05:18 pm
Thanks a lot :D uhhmm for q 33 what formula are you using to calculate 4V and 8V?I used it as A ratio..But If you are confused...
USE the formula R=V/I..
For example the current passing through X is v/R= 12/1500 = 0.008 A
To find the V in the 1000 ohm resistor : 0.008 x 1000 = 8 V...
Through Y : do the same thing..
I= 12/3000 = 0.004 A
V in 1000 ohm resistor= 0.004 x 1000=4 volts..
Post by: wstrawberries on June 06, 2011, 11:44:03 pm
I used it as A ratio..But If you are confused...
USE the formula R=V/I..
For example the current passing through X is v/R= 12/1500 = 0.008 A
To find the V in the 1000 ohm resistor : 0.008 x 1000 = 8 V...
Through Y : do the same thing..
I= 12/3000 = 0.004 A
V in 1000 ohm resistor= 0.004 x 1000=4 volts..
ohhhh okay :) thanks again!
Post by: username on June 07, 2011, 07:24:34 am
qp11sry...
7. I have trouble in this too. How does horizontal component of velocity decrease to zero? Isn't it constant in a projectile?
8. It goes up 5m, comes down 3m. Displacement = 5-3 = 2m
9. v^2 = u^2 +2as
s = v^2-u^2 / 2a
v = 0, so 0 = u^2+2as -> 2a = -u^2/s or -u^2/x
Now, the new u = U+0.2u = 1.2u
V = 0
so s = 0-1.44u^2 /2a
s = -1.44u^2/(-u^2/x) = 1.44x
11. M1V1 -M2V2 = 0
M1V1 = M2V2
V1/V2 = M2/M1
B
15. It can be either A or D (because they are closed vector triangles)
It's D, because the direction of the arrow representing the force by the cable is correct.
22. The length of the wire doesnt affect the extension. The load and cross-sectional area do. Here both are the same. So extensions of
both the wires are same (1.5 and 2), that's y total e = 1.5 + 2 = 3.5mm
29. It's upward because electrons are attracted towards +ve plate. And ofcourse, it's not a part of the circle because the path is not exactly an arc.
40. Count th4e no. of protons in Argon, it's 18. Only option with 18 protons for argon is C. So C it is
qp12
9. There's a little bit of guess work involved here.
Given it's inelastic, the rebound velocity won't be v. Say, 40% of energy is lost, say around 0.6v
Change = (-0.6v*2m) - 2mv = -3.2mv
SO the change is almost 3mv.
(Even if u take 0.7v, u won't get 4mv+)
34. When x = 0, the voltage measured is across the whole of the wire, so it has to be V.
when x decreases, and the pointer is moved at the right end of the part with resistivity p, the voltage is divided into 2 parts (p and 2p+3p)
So the voltage across the length x is now, not V, but a lil less than V because some VOltage is lost in the remaining length of the wire.
Answer: B (note the graph is not a straight line because as x decreases, the resistance doesn't decrease proportionately)
13. Ofcourse it's D, gravitational pull is constant (mg). Mass is constant and a is a constant.
37. Check 1/R = 1/R1 + 1/R2 + 1/R3
And see when u get value of R as 50.
i got the links jumbled up:(
Post by: EMO123 on June 07, 2011, 07:27:00 am
sry...so you got that or not
i got the links jumbled up:(
if not then tell i will post all in another method
Post by: physichemaths on June 07, 2011, 07:46:16 am
Post by: username on June 07, 2011, 07:47:11 am
so you got that or notumm
if not then tell i will post all in another method
i mean u answered the rong questions
http://www.xtremepapers.me/CIE/International%20A%20And%20AS%20Level/9702%20-%20Physics/9702_w10_qp_11.pdf
questions: 9,13,34,37
http://www.xtremepapers.me/CIE/International%20A%20And%20AS%20Level/9702%20-%20Physics/9702_w10_qp_12.pdf
questions: 7,8,9,11,15,22,26,27,29,40
got it mixed up:(
Post by: physichemaths on June 07, 2011, 07:59:56 am
I have trouble with question 24 june 2008? Anyone?please help me :(
Post by: username on June 07, 2011, 08:00:57 am
please help me :(answer is c
because
same steel so young modulus is the same
Post by: physichemaths on June 07, 2011, 08:08:35 am
answer is cthanks! How about ques 26?
because
same steel so young modulus is the same
Post by: physichemaths on June 07, 2011, 08:08:42 am
answer is c
because
same steel so young modulus is the same
Post by: physichemaths on June 07, 2011, 08:08:50 am
answer is c
because
same steel so young modulus is the same
Post by: Ghost Of Highbury on June 07, 2011, 08:30:36 am
umm
i mean u answered the rong questions
http://www.xtremepapers.me/CIE/International%20A%20And%20AS%20Level/9702%20-%20Physics/9702_w10_qp_11.pdf
questions: 9,13,34,37
http://www.xtremepapers.me/CIE/International%20A%20And%20AS%20Level/9702%20-%20Physics/9702_w10_qp_12.pdf
questions: 7,8,9,11,15,22,29,40
got it mixed up:(
Oh darn!
@physichemaths - 26) Ip = k1/r^2
Ip = k2*A^2
k1/r^2 = k2*(8*10^-6)^2
Iq = k1/4r^2
Iq = k1/4r^2 = k1/r^2 * 1/4 = k2*(8*10^-6)^2 *1/4 = 1.6*10^-11k2
Iq = k2A^2
A^2 = 1.6*10^-11
A = sqrt(1.6*10^-11) = 4*10^-6
Post by: username on June 07, 2011, 08:40:26 am
Oh darn!sry bout dat...
@physichemaths - 26) Ip = k1/r^2
Ip = k2*A^2
k1/r^2 = k2*(8*10^-6)^2
Iq = k1/4r^2
Iq = k1/4r^2 = k1/r^2 * 1/4 = k2*(8*10^-6)^2 *1/4 = 1.6*10^-11k2
Iq = k2A^2
A^2 = 1.6*10^-11
A = sqrt(1.6*10^-11) = 4*10^-6
Post by: physichemaths on June 07, 2011, 09:03:07 am
sry bout dat...how come the amplitude is times 2?
Post by: username on June 07, 2011, 09:04:21 am
how come the amplitude is times 2?intensity= amplitude^2
Post by: physichemaths on June 07, 2011, 09:09:49 am
intensity= amplitude^2yeah. I know. Then why is there 2 in Ip = 2kA^2? Not the square one, i mean 2 in the front?
Post by: Ghost Of Highbury on June 07, 2011, 10:23:45 am
Post by: physichemaths on June 07, 2011, 10:41:33 am
No, there are two different constants, k1 and k2. The "2" is not multiplied by A, it belongs to the constant k. k1 and k2owh. I got it. Thanks a lot! :D
Post by: JACKRABBIT on June 07, 2011, 12:11:43 pm
Quote
8. XY - 40m - 12s
XZ - 80m - 18s
80 = 18u + 0.5a*18^2
40 = 12u + 0.5a*12^2
Solve the simultaneous equations, by eliminating u and calculating a.
9. There's a little bit of guess work involved here.
Given it's inelastic, the rebound velocity won't be v. Say, 40% of energy is lost, say around 0.6v
Change = (-0.6v*2m) - 2mv = -3.2mv
SO the change is almost 3mv.
(Even if u take 0.7v, u won't get 4mv+)
15. Steady speed v, therefore weight is balanced by retarding force
mg = kv
v = mg/k
KE = 0.5mv^2 = 0.5m*(mg/k)^2 = m/2 * m^2g^2 /k^2 = m^3g^2 /2k^2
34. When x = 0, the voltage measured is across the whole of the wire, so it has to be V.
when x decreases, and the pointer is moved at the right end of the part with resistivity p, the voltage is divided into 2 parts (p and 2p+3p)
So the voltage across the length x is now, not V, but a lil less than V because some VOltage is lost in the remaining length of the wire.
Answer: B (note the graph is not a straight line because as x decreases, the resistance doesn't decrease proportionately)
---
23. frequency = 1/(time taken for 1 complete oscillation = 50m (check x-axis))
It covers 8m in 1 second (speed)
so 50m in 50/8 s = 6.25
That's its time period
Hence, frequency = 1/t = 1/6.25 = 0.16Hz
Speed = 2pi*a*f = 2pi*2*0.16 = 2.01m/s
KE = 0.5mv^2 = 4mJ
25. dsin45 = 3*lambda
d = 3lambda/sin 45 = 3sqrt(2)*lambda
maximum order of diffraction = dsin90 = n*lambda
n = d/lambda
n = (3sqrt(2)*lambda)/lambda = 3sqrt(2) = 4.24264.... ~ 4
WOW thanks man....ur good lol
AALSO...I know that a electric field with the positive plate earthed will still have a positive chargeon it...but what if a theres an electric field and the NEGATIVE plate is earthed??? will there still be a negative charge on it??? or will there be no charge on it...??
Post by: wstrawberries on June 07, 2011, 01:15:09 pm
http://www.xtremepapers.me/CIE/International%20A%20And%20AS%20Level/9702%20-%20Physics/9702_s08_qp_1.pdf
Post by: username on June 07, 2011, 01:28:56 pm
do we have to learn the frequencies and wavelenghts of colrs of light?
like red, yellow... etc.?
Post by: EMO123 on June 07, 2011, 01:37:54 pm
guysthat is simple
do we have to learn the frequencies and wavelenghts of colrs of light?
like red, yellow... etc.?
Post by: Ghost Of Highbury on June 07, 2011, 01:40:03 pm
guys
do we have to learn the frequencies and wavelenghts of colrs of light?
like red, yellow... etc.?
It's better to know the whole EM spectrum.
Post by: Ghost Of Highbury on June 07, 2011, 01:43:02 pm
Can someone please help me out with q 37
http://www.xtremepapers.me/CIE/International%20A%20And%20AS%20Level/9702%20-%20Physics/9702_s08_qp_1.pdf
COmbined resistance of VOltmeter and R2 = 50ohms
VOltage that part receives = 50/150 * 6 = 2V
Current = V/R = 2/100000 = 20*10^-6
Post by: Ghost Of Highbury on June 07, 2011, 01:43:43 pm
Quote from: JACKRABBIT
WOW thanks man....ur good lol
AALSO...I know that a electric field with the positive plate earthed will still have a positive chargeon it...but what if a theres an electric field and the NEGATIVE plate is earthed??? will there still be a negative charge on it??? or will there be no charge on it...??
I'm not sure about this, but I think it won't have a charge on it if the -ve plate is earthed.
Post by: wstrawberries on June 07, 2011, 02:35:48 pm
COmbined resistance of VOltmeter and R2 = 50ohms
VOltage that part receives = 50/150 * 6 = 2V
Current = V/R = 2/100000 = 20*10^-6
Thank you :)
Post by: Ghost Of Highbury on June 07, 2011, 03:19:41 pm
Doubt : How is the force H at the hinge greater than the Weight?
Post by: EMO123 on June 07, 2011, 03:32:54 pm
w03_qp1 Question 16in this see the forces applied
Doubt : How is the force H at the hinge greater than the Weight?
on W it is the least because it is simple to make it fall the door down
then it is H as to open the door from hinge needs more force
and the last is the T with the highest force because of the weight of the door
hope it helps
Post by: wstrawberries on June 07, 2011, 03:35:16 pm
q 20
Post by: NidZ- Hero on June 07, 2011, 05:45:19 pm
http://www.xtremepapers.me/CIE/International%20A%20And%20AS%20Level/9702%20-%20Physics/9702_w07_qp_1.pdf
Question 11, 16 , 17 , 26 , 37
http://www.xtremepapers.me/CIE/International%20A%20And%20AS%20Level/9702%20-%20Physics/9702_w08_qp_1.pdf
Question 27 , 29
http://www.xtremepapers.me/CIE/International%20A%20And%20AS%20Level/9702%20-%20Physics/9702_s07_qp_1.pdf
Question 40 , 37
http://www.xtremepapers.me/CIE/International%20A%20And%20AS%20Level/9702%20-%20Physics/9702_s08_qp_1.pdf
Question 14 , 16 , 31
THANNNNNKS
Post by: EMO123 on June 07, 2011, 06:10:44 pm
I know the answer But i want An explanation Can Some one explain11.
http://www.xtremepapers.me/CIE/International%20A%20And%20AS%20Level/9702%20-%20Physics/9702_w07_qp_1.pdf
Question 11, 16 , 17 , 26
Equal and Opposite Force Exerted on Road by the wheel so the answer is B
16.
The there is a elastic Potential Energy it is maximum at Position 3
and kinetic energy is minimum because when Potential Energy is Maximum then the K.E. is minimum.
17.
The density of the substance (x cm^3) = Density of the same substance (of unit volume)
Keeping this in mind, density = mass/volume
Mass of an atom = Mp, there Np atoms in all
Thus, mass of Np atoms = Mp*Np
Volume = unit volume
Density of P = (Mp*Np)/unit volume
Repeat the same with Q, density = (Mq*Nq)/unit volume
As density of P > density of Q
Mp*Np > Mq*Nq
(Unit volume gets cancelled)
26.
as Shown in the diagram that the field is moving from Positive to Negative
so if the electron is placed there then it will definitely the the right so
it will move towards the left
Hope it Helps
Post by: Ghost Of Highbury on June 07, 2011, 06:24:42 pm
http://www.xtremepapers.me/CIE/International%20A%20And%20AS%20Level/9702%20-%20Physics/9702_s09_qp_1.pdf
q 20
Stress = tension/area
Tension = stress*Area
stress = Y*strain
Tension = Y*Strain*Area
strain = e/l
Tension = YeA/l
Tension (P) = YeA/l
Tension (Q) = Ye0.5A/2l = YeA/4l
Tension(P)/Tension(Q) = 4/1
Post by: Ghost Of Highbury on June 07, 2011, 06:27:57 pm
I know the answer But i want An explanation Can Some one explain
http://www.xtremepapers.me/CIE/International%20A%20And%20AS%20Level/9702%20-%20Physics/9702_w07_qp_1.pdf
Question 11, 16 , 17 , 26 , 37
http://www.xtremepapers.me/CIE/International%20A%20And%20AS%20Level/9702%20-%20Physics/9702_w08_qp_1.pdf
Question 27 , 29
http://www.xtremepapers.me/CIE/International%20A%20And%20AS%20Level/9702%20-%20Physics/9702_s07_qp_1.pdf
Question 40 , 37
http://www.xtremepapers.me/CIE/International%20A%20And%20AS%20Level/9702%20-%20Physics/9702_s08_qp_1.pdf
Question 14 , 16 , 31
THANNNNNKS
w08 - 27) Distance b/w 2 antinodes = 15mm
wavelength = 15mm*2 = 30mm = 0.03m
Frequency = c/wavelength = (3*10^8)/0.03 = 1*10^10
29) THe point charge is positively charged. SO inwards. Force on X > Y , because its closer
Post by: EMO123 on June 07, 2011, 06:33:25 pm
http://www.xtremepapers.me/CIE/International%20A%20And%20AS%20Level/9702%20-%20Physics/9702_w08_qp_1.pdf27.
Question 27 , 29
The Frequency of Microwave is 1*1010 if you have got the graph then see in that
so the answer is C
29.
as the field strength is moving from positive to negative that is from center to outwards
so the and X is more near to the Center so Magnitude of force on X is Greater Than on Y
and radially inwards as explained
so the answer is B
Post by: NidZ- Hero on June 07, 2011, 06:55:24 pm
11.
Equal and Opposite Force Exerted on Road by the wheel so the answer is B
Y not A
Post by: Ghost Of Highbury on June 07, 2011, 08:02:35 pm
Post by: EMO123 on June 07, 2011, 08:03:57 pm
I'm off, good luck for tomorrow guyz. Hope you all do well. :Dthanxx and good luck to all of us
Post by: physichemaths on June 07, 2011, 09:06:35 pm
Post by: SkyPilotage on June 07, 2011, 09:32:11 pm
Hai guys. How to get the answers for nov08 ques 1?Speed of light = 3x10^8 ms^-1
So Number of wavelength/second = frequency= speed/wavelength..
Post by: EMO123 on June 15, 2011, 01:53:23 pm
Paper 04
Q-2 c)
Q-3 b)
Q-4
more doubts coming from me soon
Post by: $H00t!N& $t@r on August 28, 2011, 12:10:25 pm
1) Two satellites orbit a planet of mass 'M' with Time periods of 3 days and 4.5 days, respectively. If the orbital radius of the first satellite is 2x10^6 m, determine the radius of orbit of the other. (Answer is 2.62x10^6)
2) A mass 'm' is moved from the North pole towards the equator. If the Earth is assumed to be a perfect sphere of radius 'R' , and if the mass moved is exactly 60kg, determine the difference in weight when measured at the North pole and at the equator.
R = 6.4x10^6 m
M of Earth = 6.0x10^24 kg
G = 6.67x10-11
m = 60 kg
( Answer is 2.0 N )
3) An astronaut travels from the Earth towards the moon where he experiences no net force. Determine the distance from the surface of the moon where this phenomenon occurs. Given that,
- the radius of the moon is 1.74x10^6 m
- mass of the moon is 0.0735x10^24 kg
(Answer is 3.71x10^7 m )
Please help me out.. i desperately need the explanation for these quesitons...
Post by: tmisterr on August 28, 2011, 03:22:34 pm
Please help me out with the following questions:
1) Two satellites orbit a planet of mass 'M' with Time periods of 3 days and 4.5 days, respectively. If the orbital radius of the first satellite is 2x10^6 m, determine the radius of orbit of the other. (Answer is 2.62x10^6)
2) A mass 'm' is moved from the North pole towards the equator. If the Earth is assumed to be a perfect sphere of radius 'R' , and if the mass moved is exactly 60kg, determine the difference in weight when measured at the North pole and at the equator. ( Answer is 2.0 N )
3) An astronaut travels from the Earth towards the moon where he experiences no net force. Determine the distance from the surface of the moon where this phenomenon occurs. Given that,
- the radius of the moon is 1.74x10^6 m
- mass of the moon is 0.0735x10^24 kg
(Answer is 3.71x10^7 m )
Please help me out.. i desperately need the explanation for these quesitons...
hello :)
1. gravitational force = centripetal force. once we know that...we have solved the question.
GMm/r2=mv2/r
where M is mass of planet and m is mass of satellite. since they orbit the same planet, M is equal in both cases.
now we know that v= 2*pi*r/T where 2*pi*r is the circumference of orbit and T is the periodic time (easy, velocity=distance/speed). so next we play around with the above equation... It is not difficult to see that
M= 4*pi2*r3/GT2
Voila...so now we plug in the data values of planet 1, converting days into seconds and the value of gravitational constant G (6.67x10-11) to get the value of M as 7.04....x1019
Ok so now we play around with the equation for M to make r the subject of the formula...easy enough.
We know T for the second planet is 4.5 days (don't forget to convert to 4.5*24*3600 seconds) and we now know that M is 7.04...x1019. simple arithmetic............r is (to be exact lol) 2620741.394 m but due to rules of accuracy we quote it to 3 s.f hence 2.62x106[/sup
Post by: tmisterr on August 28, 2011, 03:31:51 pm
all we need to see is this:
M=4*pi2*r3/GT2
and since M is the same for both planets, we can replace this equation directly into the equation of R (radius of second planet)
to get R=cube root( T2r3/t2)
try deriving for yourself
where T is the period of planet 2 (4.5 days converted to seconds), r is the radius of planet 1 (2x106) and t is the period of planet 1 (3 days converted to seconds)
I'm sure you'll find this saves a lot of time :)
Post by: $H00t!N& $t@r on August 28, 2011, 04:33:09 pm
Post by: $H00t!N& $t@r on August 30, 2011, 09:48:19 am
Post by: Master_Key on August 30, 2011, 01:55:42 pm
Please help me out with the following questions:2)
1) Two satellites orbit a planet of mass 'M' with Time periods of 3 days and 4.5 days, respectively. If the orbital radius of the first satellite is 2x10^6 m, determine the radius of orbit of the other. (Answer is 2.62x10^6)
2) A mass 'm' is moved from the North pole towards the equator. If the Earth is assumed to be a perfect sphere of radius 'R' , and if the mass moved is exactly 60kg, determine the difference in weight when measured at the North pole and at the equator.
R = 6.4x10^6 m
M of Earth = 6.0x10^24 kg
G = 6.67x10-11
m = 60 kg
( Answer is 2.0 N )
3) An astronaut travels from the Earth towards the moon where he experiences no net force. Determine the distance from the surface of the moon where this phenomenon occurs. Given that,
- the radius of the moon is 1.74x10^6 m
- mass of the moon is 0.0735x10^24 kg
(Answer is 3.71x10^7 m )
Please help me out.. i desperately need the explanation for these quesitons...
I didn't get the exact answer but i tried.
F = GM/r2
F = 6.67x10-11 * 6x1024 / (6.4x106)2
F = 9.77/kg
F = 586.2
Gravitational field = 9.8
Weight = 60*9.8 = 588 N.
Diff = 588-586.2 = 1.8 N.
The question mentioned Earth to be perfect sphere, when you do with the actual values you may get the exact answer. I tried and got this.
Post by: tmisterr on August 31, 2011, 08:52:47 am
Please help me out with the following questions:
1) Two satellites orbit a planet of mass 'M' with Time periods of 3 days and 4.5 days, respectively. If the orbital radius of the first satellite is 2x10^6 m, determine the radius of orbit of the other. (Answer is 2.62x10^6)
2) A mass 'm' is moved from the North pole towards the equator. If the Earth is assumed to be a perfect sphere of radius 'R' , and if the mass moved is exactly 60kg, determine the difference in weight when measured at the North pole and at the equator.
R = 6.4x10^6 m
M of Earth = 6.0x10^24 kg
G = 6.67x10-11
m = 60 kg
( Answer is 2.0 N )
3) An astronaut travels from the Earth towards the moon where he experiences no net force. Determine the distance from the surface of the moon where this phenomenon occurs. Given that,
- the radius of the moon is 1.74x10^6 m
- mass of the moon is 0.0735x10^24 kg
(Answer is 3.71x10^7 m )
Please help me out.. i desperately need the explanation for these quesitons...
3.
ok so the only way I see for solving this question is using knowledge not included in this question....because the radius and mass of moon are not enough (in my opinion). We know that the moon orbits the earth in 28 days, and that the mass of the earth (from question 2) is 6x1024.
We now have to find the distance between the earth and the moon, we shall assume that the earth and the moon are perfect spheres with all their mass at the cores and that the orbit is a perfect circle so that we can use circular motion and theory of gravitation
G*Me*Mm/r2=Mm*v2/r
here Me is mass of earth and Mm is mass of moon.
from my explanation of question 1, you should then see that r(distance between centers of earth and moon)=cube root(G*Me*T2/4*pi2). Replacing T as 28 days (converted to seconds, 28*24*3600), we get r to be 3.9*108m.
So now we have enough information to solve the question.
since there is no force on the astronaut, he is being pulled with equal but opposite forces from the earth and the moon.
so G*Me*a/R2=G*Mm*a/r*2
here I use a to be the mass of the astronaut, R to the the distance from the earth and r to be the distance from the moon,
From the first part of calculations, we see that the distance from the earth to the moon is about 3.9x108m,
let say the astronaut is x m from the center of the moon, then he will be (3.9x108-x)m from the earth, therefore
Me/(3.9x108-x)2=Mm/x2
ok so simplify the equation,
square root (Me/Mm)=3.9x108-x/x
solve for x to get x=38863670
but we must remember that this is the distance from the core of the moon. we must subtract the radius of the moon to get the distance from the surface of the moon.
Therefore, 38863670-1.74*106=37123670=3.17*107m
If I think of away of solving the question without adding in the extra data I will post it, but I hope this gives you an idea of what is required.
Post by: Ghost Of Highbury on October 02, 2011, 11:54:03 am
Nov 08 p4 No 1
a)(iii) The normal reaction exerted by a planet on a mass is equal to the weight of the mass.
Normally for an object undergoing circular motion,
Resultant towards centre = Centripetal force
Resultant force towards centre of the planet will be : Gravitational force - Weight
Hence
Gravitational force - Weight = Centripetal force ---> Weight = GMm/R2 - mR2
Since Weight = Normal reaction = GMm/R2 - mR
Jun 07 p4 No 1
b)(i) Gravitational potential energy is given by the formula = -GMm/R
In the question you are asked to find the change in gravitational potential.
Let ^Ep = Change in gravitational potential energy
Uf = Final gravitational potential energy = -GMm/3R
Ui = Initial gravitational potential energy = -GMm/2R
Hence ^Ep = Uf - Ui
^Ep = (-GMm/3R) - (-GMm/2R)
Simplify this equation and you'll get your required solution which is ^Ep = GMm/6R
Why is gravitational force - weight = centripetal force??
Post by: astarmathsandphysics on October 02, 2011, 10:33:07 pm
31.F=qE=(5*1.6*10^(-19)*5000/0.008 not E=electric field =V/d
Post by: ashwinkandel on October 05, 2011, 03:11:48 pm
1.Calculate the total charge passing through the electrolyte in this time
Post by: astarmathsandphysics on October 05, 2011, 03:30:11 pm
the above multiples the number of ions moved by the charge on each ion - note that each Cu^2+ has a charge of 2e
Post by: ashwinkandel on October 05, 2011, 04:26:39 pm
i. Calculate the current it will take from the mains supply. =>P=VI=>I=41.3A
ii. Suggest why the shower requires a separate circuit from the other appliances.
iii. Suggest a suitable current rating for the fuse in the circuit.
I need help for part ii. and iii.
Post by: astarmathsandphysics on October 06, 2011, 11:03:20 am
iii. Suggest 50A. Smaller and it might blow too often.
Post by: ashwinkandel on October 07, 2011, 08:21:32 am
1.
(http://i56.tinypic.com/2a6rnzs.jpg)
i.Use Kirchoff's first law to find the current through the 4.00 ohm and 8.00 ohm resistors
ii. Calculate the emf E1
iii. Calculat the emf E2
iv. Calculate the current through the 12.00 ohm resistor
2.
A student builds the circuit in the diagram, using a battery of negligible internal resistance. The reading on the voltmeter is 9.0V.
(http://i55.tinypic.com/ic5vgi.jpg)
i.The voltmeter has a resistance of 1200 ohm. calculate the emf of the battery.
ii. The student now repeats the experiment using a voltmeter of resistance 12 Kiloohm. Show that the reading on this voltmeter would be 9.5V
iii. Refer to your answers in i. and ii. and explain why a voltmeter should have as high a resistance as possible.
Post by: astarmathsandphysics on October 07, 2011, 09:29:08 am
Post by: astarmathsandphysics on October 07, 2011, 09:43:10 am
My printer is out of toner and I cant't see the picture or print it.
Post by: ashwinkandel on October 07, 2011, 09:53:35 am
Post by: astarmathsandphysics on October 07, 2011, 10:51:21 am
Post by: astarmathsandphysics on October 07, 2011, 10:53:48 am
Post by: ashwinkandel on October 07, 2011, 11:21:37 am
Post by: astarmathsandphysics on October 07, 2011, 11:24:01 am
Where is the voltmenter reading to answer this question>?
Post by: ashwinkandel on October 07, 2011, 11:33:31 am
Post by: astarmathsandphysics on October 07, 2011, 12:12:39 pm
Post by: astarmathsandphysics on October 07, 2011, 12:52:07 pm
Post by: ashwinkandel on October 07, 2011, 01:16:08 pm
Post by: astarmathsandphysics on October 07, 2011, 09:29:39 pm
A voltmeter is just a very high resistance ammeter, scaled to read voltages. Both ammeter and voltmenter actually detect the current passing through them. The voltmeter is scaled for voltage using V=IR, and since it is always in parallel, it actually gives the voltage across that paralleled component.
Post by: ashwinkandel on October 08, 2011, 03:45:07 am
Post by: ashwinkandel on October 08, 2011, 07:02:36 am
a. Calculate the thickness of the piece of silicon.[3]
b. Calculate the current which would pass through the silicon if a potential difference of 12V were applied across P and Q.[2]
c. Discuss how the current would change if it were large enough to cause the silicon to become significantly warmer.[3]
Post by: ashwinkandel on October 08, 2011, 07:11:06 am
a.
i. Sketch the graph to show how the current through the diode would vary as the voltage across it is increased from 0 to 1.0 V [1]
ii.The supply is now connected in the reverse direction and once more the potential difference across the diode is increased from 0V to 1.0V. Complete the I-V graph . [1]
b.Suggest why the safety resistor is required.[2]
c.When the potential difference across the safety resistor is 1.4V the current through it is 20mA. Calculate the resistance of the safety resistor. [2]
Post by: astarmathsandphysics on October 08, 2011, 08:10:20 am
Post by: astarmathsandphysics on October 10, 2011, 08:54:03 am
Post by: ashwinkandel on October 14, 2011, 05:57:49 am
In order to switch on a heater when the temperature falls below a set value a potential divider is connected to a switching circuit. When the input voltage to the switching circuit falls below 0.5 V it switches the heater on.
a. Copy the Fig 13.16 and add a suitable potential divider circuit to trigger the switching circuit.
b. Explain how the operator could lower the temperature at which the heater is switched on.
Post by: ashwinkandel on October 14, 2011, 06:22:31 am
A student is asked to compare the emf.s of a standard cell and test cell. He sets up the circuit as shown in Fig. 13.18 using the test cell.
i. He achieves balance when the distance AB is 22.5 cm. He repeats the experiment with the standard cell of emf of 1.434 V. The balance point using this cell is at 34.6 cm. Calculate the emf of the cell
Post by: ashwinkandel on October 14, 2011, 06:25:12 am
Post by: ashwinkandel on October 14, 2011, 06:37:11 am
a. Explain one advantage circuit 1 has over circuit 2. [2]
b. One problem with circuit 1 is that there is always a current in the resistor. Explain how this problem could be rectified. [2]
c.
i. The lamp is rated at 60W at 240V. Calculate the resistance of the lamp filament at its normal operating temperature. [2]
ii. State and explain how the resistance of the filament at room temperature would compare with the value calculated in c i. [2]
Post by: astarmathsandphysics on October 14, 2011, 08:14:30 am
Post by: astarmathsandphysics on October 14, 2011, 09:05:24 am
b) put a diode between the resistor and the bottom junction
c)P=V^2/R
60=240^2/R so R=240^2/60=960 ohms
It will be lower since resistance decreases with decreasing temp
Post by: ashwinkandel on October 14, 2011, 09:39:44 am
Post by: astarmathsandphysics on October 16, 2011, 09:42:53 pm
Post by: astarmathsandphysics on October 16, 2011, 10:08:42 pm
Post by: cs on October 17, 2011, 12:33:43 pm
and the next attachment, i dont get why is Krypton-90 placed after "S"
and need explanation for the 3rd attachment,
what is the "proper" working for attachment 4
thank you :)
Post by: ashwinkandel on October 18, 2011, 04:48:54 pm
b. Explain why the measured voltage falls.[2]
c.i. Calculate the current in the circuit.[2]
ii. Calculate the internal resistance of the cell.[2]
iii. State any assumptions you made in your calculations.[2]
Post by: astarmathsandphysics on October 18, 2011, 10:27:26 pm
Post by: astarmathsandphysics on October 19, 2011, 09:00:17 am
Post by: astarmathsandphysics on October 19, 2011, 09:13:24 am
Post by: cs on October 19, 2011, 09:55:23 am
another doubt i have, part c
Post by: ashwinkandel on October 19, 2011, 12:08:53 pm
Q1. Figure 13.17 shows a circuit used to monitor the variation of light intensity in a room.
a. Identify the component X and describe how the circuit works.
b. Suggest the reason for including the variable resistor in the circuit.
Q2. The diagram shows a potential divider. The battery has negligible internal resistance and the voltmeter has infinite resistance.
a. State and explain how the reading on the voltmeter will change when the resistance of the variable resistor is increased.
b.Resistor R2 has a resistance of 470 ohm. Calculate the value of the variable resistor when the reading on the voltmeter is 2.0V
c.The voltmeter is now replaced with one of resistance 2Kohm. Calculate the reading on this voltmeter.
d. R2 is replaced with a light dependent resistor. Explain how the reading on the voltmeter changes as the light levels are reduced.
Waves
Q1.The diagram shows the traces of two sound waves displayed on an oscilloscope screen.
i. Explain whether or not the waves are in phase.
P.S: When are waves said to be in phase?
ii. State how their wavelength compare.
iii. Calculate the ratio of the intensities of the two waves
Post by: astarmathsandphysics on October 19, 2011, 02:42:13 pm
Post by: astarmathsandphysics on October 19, 2011, 04:12:23 pm
Post by: ashwinkandel on October 19, 2011, 04:20:50 pm
Post by: astarmathsandphysics on October 19, 2011, 04:55:01 pm
Post by: cs on October 20, 2011, 03:49:10 pm
Post by: NidZ- Hero on October 20, 2011, 06:46:47 pm
1(B)
How is it 25 cm??
Post by: astarmathsandphysics on October 20, 2011, 07:00:54 pm
Post by: astarmathsandphysics on October 20, 2011, 07:07:28 pm
Post by: cs on October 21, 2011, 02:08:10 am
here
thank you sir :)
Post by: cs on October 21, 2011, 06:11:03 am
Post by: astarmathsandphysics on October 21, 2011, 11:23:41 am
Post by: ashwinkandel on October 22, 2011, 06:32:24 am
Q1.Explain why in the remote mountainous regions, such as the Hindu Kush, radio signals from terrestrial transmitters can be received, but television reception can only be received from satellite transmission.
Q2.Damita and Jamal are organizing a disco. Damita suggests that feeding the sound from the music center to a second loudspeaker will increase the loudness of the music. Jamal says it won't work as there will be places where the sound will be loud, due to constructive interference, and places where it will be much quieter due to destructive interference. State who is correct and explain your working.
Post by: cs on October 22, 2011, 07:38:54 am
-weather monitoring
-communication
is it the same for both of them?
Post by: astarmathsandphysics on October 22, 2011, 07:52:33 am
Post by: astarmathsandphysics on October 22, 2011, 07:57:46 am
Q2.I think the second is correct. both types of interference will occur alongs lines starting midway between the sources, but if they are placed next to each other they will act more like a single loudspeaker and this will be reduced.
Post by: astarmathsandphysics on October 22, 2011, 08:05:25 am
Post by: minicooper on October 26, 2011, 08:55:28 am
AS Physics doubts:
What do I need to know about viscuous drag for AS level? What is its relation to upthrust and weight and is it equal to air resistance?
How do you derive the equation E=1/2mv*2 (I don't know how to write v squared here)?
Also, how do you "Derive, using Kirchhoff's laws, a formula for the combined
resistance of two or more resistors in series and in parallel"?
Thanks, would really appreciate it. :)
Post by: astarmathsandphysics on October 26, 2011, 09:55:52 pm
It depends on the board, but you may need to know Stoke's law.
Drag=resistance
upthrust=weight for a floating body - this is Archimedes principle
F=ma=m(dv/dt)=m(dx/dt)(dv/dx)=mv(dv/dx)
hence Fdx=mvdv
integrate both sides to get Fx = mv^2/2
For Kirchoff's laws the proof is here
http://www.astarmathsandphysics.com/a_level_physics_notes/electricity/a_level_physics_notes_resistivity_and_resistance.html
Post by: minicooper on October 28, 2011, 04:53:50 pm
Post by: astarmathsandphysics on October 28, 2011, 05:21:54 pm
Post by: cs on October 31, 2011, 11:35:39 am
Post by: astarmathsandphysics on October 31, 2011, 03:48:31 pm
Post by: astarmathsandphysics on October 31, 2011, 10:19:25 pm
Post by: ashwinkandel on November 02, 2011, 09:37:35 am
Q.
Figure 1 shows the set up used to analyse the spectrum of a sodium discharge lamp using a diffraction grating with 5000 lines/cm . Figure 2 shows the developed photographic film showing the spectral lines observed.
Look Fig 1 and 2 and answer
i. Explain why two spectra are observed.
ii. Study the two spectra and describe two differences between them.
iii. The green maximum near end A is at an angle of 15.5 degree. Calculate the wavelength of green light.
iv. Calculate the angle produced by second green light.[stream=475,325][/stream]
Post by: astarmathsandphysics on November 02, 2011, 12:34:46 pm
For each wavelength multiple spectra are produced n=1.2.3 etc
ii) one is more spread out than the other. The spectrum on the left is brighter than the one on the right.
iii)sin theta=n*wavelength/a
sin 15.5=1*wavelength/(0.01/5000) so wavelength=sin (15.5)*0.01/5000=5.34*10^-7 m
iv)sin theta = 2*5.34*10^-7 /(0.01/5000) =0.53 so theta =sin^-1 0.53 =32.3 degrees
Post by: ashwinkandel on November 02, 2011, 02:21:34 pm
Post by: astarmathsandphysics on November 02, 2011, 03:36:58 pm
Post by: ashwinkandel on November 05, 2011, 12:44:49 pm
Post by: SZM on November 05, 2011, 12:59:16 pm
here ashwinkandel, since u r doing alevel cambridge. i am sure, u must have small experience towards alevel stuffs. U r from where?? I am from Sri-Lankan.
hve u heard anything about tmisterr and MKH?? Pls let me know if u hve heard about them??? becoz after so many days, i havent seen their online/// why is dat??? ow cn i contact them?? becoz i found their helpful really useful, thaty. and i need to discuss with them some small sort of problem. So anyone pls let me know what has happened these two members?? I was looking or their online. but unfortunately they r offlined?? pls tel me , ow to contact them?
I am doing alevel edexcel privately not schooling, so pls help me.
i thought of doing self study, but i am not studying anything, becoz i am thinking all the time, how to start and where to start and how to speed up, which is said to be nonsense. I am being really stupid . But i like to self study, but it doesnt come. I have potential to self study, but it still doesnt work or doesnt come naturrally. i dont know why?
I have read different thread, where some member has mentioned, that to do exam well and get best grade, answering techniques are very very important and we must improve that very well. Then i thought, if i do self study and after finish studying, whn i do pastpaper, then i will go thru markschemes and then mark . But that is not enough, becoz marking scheme gives only point form. SO i hve to go for classes to ask teacher how to start the answer and how to end up and to start and end, there must be the meaning and there should be a logical sequence. So these kind of style of writing is very important in the exam. Am i correct??? Teachers always tel us, that the best answers are short and precise. wat does that mean?? why cant i write long, if i want to explain further, wiil the examiner get angry or wat?? how to improve that?? for example the suggest questions, i cannot write short and precise, i will write a bit more// is that gud or bad??? pls tel me dear. i need ur reply tho this matter.
1.My mother told me it is not necessary to attend the class ALWAYS... She told me at the beginning, it is a must to attend, after some few classes, like each subject 3 /4 times attending classes, u will get the potential to do self study after having idea from the tutor how to manage with//// Is dat true or will that work>>>
2.my idea is , i collect the concepts , that i found difficult to understand in all units and then whn i go for class, i cn ask the tutor to explain in details, the doubts i have inorder to clear them within 2 and half hours. wat do u think??
3.or should i attend the class from the beginning till i complete the syllabus. till dat i have to attend the classes. is dat so or not? pls tel me.. I am worried, becoz with difficulty i found the classes and make the time. afterall if i dont attend, then they wont take me ,whn i need to go for classes. SO i dont want to waste the time and i dont want to get stuck at the last moment or last minute. I should be very careful. so therefore pls tel me ur suggestion to this matter... I am really worried about that. Thats all i want from ur suggestion .pls help me dear.
But i must not waste the time even for a minute.
waitn for ur reply. need ur reply urgenlty.
Post by: ashwinkandel on November 05, 2011, 01:23:47 pm
Post by: SZM on November 05, 2011, 01:31:45 pm
By the way nice to hear about u. When are u going to do final exam ???? Both AS and A2????
I just want from ur suggestions, is self study suitable for alevel???? I mean 75 percent self study and 25 percent from teachers Help..
wat do u think??
need ur reply urgrnlty
Post by: ashwinkandel on November 05, 2011, 01:39:34 pm
Post by: SZM on November 05, 2011, 01:47:09 pm
wat do u think??/
here, u said u r giving AS exam now, which means u will be doing exam in october??? am i correct??? wat do u mean by 4 given up 6 remaining??? cant understand very well????
pls explain me in details
need ur reply urgently
Post by: SZM on November 05, 2011, 01:54:54 pm
here, u said, Self study is perfectly suitable for theory part. Ow cn u say??? r u sure about that 100 percent???
dear tel me ow to achieve A* for math, phy, chem and bio??? wat is the successful method to achieve this grade???/
which materials should i start with??? ow to cover up everything, becoz i will be doing both AS and A2 exam in May.. So i wil have 5 months more.
i think i have t spend 3 months to study the concepts in all units well along with some questions and another 2 months , i ahve to spend to do loads of pastpaper// ow wat should i do????
Post by: ashwinkandel on November 05, 2011, 05:02:58 pm
Post by: SZM on November 05, 2011, 05:12:42 pm
waitin for ur reply
Post by: SZM on November 05, 2011, 05:24:40 pm
if dat is gud, cn i follow for other subjects such as phy,chem and bio??? will that method suitable??
need ur reply urgrntly
Post by: astarmathsandphysics on November 06, 2011, 09:39:42 am
Post by: ashwinkandel on November 06, 2011, 11:20:22 am
Post by: astarmathsandphysics on November 06, 2011, 08:15:39 pm
(http://diffractiongrating.net/wp-content/uploads/2011/10/what-is-diffraction-grating.jpg)
within each bright spot the cours spread out tp form a rainbow which is why you get different coloured fringes.
Neaer the edges, since we are projecting onto a flat surface, the image spreads out.
Post by: ashwinkandel on November 12, 2011, 04:18:50 am
Q.Approximate values for the radius of a gold atom and the radius of a gold nucleus are 10^-10 m and 10^-15 m respectively.
a. Estimate the ratio of volume of a gold atom to the volume of a gold nucleus. [2]
b. The density of gold is 19000 kg/m^3. Estimate the density of a gold nucleus stating any assumptions that you make in your answers.[3]
Post by: ashwinkandel on November 12, 2011, 06:28:10 am
Questions from States of Matter:
1.
a. Explain how a manometer may be adapted to measure large pressures.
b. Explain why the pressure measured by manometer does not depend on the cross-sectional area of the tube used in the manometer.
2. A boy stands on a platform of area 0.050 m^2 and a manometer measures the pressure created in a flexible plastic container by the weight W of the boy. The density of water is 1000 kg/m^3/ Determine
i. the pressure difference between the inside of the plastic container and the atmosphere outside. [2]
ii. the weight W of the boy. [2]
3.The diagram shows a U-tube, open at both ends, which contains two different liquids X and Y that do not mix. The numbers on the metre rule are distances in centimetres. The density of liquid Y is 800 kg/m^3
i.Explain how diagram shows that liquid Y has a greater density than liquid X [2]
ii. Calculate the density of X. [3]
iii. Explain why the pressure in the U-tube is the same on the both sides of the manometer at the level L. [1]
iv. Calculate the pressure caused by liquid in the U-tube at level L. [2]
Post by: astarmathsandphysics on November 12, 2011, 09:44:14 pm
Post by: astarmathsandphysics on November 12, 2011, 10:04:22 pm
Post by: ashwinkandel on November 17, 2011, 11:53:38 am
Post by: astarmathsandphysics on November 17, 2011, 02:10:29 pm
Post by: astarmathsandphysics on November 17, 2011, 02:12:21 pm
Post by: astarmathsandphysics on November 17, 2011, 02:14:31 pm
time =distance/speed=4000/16.67= 240 seconds=4 minutes
Post by: astarmathsandphysics on November 17, 2011, 02:17:31 pm
Post by: astarmathsandphysics on November 17, 2011, 02:21:48 pm
number wavelengths in 1m is 1/(500*10^-9} =2*10^6
Post by: ashwinkandel on November 18, 2011, 10:47:57 am
Only one question today
Post by: astarmathsandphysics on November 18, 2011, 12:18:09 pm
Post by: ashwinkandel on November 20, 2011, 01:18:13 pm
Post by: ashwinkandel on November 20, 2011, 01:25:53 pm
Post by: astarmathsandphysics on November 20, 2011, 11:06:44 pm
For second question apply F=ma to each mass
For 2 kg 2g-T=2a
for 8 kg mass
T-6=8a
Add these to get 2g-6=10a soa=(2g-6)/10=1.4m/s^2
for third question use paralel resistor formula 1/R=1/10+1/10+1/10+1/10+1/10+1/10+1/100=0.61 so R=1/0.61 =1.6 ohms
4th quest P=IV*0.8
4000000=25000*I*0.8 so I=200A
Post by: astarmathsandphysics on November 20, 2011, 11:18:32 pm
sin theta =n*wavelength/d
d=1/300000, the width of 1 slit so sin theta =n*450*10^-9/(1/300000) =n*0.135
sin theta is at most 1 so n<=1/0.135=7.4 but since n is whole number it can be at ost 7
6th question F*0.6+F*0.6=900*0.2
F=150N
7th question half the water falls by h/2 so loss in gpe =mgh/4
8th question The excess pressure is rhog(h_2-h_1)=rho g (2h)
9th question all kinetic energy of electron is changed into potential energy of the field 1/2 mv^2 =qV=qEx so x=(1/2 mv^2)/(qE)
Post by: angel_ak on November 21, 2011, 11:17:08 am
C = aT + bT3
(a) What is the unit of C?
(b) What are the units of a and b?
i found the units of C.... kgm2s-2K-1
but how to find that of a and b?
i know its easy but silly me :(
Post by: angel_ak on November 21, 2011, 01:35:22 pm
i did part a..
but din understand part b at all :(
Post by: ashwinkandel on November 21, 2011, 01:37:55 pm
If the equation is valid then it is homogeneous. That means every terms should have same unit.
so unit of at= unit of c which implies that unit of a is unit of c divided by unit of T
Similarly, Find the unit for b
Post by: angel_ak on November 21, 2011, 01:43:30 pm
that was quite simple.. i don know why i din come up wid the right sol :'(
Post by: astarmathsandphysics on November 21, 2011, 02:19:53 pm
Post by: angel_ak on November 21, 2011, 02:31:04 pm
A. 9 x 10-8 m3
B. 8.8247 x 10-8 m3
C. 8.825 x 10-8 m3
D. 8.8 x 10-8 m3
Post by: ashwinkandel on November 21, 2011, 02:37:32 pm
Post by: astarmathsandphysics on November 21, 2011, 02:47:47 pm
Post by: angel_ak on November 21, 2011, 06:23:05 pm
Post by: astarmathsandphysics on November 21, 2011, 10:38:23 pm
r=1.06/2=0.53mm
V=pi r^2 l= 3.14*(0.53*10^-3)^2*0.1=8.82*10^-8 m^3
Since the diameter is +-about 2.5% take only 1 sig fig A
Post by: silvercameron on November 22, 2011, 06:01:44 pm
Post by: angel_ak on November 22, 2011, 06:09:53 pm
but m confused why we are considering the ans only in 1 sf? :s
Post by: angel_ak on November 22, 2011, 06:13:04 pm
but the book says ans is D ie 8.8 x 10-3
Post by: astarmathsandphysics on November 22, 2011, 08:15:24 pm
Post by: silvercameron on November 22, 2011, 10:02:17 pm
Post by: astarmathsandphysics on November 22, 2011, 10:57:05 pm
Post by: silvercameron on November 23, 2011, 08:17:53 pm
I tried this today and did it like this:
resistance of aluminium = resistance of copper.
So : (rho*l)/A of aluminium= (rho*l)/ A of copper.
Radius of aluminium = 1.2/2 = 0.6 mm = 6*10-4 m
Area of aluminium= pi* r2 = pi* (6*10-4)2 = 1.131*10-6
therfore : ((2.6*10-8)*l)/(1.13*10-6)=((1.7*10-8)*l)/pi*r2
l and l gets canceled cos they are the same.
(2.6*10-8)/(1.131*10-6)= (1.7*10-8)/(pi*r2)
we cross-multiply, get the radius, multiply it by 2 and finally get diameter= 9.72*10 -4 m
^^is my method correct?
Post by: astarmathsandphysics on November 23, 2011, 09:45:50 pm
Post by: ashwinkandel on December 05, 2011, 02:05:46 pm
1.
Figure shows a wire XY which carries a direct current. Plotting compass R, placed alongside the wire, points due north. Compass P is placed below the wire and compass Q is placed above the wire.
a. State the direction of the current in the wire.
b. State in which direction compass P points
c. State in wihch direction compass Q points if the current in the wire is reversed.
2.
At a point on the Earth's surface the horizontal component of the Earth's magnetic field is 1.6*10^-5 T. A piece of wire 3.0 m long and of weight 0.020N lies in the east-west direction on a labortary bench. When a large current flows in the wire, the wire just lifts off the surface of the bench.
i. State the direction of the current in the wire.[1]
ii. Calculate the minimum current needed to lifet the wire from the bench. [3]
Post by: astarmathsandphysics on December 05, 2011, 02:40:46 pm
b. same as Q
c.North West
2. by flemings left hand rule current is from west to east
ii. F=BIl
0.02=1.6*10^-5*I*3 so I=0.02/(1.6*10^-5*3)417A
Post by: ashwinkandel on December 05, 2011, 03:02:39 pm
Post by: astarmathsandphysics on December 06, 2011, 12:30:11 pm
Post by: ashwinkandel on December 06, 2011, 01:48:47 pm
Q1.
Explain why the toy falls off when the speed of the turntable is increased.
Q2.
A teacher swings a bucket of water, of total mass 5.4 kg, round in a vertical circle of diameter 1.8m. [3]
i.Calculate the minimum speed which the bucket must be swung at so that the water remains in the bucket at the top of the circle. [2]
ii. Assuming that the speed remains constant, what will be the force on the teacher's hand when the bucket is at the bottom of the circle? [1]
Q3.
In a training, military pilots are given various tests. One test puts them in a seat on the end of a large arm which is then spun round at high speed, as shown in the diagram.
i. Describe what the pilot will feel and relate this to the centripetal force.[3]
ii. at top speed the pilot will experience a centripetal force equivalent to six times his own weight (6 mg)
a. Calculate the speed of pilot in this test. [3]
b. Calculate the number of revolutions of the pilot per minute. [2]
iii. Suggest why it is necessary for pilots to be able to be able to withstand forces of this types [2]
Q4.
The diagram shows a centrifuge used to separate solid particles suspended in a liquid of lower density.
a.The container is spun at a rate of 540 revolutions per minute
i.Calculate the angular velocity of the container. [2]
ii. Calculate the centripetal force on the particle of mass 20mg at the end of the test tube. [2]
b. An alternative methods of separating the particles from the liquid is to allow them to settle to the bottom of a stationary container under gravity. By comparing the forces involved, explain why the centrifuge is more effective method of separating the mixture. [2]
The diagram for questions 1 , 3 and 4 are attached below.
Post by: astarmathsandphysics on December 06, 2011, 08:10:54 pm
The centrifugal force is F=mv^2/r
When mv^2/r >mu mg the toy starts to slip
Post by: astarmathsandphysics on December 06, 2011, 08:33:06 pm
Post by: Naruto123 on December 10, 2011, 05:49:20 am
b) Calculate the displacement at which the energy of the particle has equal amounts of kinetic and potential energies
thanku in advance :)
Post by: astarmathsandphysics on December 10, 2011, 07:44:52 pm
x=Acos wt so x/A=cos wt=sqrt(1-sin^2 wt)=sqrt(1-(1/2)^2)=(sqrt(3))/2
b)1/2mv^2=1/2msin^2(wt)=1/2kx^2=1/2kcos^2 wt
tan^2 wt =k/m so tan wt =sqrt(k/m) and sec^2 wt=1/(cos^2 wt)=1+tan^2 wt =1+k/m so cos^2 wt=1/(1+k/m)
x=1/2k cos wt=1/2 k sqrt(1/(1+k/m))
Post by: ashwinkandel on December 12, 2011, 01:26:18 pm
1.
The diagram shows the path of an electron as it travels in air. The electron rotates clockwise around a uniform magnetic field into the plane of the paper, but the radius of the orbit decreases in size.
a i. Explain the origin of the force that causes the electron to spiral in this manner.[2]
ii. Explain why the radius of the ircle gradually decreases. [2]
b. At one point in the path, the speed of the electron is 1.0*10^7 m/s and the magnetic flux density is 0.25T.Calculate:
i.the force on an electron at this point due to the magnetic field.[2]
ii. the radius of the path at this point. [2]
2.
The diagram shows an arrangement to deflect protons from a source to a detector using a magnetic field. A uniform magnetic field exists only within the area shown. Protons move from the source to the detector in the plane of the paper.
a
i.Copy the diagram and sketch the path ofa proton from the source to the detector.
Draw an arrow at two points on the path to show teh direction of the force on the proton produced by the magnetic field. [3]
ii.State the direction of the magnetic field within the area shown.[1]
b.The speed of a proton as it enters the magnetic field is 4.0*10^6 m/s. The magnetic flux density is 0.25T. Calculate:
i. the magnitude of the force on the proton caused by the magnetic field. [1]
ii. the radius of curvature of the path of the proton in the magnetic field. [2]
c.Two changes to the magnetic field in the area shown are made. These changes allows an electron with the same speed as the proton to be deflected along the same path as the prton. State the two changes made. [2]
3.
The diagram shows an electron tube. Electrons emitted from the cathode accelerate towards the anode and then pass into a uniform electric field created by two oppositely charged parallel metal plates.
a i.Explain why the beam curves upward. [2]
ii. Explain how the pattern formed on the fluorescent screen shows that all the electrons have the same speed as they leave the anode.[2]
b. Write down an equation relating the speed of the electrons v to the potential difference V(ac) between the anode and the cathode.[1]
c. The deflection of the beam upward can be cancelled by applying a suitable uniform magneti field in the space between the parallel plates.
i.State the direction of magnetic field for this to happen. [1]
ii. Write down an equation relating the speed of the electrons, v , the electri field E that exists between the plates and the magnetic flux density B needed to make the electrons pass undeflected between the plates. [2]
iii. Calculate the value of B required, using the appratus shown in the diagram, given that the specific charge on an electron e/m is 1.76*10^11 C/kg [2]
4.Protons and helium nuclei from the Sun pass into the Earth's atmosphere above the poles, where the magnetic flux density is 6.0*10^-5 T. The particles are moving at a speed of 1.0*10^6 at right angles to the magnetic field in this region. The magnetic field can be assumed to be uniform.
a.Calculate the radius of the path of a proton as it passes avoce the Earth's pole.[3]
b.Draw a diagram to show the deflection caused by the magnetic field on a proton and on a helium nucleus which both have teh same initial velocity as they enter the magnetic field. State on the diagram the radius of the path of each particle.
Mass of helium nucles=6.8*10^-27kg;Charge on a helium nucleus = 3.2*10^-19 C [2]
5. In Milkan's oil drop experiment, an oil drop of weight 1.5*10^-14 N is held stationary between plates 10 mm apart by a p.d between the plates of 470 V.
a.Draw a digram of the apparatus and explain why the oil drop remains stationary.[3]
b. i.Calculate the charge on the oil drop. [3]
ii. Explain what is meant by the quantanisation of charge. [3]
c. When the charge on the oil drop is changed, the p.d needed to keep the drop stationary also changes. Values of 940 V and 313 V are also obtained.
i.Describe how the charge on the oil drop can be changed while the drop remains between the plates.[1]
ii. Explain why only certain values of p.d are found in this experiment and predict another value of the p.d that mey be required when the charge on the oil drop is changed.[3]
The diagrams for questions 1,2 and 3 are attached below.
Post by: astarmathsandphysics on December 12, 2011, 04:32:13 pm
Post by: astarmathsandphysics on December 13, 2011, 01:29:04 pm
Post by: Most UniQue™ on December 16, 2011, 08:25:24 am
Thanks in advance :)
Post by: astarmathsandphysics on December 16, 2011, 08:49:50 am
%uncertainty in R is 1/2 (%uncertainty in V + %uncertainty in L)
because of the root
%=1/2 (0.5/15 +0.1/20) *100 =1.92%
Post by: ashwinkandel on December 22, 2011, 03:55:55 pm
Q1. Ganymede is teh largest of Jupiter's moons with a mass of 1.48 * 10^23 kg. It orbits Jupiter with an orbital radius of 1.07*10^6 km and it rotates on its own axis with a period of 7.15 days. It has been suggested that to monitor an unmanned landing craft on the surface of Ganymede a geostationary satellite should be placed in orbit around Ganymede.
a. Calculate the orbital radius of the proposed geostationary satellite.
b. Suggest a difficulty that might be encountered in achieving a geostationary orbit for this moon.
Q2.The planet Mars has a mass of 6.4 * 10^23 kg and a diameter of 6790 km.
a
i.Calculate the acceleration due to gravity at the planet's surface.[2]
ii. Calculate the gravitational potential at the surface of the planet. [2]
b.
A rocket is to return some samples of Martian material to Earth. Write down how much energy each kilogram of matter must be given to escape completely from Mar's Gravitational Field. [1]
c. Use your answer to b to show that the minimum speed that the rocket must reach to escape from teh gravitational field is 5000 m/s.
d. Suggest why it has been proposed that for a successful mission to Mars that the craft that takes the astronauts to Mars will be assembled at the space station and launched from there, rather than from the Earth's Surface [2]
Q3.
a. Explain what is meant by gravitational potential at a point. [2]
b. The diagram shows the potential near a planet of mass M and radius R.
On a copy of diagram, draw similar curves.
i. for a planet of same radius but of mass 2M- label this i. [2]
ii. for a planet of same mass but of radius 2R- label this ii. [2]
c. Use the graph to explain from which of the three planets it would require the least energy to escape [2]
d. Venus has a diameter of 12100 km and a mass of 4.87*10^24 kg.
Calculate the energy needed to lift one kilogram from the surface of Venus to a space station in orbit 900 km from the surface. [4]
Q4.
a. Explain what is meant by the gravitational field strength at a point. [2]
The diagram shows the dwarf planet, Pluto, and its moon. These can be considdered to be a double planetary system orbiting each other about their joint centre of mass.
b. Calculate teh gravitational pull on Charon due to Pluto. [3]
c. Use your result to b to calculate Charon's orbital period. [3]
d. Explain why Pluto's orbital period must be the same as Charton's [1]
Q5.
a. Determine the gravitational field strength at a height equal to 2R above the Earth's surface, where R is teh radius of the Earth. [1]
b. A satellite is put into an orbit at this height. State the centripetal acceleration of the satellite. [1]
c. Calculate the speed at which the satellite must travel to remain in this orbit. [2]
d
i. The diagram shows the orbital path of the satellite. Frictional forces mean that the satellite gradually slows down.
Copy the diagram and show on your copy the resulting path of the satellite. [1]
ii. Suggest and explain why there is not a continuous bombardment of old satellites colliding with the Earth.[2]
Figure are attached below[Gravitation Field>> q no. 3,4 and 5
Post by: ashwinkandel on December 22, 2011, 04:03:16 pm
Q1.
a. Define electric field strength. [2]
b. Two charged conducting spheres each of radius 1.0 cm are placed with their centres 10 cm apart, as shown in the diagram.
Sphere A carries a charge of +2*10^-9
The graph shows how the electric field strength between the two spheres varies with distance x.
i. Calculate the field produced by sphere A at the 5.0 cm mark.[2]
ii. Use your result to b i. to calculate the charge on sphere B. [3]
c.
i. Sphere B is now removed. Calculate the potential at the surface of sphere A. [2]
ii. Suggest and explain how the potential at the surface of sphere A would compare before and after sphere B was removed. [2]
Q2.
An alpha particle emitted in the radioactive decay of radium has an energt of 8.0*10^-13 J
a
i. Calculate the potential difference thata an alpha particle, initially at rest, would have to be accelerated through to gain this energy. [2]
ii. Calculate the speed of the alpha particle at this energy. [3]
b. The diagram shows the path of the alpha particle of this energt as it approaches a gold nucleus.
i. State the speed of the alpha particle at its point of closest approach to the gold nucleus. [1]
ii. Write down the kinetic energy of the alpha particle at this point. [1]
iii. Write down the potential energy of the alpha particle at this point.[1]
c. Use your answer to b iii. to show that the alpha particle will reach a distance of 4.5*10^-14 m from the centre of the gold nucleus. [1]
d. Suggest and explain what this information telles us about the gold nucleus. [2]
(Mass of an alpha particle= 6.65*10^-27 kg, charge on an alpha particle=+2e, charge on an gold nucleus= +79e)
Q3.
a.Define electric potential at a point. [2]
b. The graph shows the potential well near a hydrogen nucleus.
The first electron orbital can be considered to be a circle of diamter 1.04*10^-10 m.
i. Determine the potential at a point on this orbital. [2]
ii. Calculate the energy required to ionise the atom. [2]
c. Use the graph to estimate the electric field strength at a distance of 1.0 * 10^-10 m from the cetnre of the nucleus. [2]
4.The diagram shows a conducting sphere of radius 0.80 cm carrying a charge of +6.0*10^-8 C resting on a balance.
a. Calculate the electric field at the surface of the sphere. [2]
b.A identical sphere carrying a charge of -4.5*10^-8 C held so that its centre is 5.0 cm vertically above the centre of the first sphere.
i. Calculate the electric force between the two spheres.[2]
ii. Calculate the new reading on the balance. [1]
c. The second sphere is moved vertically downwards through 1.5 cm. Calculat the work done against the electric field, in moving the sphere. [3]
Post by: astarmathsandphysics on December 22, 2011, 08:42:34 pm
Post by: astarmathsandphysics on December 23, 2011, 10:01:21 am
Post by: astarmathsandphysics on December 25, 2011, 09:54:40 am
Post by: ashwinkandel on December 28, 2011, 04:53:59 pm
Capacitance:
Q1.
a. State one use of a capacitor in a simple electric circuit. [1]
b. The diagram shows the change in current with time when the capacitor is discharged.
i. Deduce the resistance R of the resistor.[2]
ii. Explain why the current decreases as the capacitor discharges. [2]
iii. The charge on the capacitor is equal to the area under the graph. Estimate the charge on the capacitor when the potential difference across it is 9.0 V. [2]
iv. Calculate the capacitance of the capacitor. [2]
Q2. The spherical dome on a Van de Graff generator is placed near an earthed metal plate. The dome has a diameter of 40 cm and the potential at its surface is 54 kV.
a
i.Calculate the charge on the dome.[2]
ii. Calculate the capacitance of the dome. [2]
The metal plate is moved slowly towards the dome and it discharges through the plate. The graph in the diagram shows how the potential at the surface of the sphere changes during the discharge.
b.Calculate the energy that is dissipated during the discharge.[4]
c.Suggest why the discharge ceases while there is still some charge on the dome.[2]
Q3.
a.Show that the capacitance C of an isolated conducting sphere is given by the formula:
C=4*pi*EpsilonNought*r
The diagram shows two identical conducting brass spheres of radius 10 cm mounted on insulating stands. Sphere A has a charge of +5.0*10^-8 C and sphere B is uncharged.
b
i.Calculate the potential at the surface of sphere A.[2]
ii. Calculate the energy stored on sphere A.[2]
Sphere B is brought up to sphere A and is touched on it so that the charge is shared between the two spheres, before being removed to its original position.
c
i.Calculate the energy stored on each sphere.[3]
ii.Suggest why there is a charge in the total energy of the system. [1]
Q4.
a. Define the term capacitance of a capacitor. [2]
b. The diagram shows a circuit which can be used to measure the capacitance of a capacitor.
The reed switch vibrates back and forth at a frequency of 50Hz. Each time it makes contact with A the capacitor is charged by the battery so that there is p.d of 12V accross it. Each time it makes contact with B it is fully discharged through the resistor.
i. Calculate the charge that is stored on the capacitor when there is a p.d of 12V accross it.[2]
ii.Calculate the current in the resistor. [2]
iii. Calculate the power in milliammeter. [3]
c. A second capacitor of the same value is connected in series with the first capacitor.Discuss the effect on both the current recorded and the power dissipated in the resistor. [4]
Post by: ashwinkandel on December 28, 2011, 04:59:59 pm
1
a. Explain what is meant by a magnetic flux linkage of 1Wb.[2]
b. The diagram shows how the magnetic flux density through a 240 turn coil with a cross-sectional area 1.2*10^-4 m^2 varies with time.
i.Determine the maximum rate of change of flux in the coil. [2]
ii. Determine the induced e.m.f in the coil. [2]
iii. Sketch a diagram to show the induced e.m.f. varies with time. Mark values on both the e.m.f and time axes. [2]
2
The diagram shows a square coil about to enter a region of uniform magnetic field of magnetic flux density 0.30T. The magnetic field is at right-angles to the plane of the coil.The coil has 150 turns and each side is 2.0 cm in length. The coil moves at a constant speed of 0.50 m/s
a
i. Calculate the time taken for the coil to enter completely the region of magnetic field. [1]
ii. Determine the magnetic flux linkage through the coil when it is all within the region of magnetic field. [2]
b. Explain why the induced e.m.f. is constant while the coil is entering the magnetic field. [1]
c. Use your answer in a to determine the induced e.m.f across the ends of the coil. [4]
d. What is induced e.m.f across the ends of the coil when it is completely within the magnetic field? explain your answer. [2]
e. Sketch the graph to show the variation of the induced e.m.f with the time from the instant that the coil enters the magnetic field. Your time axis should go from 0 to 0.08 s. [2]
3.
a. State Faraday's law of electromagnetic induction. [2]
b. A circular coil of diameter 200mm has 600 turns. It is placed with its plane perpendicular toa horizontal magnetic field of uniform flux density 50 mT. The coil is then rotated through 90 degree about a vertical axis in a time of 120 ms.
Calculate:
i. the magnetic flux passing through the coil through the coil before the rotation. [2]
ii. the change of magnetic flux linkage produced by the rotation. [2]
iii. the average e.m.f. induced in teh coil during the rotation. [2]
4.
a. State Len'z law and explain how you would use a coil and a magnet to demostrate the law. Make clear any other appratus that you use.[4]
b. A vehicle brake consists of aluminium disc attached to a car axle. Electromagnets cause an e.m.f to be induced in the disc.
i. Explain how the induction of an e.m.f causes the vehicle to slow down. [3]
ii. Explain why the braking effect increases when the speed of car increases. [2]
5. A bicycle wheel is mounted vertically on a metal axle in a horizontal magnetic field. Sliding connections are made to the metal edge of the wheel and to the metal axle.
a
i.Explain why an e.m.f. is induced when the wheel rotates.[2]
ii. state and explain two ways in which the e.m.f. can be increased. [2]
b. The wheel rotates five times per second and has a radius of 15 cm. The magnetic flux density may be assumed to be uniform and of value 5.0*10^-3 T.
Calculate:
i. the area swept out each second by one spoke. [2]
ii. the induced e.m.f between the contacts. [2]
Post by: astarmathsandphysics on December 28, 2011, 11:21:52 pm
Post by: astarmathsandphysics on December 31, 2011, 09:21:10 am
Post by: NidZ- Hero on January 07, 2012, 06:39:00 am
Question 9
With Working Pls
tk
Post by: astarmathsandphysics on January 07, 2012, 11:25:08 pm
Post by: astarmathsandphysics on January 07, 2012, 11:28:03 pm
Post by: raamishstudent on February 07, 2012, 02:51:13 pm
or
Pg 43 qs 29 of unsolved pastpapers
Plzzz help
Post by: raamishstudent on February 07, 2012, 03:03:00 pm
A. 0.41 m/s in a north-easterly direction.
B. 1.00 m/s in a northerly direction.
C. 1.41 m/s in a north-westerly direction.
D. 2.24 m/s in a direction 63.4degree east of north.
E. 2.41 m/s in a north-easterly direction
Post by: Most UniQue™ on February 07, 2012, 03:39:27 pm
A body of mass 2kg is moving on a horizontal surface with a speed of 1.41ms in a north-easterly direction.A force of 0.2 N acting in a westerly direction is applied to the body for 10s. If friction is negligible the body is then moving with a speed of
A. 0.41 m/s in a north-easterly direction.
B. 1.00 m/s in a northerly direction.
C. 1.41 m/s in a north-westerly direction.
D. 2.24 m/s in a direction 63.4degree east of north.
E. 2.41 m/s in a north-easterly direction
Whats the right answer? I wanna check
Post by: astarmathsandphysics on February 09, 2012, 02:43:28 pm
Post by: Most UniQue™ on February 18, 2012, 06:55:13 pm
temperature rise from 40 °C to 100 °C.
What is the percentage uncertainty in the measurement of the temperature rise?
A 0.5 % B 0.8 % C 1.3 % D 1.7%
2. In an experiment, a radio-controlled car takes 2.50 ± 0.05 s to travel 40.0 ± 0.1 m.
What is the car’s average speed and the uncertainty in this value?
A 16 ± 1 m s–1
B 16.0 ± 0.2 m s–1
C 16.0 ± 0.4 m s–1
D 16.00 ± 0.36 m s–1
Post by: astarmathsandphysics on February 19, 2012, 12:23:56 pm
2. speed is 40/2.5 =16 but could be as low as 39.9/2.55=15.65 or as high as 40.1/2.45=16.37
D
Post by: Most UniQue™ on February 19, 2012, 12:42:32 pm
1. 1.7% the inituial tempe could be as high as 40.5 and the final temp could be as low as 40.5 ie a reading as low as 59 compared to an actual rise of 60. The difference is 1 so the % is 1/60*100=1.7%
2. speed is 40/2.5 =16 but could be as low as 39.9/2.55=15.65 or as high as 40.1/2.45=16.37
D
For question 2 , I got the same answer as yours but the actual answer is C .
Post by: astarmathsandphysics on February 19, 2012, 12:56:41 pm
Post by: Most UniQue™ on February 19, 2012, 01:10:31 pm
Then they are rounding up extremely.
Yah Thats what I had been guessing...
Post by: emzie on March 02, 2012, 04:41:27 pm
I just have this doubt in equilibrium
Is the Kp value affect by change in pressure at constant temperature, if not then why?
Thank in advance!
Post by: Most UniQue™ on March 02, 2012, 04:49:59 pm
Hello! Hope you all doing great :)
I just have this doubt in equilibrium
Is the Kp value affect by change in pressure at constant temperature, if not then why?
Thank in advance!
Isnt this is related to chemistry cuz I have seen it there.
Post it here:
https://studentforums.biz/sciences-149/chemistry-doubts-10366/
Post by: emzie on March 02, 2012, 05:27:22 pm
I was going to post it there, dunno how i can here :o
Anyway Thanks
Post by: astarmathsandphysics on March 02, 2012, 08:25:57 pm
When I realised my mistake, I told them, "I've had so much fun. If I ever end up in the wrong place again I will make sure it's this one".
Post by: astarmathsandphysics on March 03, 2012, 11:19:49 pm
Mass of 1 mol of gold atoms=197.97*1.67*10^-27*6.02*10^23=198g
no gold atoms needed=0.2/198*6.02*10^23=6.08*10^20
Nogold atoms per sec =6.08*10^20/(3*3600)=5.63*10^16
so current=5.63*10^16*1.6*10^-19=9*10^-3A
Post by: ashwinkandel on March 06, 2012, 01:51:26 pm
Post by: astarmathsandphysics on March 06, 2012, 02:56:14 pm
net forec=F_1-F_2=2kx
Post by: Most UniQue™ on March 06, 2012, 04:30:41 pm
force due to one spring is F_1 =kx and force due to other is F_2=-kx cos in opposite direction
net forec=F_1-F_2=2kx
Wow sir your really good in physics!
Post by: astarmathsandphysics on March 07, 2012, 02:27:22 pm
Post by: ashwinkandel on March 09, 2012, 03:16:55 am
Post by: astarmathsandphysics on March 09, 2012, 10:25:55 am
Post by: ashwinkandel on March 14, 2012, 04:31:51 pm
I have attached the questions
Post by: astarmathsandphysics on March 18, 2012, 08:02:23 pm
b)helium nuclei have twice the carge of protons so repel even more strongly, making the activation renergy higher
c)mass of reactants=3*4.001506=12.004518u
mass of product=12u
difference=0.004518u
mass=0.004518*1.57*10^-27kg
E=mc^2=0.004518*1.57*10^-27*(3*10^8)^2=6,79*10^-13J
Post by: astarmathsandphysics on March 18, 2012, 08:12:24 pm
see attachment
Post by: astarmathsandphysics on March 18, 2012, 08:26:53 pm
b)see attachment
Post by: ashwinkandel on March 24, 2012, 08:00:52 am
I have attached four questions below.
Post by: astarmathsandphysics on March 24, 2012, 08:26:09 am
Post by: astarmathsandphysics on March 25, 2012, 08:32:28 am
Will also make some videos
Post by: ashwinkandel on March 26, 2012, 02:05:22 pm
Post by: astarmathsandphysics on March 26, 2012, 02:14:28 pm
Post by: astarmathsandphysics on March 26, 2012, 02:34:46 pm
b)gradient is negative and constant so a=-kx
c)k=omega^2=10/0.02=500 so omega=sqrt(500)=22.4
f=omega/2pi=22.4/6.28=3.6 Hz
Post by: silvercameron on April 02, 2012, 01:27:55 pm
Post by: astarmathsandphysics on April 02, 2012, 09:11:17 pm
e=(separation speed)/(approach speed)=(v2-v1)/(u1+u2)
e=1 for a perfectly elastic collision so v2-v1=u1+u2
answer is D
Post by: astarmathsandphysics on April 02, 2012, 09:14:47 pm
Use Newtons Law of restitution
e=(separation speed)/(approach speed)=(v2-v1)/(u1+u2)
e=1 for a perfectly elastic collision so v2-v1=u1+u2
answer is D
q11
Applying F=ma for 2 kg mass 2g-T=2a
applying F=ma for 8kg mass T-6=8a
add these to get 2g-6=10a so a=(2g-6)/10=1.36
Post by: Physics09 on April 03, 2012, 12:11:42 pm
Explanation to the answer of MCQ 27, urgently please!!
Post by: astarmathsandphysics on April 03, 2012, 11:33:42 pm
Post by: Physics09 on April 04, 2012, 10:01:48 am
Also, MCQ 34 and 27
http://www.xtremepapers.com/CIE/International%20A%20And%20AS%20Level/9702%20-%20Physics/9702_s10_qp_12.pdf
Thanks again!
Post by: SkyPilotage on April 04, 2012, 03:01:23 pm
Q34 - Total Resistance is 1.5R ---> so I^2 = P / 1.5R = 8 A^2 ---> I = 2.83 A
Here I took the value of R as 1 ohm, you can take it as any value you want as long as you are cosistent with all the resistors P,Q and R.
So P across P = (2.83)^2 x 1 = 8 Watts
P across Q = (2.83 / 2 ) ^2 x 1 = 2 Watts
P across R = P across R2 = 2 Watts.
Answer is A
There you have it!
Post by: SkyPilotage on April 04, 2012, 03:09:06 pm
-Is it a rule for all op-amps?
-why does the non-inv input have to be the same as the inv input? i.e why does the diff have to be zero?
Q2.How does reducing the gain cause less distortion, increased bandwidth and stability ?
Thank you in advance.
Post by: Physics09 on April 04, 2012, 06:56:30 pm
But just one thing, how is the total resistance 1.5?
Post by: SkyPilotage on April 05, 2012, 12:31:27 pm
Resistor P is in series with the equivalent of both resistors Q and R so The total resistance of the circuit is 1R + 0.5R = 1.5R
Hope it cleared it up.
Post by: Physics09 on April 05, 2012, 01:16:19 pm
Thank you!
Post by: Rvel Zahid on April 08, 2012, 04:09:58 am
http://www.xtremepapers.com/CIE/International%20A%20And%20AS%20Level/9702%20-%20Physics/9702_w10_qp_11.pdf
24 and 33.
Post by: SkyPilotage on April 08, 2012, 06:30:59 am
So each of the points A, B , C and D will move up and down ONLY, so do not get tricked that they will move to the right.
First of All, B and D are wrong because they are the maximum displacement , so obviously, the acceleration is max and speed is zero.
To get the right answer you have to image the wave moving. If you just shift this wave to the right abit, you will notice that A will be at the bottom of the shifted wave and C will be at the top. Thus, C is the point with the maximum speed going upwards.
Q-33 Resistance R is proportional to L and 1/A.
If L is doubled, then R is doubled ----> 2R
If the radius is doubled, then the area (Pie)(2r)^2 = 4*pie*r^2, Resistance is divided by 4 ----> R/4
Therefore, The total resistance is 2R/4 = 0.5 R
Hope that helped, and feel free to post any more Qs :)
Post by: Rvel Zahid on April 08, 2012, 08:41:03 am
Q-25 Alright for this question, you need to identify water waves as transverse waves. For transverse waves, point particles move up and down and NOT left and right.Thankyou so much man. Hats off to your explanation specially in the 1st mcq, i was just not getting the point.
So each of the points A, B , C and D will move up and down ONLY, so do not get tricked that they will move to the right.
First of All, B and D are wrong because they are the maximum displacement , so obviously, the acceleration is max and speed is zero.
To get the right answer you have to image the wave moving. If you just shift this wave to the right abit, you will notice that A will be at the bottom of the shifted wave and C will be at the top. Thus, C is the point with the maximum speed going upwards.
Q-33 Resistance R is proportional to L and 1/A.
If L is doubled, then R is doubled ----> 2R
If the radius is doubled, then the area (Pie)(2r)^2 = 4*pie*r^2, Resistance is divided by 4 ----> R/4
Therefore, The total resistance is 2R/4 = 0.5 R
Hope that helped, and feel free to post any more Qs :)
Yes! now you have triggered a bomb i will be coming up with more Qs. :D
Post by: SkyPilotage on April 08, 2012, 09:06:45 am
Thankyou so much man. Hats off to your explanation specially in the 1st mcq, i was just not getting the point.
Yes! now you have triggered a bomb i will be coming up with more Qs. :D
You dont have to take off your hat *doubt you were wearing one* , I'm more than glad to help!
Plus its always a challenge to defuse bombs, So Bring it on! :D
Post by: Rvel Zahid on April 08, 2012, 09:13:50 am
mcq 13, 15
Post by: SkyPilotage on April 08, 2012, 09:53:33 am
Moment due to the weight of the block = 900x0.2 Nm
Therefore F = (900x0.2) / 1.2 = 150 N ---> B
Q-15 Usually in g.p.e problems they just give you the height and you just use the number given.
But you have to know it is the height the Centre of gravity of the object moved.
So in this problem p.e is mxgx(h/2) when the tap is closed because the centre of gravity is 1/2 h from the bottom.
When the tap is opened, the new p.e of the water is mxgx(h/4) because the centre of gravity is 1/4 h from the bottom.
Therefore the loss is m.g.h/4-----> B
Post by: Rvel Zahid on April 08, 2012, 07:20:12 pm
still man thanks for prompt and comprehensive response.
i didn't really got the 15th mcq,
the gravitational force shouldn't be constant on the liquid? :/ or if we r talking bout liquids we always have to consider centre of gravity while calculating G.P.E
another mcq: http://www.xtremepapers.com/community/attachments/physix1-jpg.6355/
Post by: SkyPilotage on April 08, 2012, 07:46:50 pm
For that other mcq, you have to draw out several lines from the origin to each of the points A/B/C/D.
So you will notice that the line is steepest i.e highest gradient when it connects the origin and point C. Gradient is determined by V/I.
Since R is I/V so at point C , it is the lowest resistance.
A common error is drawing tangents to the points on the graph so be careful.
Post by: Rvel Zahid on April 09, 2012, 06:17:09 am
Q. 34
http://www.xtremepapers.com/CIE/International%20A%20And%20AS%20Level/9702%20-%20Physics/9702_w09_qp_11.pdf
q.9 and q. 14
Post by: ashwinkandel on April 09, 2012, 10:12:36 am
Post by: SkyPilotage on April 09, 2012, 10:18:44 am
Q1.Can someone shed light on the virtual earth approximatio in operatioal amplifiers?
-Is it a rule for all op-amps?
-why does the non-inv input have to be the same as the inv input? i.e why does the diff have to be zero?
Q2.How does reducing the gain cause less distortion, increased bandwidth and stability ?
Q3.Are the guard bands included in the bandwidth or only the sidebands?
Thank you in advance.
If anyone would be kind enough :)
Post by: ashwinkandel on April 09, 2012, 10:33:10 am
Post by: SkyPilotage on April 09, 2012, 10:55:59 am
There tends to be confusion between Vin and the Diff. Input volt, where Vin is the voltage coming before the Rin Resistor. While, Diff input voltage is V+-V- which has to be equal to zero. I am not sure why it has to but I think it has -like you said- to with the open-loop gain which has to be infinite, but ill try makin sure.
Thanks for your help mate! ;D
Post by: ashwinkandel on April 09, 2012, 11:31:24 am
Post by: Rvel Zahid on April 09, 2012, 03:47:16 pm
Rvel Zahid I have attached solutions to your questions:Bombastic Ashwin! Thank you soooooo much, makes things very clear. :)
one more questionhttp://www.xtremepapers.com/CIE/International%20A%20And%20AS%20Level/9702%20-%20Physics/9702_w09_qp_11.pdf
mcq 14 how it is A?
Post by: ashwinkandel on April 09, 2012, 04:16:19 pm
Post by: Rvel Zahid on April 09, 2012, 04:52:07 pm
I was a bit unclear last time for q.n 14. Here i have attached another versionyeah this mcq was hard to grasp...Thanks :)
one more :D
i can't understand how to calculate phase difference between two waves...
Post by: SkyPilotage on April 09, 2012, 07:40:59 pm
Post by: SkyPilotage on April 09, 2012, 11:02:17 pm
yeah this mcq was hard to grasp...Thanks :)
one more :D
i can't understand how to calculate phase difference between two waves...
Its like you find the ratio of the wave lag to the wavelength. Here it is 1/2 lamda so thats the phase diff. and destructive interference will occur where they cancel each other out.
Post by: Rvel Zahid on April 10, 2012, 11:12:49 am
Its like you find the ratio of the wave lag to the wavelength. Here it is 1/2 lamda so thats the phase diff. and destructive interference will occur where they cancel each other out.ok thanks a bunch.
some more. attached the files below. take ur time not urgent. Thankyou. :)
Post by: ashwinkandel on April 10, 2012, 01:45:54 pm
Post by: Rvel Zahid on April 10, 2012, 04:12:16 pm
Rvel Zahid Unless someone else answers your questions , I will answer it tomorrow.Ok brother dats so nice of you. i don't need urgent answers, so its ok no problem. Take care :)
Post by: Rvel Zahid on April 11, 2012, 08:33:08 pm
need help in 11 and 32
Post by: haris94 on April 11, 2012, 09:18:46 pm
http://www.xtremepapers.com/papers/CIE/Cambridge%20International%20A%20and%20AS%20Level/Physics%20(9702)/9702_w08_qp_1.pdf
need help in 11 and 32
q11:
F=ma
For the 2kg box 20-T=2a
20 is the weight of the box ; T is the tension in the string ; 2 is the mass of the box ; a is the acceleration
For the 8kg box T-6=2a
6 is the frictional force
Solve both simultaneously and you get a=1.4
Answer is A
Post by: haris94 on April 11, 2012, 09:36:21 pm
http://www.xtremepapers.com/papers/CIE/Cambridge%20International%20A%20and%20AS%20Level/Physics%20(9702)/9702_w08_qp_1.pdf
need help in 11 and 32
q32:
i need someone to please confirm this.
here's my theory - the steel cable will not be responsible for the resistance of the power cable since it is enclosed within the copper cables.
Now all the 6 copper cables are in a circle so i think it should be taken as if they are all parallel to each other.
So 1/R = (1/10)*6
R=1.67
Answer is B
Post by: SkyPilotage on April 11, 2012, 10:01:57 pm
1- Answer will be 1.64 i.e closer to B
2- They wouldnt have given you the value, never seen CIE do sth like that from year 2000 Examinations :P
What do you think?
Post by: haris94 on April 11, 2012, 11:03:30 pm
2- They wouldnt have given you the value, never seen CIE do sth like that from year 2000 Examinations :P
What do you think?
Valid point.
I may have overthought this a bit too much. Thanks :P
Post by: SkyPilotage on April 11, 2012, 11:13:55 pm
mess up all the easy questions and solve the hard ones fluently :D
Post by: galaxy on April 12, 2012, 05:52:58 pm
I need help with Q:7 a of June 05 Paper 4...plz
and also Q:6 bof the same paper
JazakAllah
Post by: Most UniQue™ on April 12, 2012, 08:15:11 pm
ok thanks a bunch.
some more. attached the files below. take ur time not urgent. Thankyou. :)
Attachment 11.JPG:
I= Q/T
Q=ne
Therefore I= N(e)/t => I/e = N/t
Since rate of flow = N/t , I/e = N/t
Rate of flow = I/e
Rate of flow = 4.8/ 1.6x10^-19 = 3.0 x 10^-19 s^-1
Electrons flow from negative to positive terminal so there flow of electron is in the direction Y to X
Answer = C
-----------------------------------------------------------------------------------------------------------
Attachment 5.JPG
Anticlockwise moment = Clockwise moment
10 x 0.6 + 100 x 0.1 = 20 x 0.4 + 20 x d
d = 0.4 m from pivot so answer is D
-----------------------------------------------------------------------------------------------------------
Will solve the others later...
Post by: Rvel Zahid on April 13, 2012, 08:36:34 am
Attachment 11.JPG:
I= Q/T
Q=ne
Therefore I= N(e)/t => I/e = N/t
Since rate of flow = N/t , I/e = N/t
Rate of flow = I/e
Rate of flow = 4.8/ 1.6x10^-19 = 3.0 x 10^-19 s^-1
Electrons flow from negative to positive terminal so there flow of electron is in the direction Y to X
Answer = C
-----------------------------------------------------------------------------------------------------------
Attachment 5.JPG
Anticlockwise moment = Clockwise moment
10 x 0.6 + 100 x 0.1 = 20 x 0.4 + 20 x d
d = 0.4 m from pivot so answer is D
-----------------------------------------------------------------------------------------------------------
Will solve the others later...
Thankyou so much Muscle Nerd :)
Post by: mathy on April 13, 2012, 08:45:32 am
anyone ?
I need help with Q:7 a of June 05 Paper 4...plz
and also Q:6 b of the same paper
JazakAllah
also Q:6a i of june 06
Post by: SkyPilotage on April 13, 2012, 07:36:52 pm
Assalamoalaikum wr wb!
I need help with Q:7 a of June 05 Paper 4...plz
and also Q:6 bof the same paper
JazakAllah
Wealykumoessalam warahmatullah webarakatuh.
Q7 a) in attachement.
Post by: SkyPilotage on April 13, 2012, 07:43:59 pm
Wealykumoessalam warahmatullah webarakatuh.
Q7 a) in attachement.
Q6 b)
Post by: SkyPilotage on April 14, 2012, 09:23:58 am
Assalamoalaikum wr wb!
anyone ?
also Q:6a i of june 06
Q6a J06
Post by: mathy on April 14, 2012, 09:45:43 am
ok can u please explain me the reason for Q6 b)..why is the first (top one) like that? what;s the relation ship to be used?
and for 6 a for j06y, wud u plz explain, me the direction of B....like how did u get it downwards?
By the way Thanks a lot!!!!!!! it helped me...:)
P.S. I tried quoting the post but was unable to do..?! i dunno y?? :s
Post by: SkyPilotage on April 14, 2012, 10:12:46 am
jazakAllah khairen!!!!!!!!!!!
ok can u please explain me the reason for Q6 b)..why is the first (top one) like that? what;s the relation ship to be used?
and for 6 a for j06y, wud u plz explain, me the direction of B....like how did u get it downwards?
By the way Thanks a lot!!!!!!! it helped me...:)
P.S. I tried quoting the post but was unable to do..?! i dunno y?? :s
June 2005
Question 6 B)
I) Magnetic Flux is proportional to the change in current so its the same as the current graph given in the question.
II) The e.m.f is the negative of the gradient of the magnetic flux graph by Faraday Lenz Law that Emf= - N(flux)/time so it is the negative of the rate of change of flux. So when the graph of Phi is steepest, emf is maximum
June 2006
Question 6 A)
The magnetic field around point Y due to XY is circular, Therefore at Point Q, the direction of the magnitic field is tangential to the magnetic field at Y. Which is downwards.
Hope it helped you, brother!
--------If you need to quote, just click quote which is at the top right of that post. Then you can type what you want under it :)-------------
Post by: mathy on April 14, 2012, 08:44:25 pm
June 2005It definitely did... :) JazakAllah khairen!! :)
Question 6 B)
I) Magnetic Flux is proportional to the change in current so its the same as the current graph given in the question.
II) The e.m.f is the negative of the gradient of the magnetic flux graph by Faraday Lenz Law that Emf= - N(flux)/time so it is the negative of the rate of change of flux. So when the graph of Phi is steepest, emf is maximum
June 2006
Question 6 A)
The magnetic field around point Y due to XY is circular, Therefore at Point Q, the direction of the magnitic field is tangential to the magnetic field at Y. Which is downwards.
Hope it helped you, brother!
--------If you need to quote, just click quote which is at the top right of that post. Then you can type what you want under it :)-------------
thanks a lott!!!!!!
and By the way I did know how to quote..I guess, I had the quick reply feature on, bcoz of which that wasn't working, I turned it off, and it works now.. :D
Post by: SkyPilotage on April 14, 2012, 09:15:24 pm
What do you guys think about why the shock waves frequency has to be higher than the natural frequency of the seismometer?
Post by: Physics09 on April 18, 2012, 08:08:37 pm
http://www.xtremepapers.com/papers/CIE/Cambridge%20International%20A%20and%20AS%20Level/Physics%20(9702)/9702_s11_qp_11.pdf
MCQs 4,9,10,13 and 14
Thank you!
Post by: raamishstudent on April 20, 2012, 06:28:43 am
9D formula will be m2v2 for momentum after collision so (m+m)v = 2mv. K.e energy will be 1/2mv2 * 1/2mv2 = mv2
10D direction is changing so negative sign. So change will be -mv -(mv) = -2mv
And i dont know how to do 13 and 14 so plzz somebdy do them :)))
Post by: Rvel Zahid on April 20, 2012, 03:27:44 pm
some past paper problems:-
http://www.xtremepapers.com/papers/CIE/Cambridge%20International%20A%20and%20AS%20Level/Physics%20(9702)/9702_w07_qp_1.pdf
23 , 24 and 25
http://www.xtremepapers.com/papers/CIE/Cambridge%20International%20A%20and%20AS%20Level/Physics%20(9702)/9702_w06_qp_1.pdf
q. 27
Post by: Physics09 on April 21, 2012, 09:29:34 pm
Fringe separation is indirectly proportional to distance b/w the slits.
and directly proportional to the distance b/w the slits and the screen.
Hope it helps!
Post by: Physics09 on April 21, 2012, 09:31:09 pm
Post by: Most UniQue™ on April 22, 2012, 11:25:47 am
Maximum speed of P=2 x 3.14 x 2 x 0.16=2.01
Max K.E.=1/2 x 0.002 x (2.01)^2=4.0mJ
B is answer.
25. d sin (theta)=nlamda
Equation 1: d sin 45=3l Equation 2: d sin(90)=nl
Divide Equation 1 by 2: sin 45=3/n
so n=4.2426..... and you have to round down to the nearest integer which is 4.
Answer is B.
Post by: Rvel Zahid on April 24, 2012, 08:22:55 pm
http://www.xtremepapers.com/papers/CIE/Cambridge%20International%20A%20and%20AS%20Level/Physics%20(9702)/9702_s09_qp_2.pdf
paper21 : Q.5
Post by: LekhB on April 26, 2012, 05:21:23 am
Design an experiment to investigate how the intensity of light from a source varies with time.
Post by: Rvel Zahid on April 27, 2012, 04:09:16 am
Can anyone help with that?hey check out this thread! this might help you. :)
Design an experiment to investigate how the intensity of light from a source varies with time.
http://www.physicsforums.com/showthread.php?t=166961
Post by: Ghost Of Highbury on May 08, 2012, 03:55:31 am
In Physics/Chemistry paper 4s, paper 5s and Math P3?
For example, I recently solved a Communication system sums and got the answer as 5.3/12 *1000 = 441.6666..7 m
Look what the markscheme says
"length = 5.3 / 12 km
= 440 m"
I could've rounded it off to 3 s.f and made it 442m. Why does it say 440m then? How do you go about this? Will the deduct marks if the answer
isn't rounded off to the appropriate number of s.f or d.p?
Thanks
Post by: SkyPilotage on May 08, 2012, 11:50:03 am
Can anyone throw some light on the number of significant figures or decimal places we should round off the final answer to?
In Physics/Chemistry paper 4s, paper 5s and Math P3?
For example, I recently solved a Communication system sums and got the answer as 5.3/12 *1000 = 441.6666..7 m
Look what the markscheme says
"length = 5.3 / 12 km
= 440 m"
I could've rounded it off to 3 s.f and made it 442m. Why does it say 440m then? How do you go about this? Will the deduct marks if the answer
isn't rounded off to the appropriate number of s.f or d.p?
Thanks
I have the same problem too, I dont know if they will deduct marks, but nothing is written about it in the marking scheme or the examiner report.
Sometimes marking scheme says deduct 1 mark if its not 3 s.f.
But I think the reason it is 420 is because 5.3 and 12 are 2 s.f so the answer must be the least s.f of both of them. which is 2s.f = 420 m :)
Post by: Ghost Of Highbury on May 08, 2012, 03:01:27 pm
I have the same problem too, I dont know if they will deduct marks, but nothing is written about it in the marking scheme or the examiner report.
Sometimes marking scheme says deduct 1 mark if its not 3 s.f.
But I think the reason it is 420 is because 5.3 and 12 are 2 s.f so the answer must be the least s.f of both of them. which is 2s.f = 420 m :)
420m? 2s.f?
3 s.f no?
Post by: SkyPilotage on May 09, 2012, 02:35:16 am
Since , 3 s.f are not used in the question. Dont use it, but Im not sure if they will deduct marks or not.
Post by: Naruto123 on May 11, 2012, 11:20:12 am
question 6b why is it 90 degrees
Can any1 summarize all the graphs for primary current voltage flux and secondary flux current voltage .
Post by: Ghost Of Highbury on May 13, 2012, 09:27:46 am
The flux changes according to the current, so same shape.
But you know, the field strength = -potentiala gradient of the flux.
Or E = d(flux)/dt (faraday's law)
so if the shape of the flux against time is like a sine curve. d/dt (sin t) = cos t.
So the E will be like a cos t starting from (0,1)
And as you know the phase difference b/w sine and cos curve is 90 degrees.
Post by: Ghost Of Highbury on May 13, 2012, 10:19:56 am
At t=0, will the signal be 0 or 2?
question 7b)
Why does the damping decrease when the resistance increases?
Post by: SkyPilotage on May 21, 2012, 04:56:27 am
Summer 2007, question 10a)How are you studying for paper 5?
At t=0, will the signal be 0 or 2?
question 7b)
Why does the damping decrease when the resistance increases?
Are you studying experiments? cz I dont know where to find them :S
Didnt go into school labs .. soo If you could share some resources I would appreciate it!
Post by: Ghost Of Highbury on May 23, 2012, 05:30:29 pm
How are you studying for paper 5?
Are you studying experiments? cz I dont know where to find them :S
Didnt go into school labs .. soo If you could share some resources I would appreciate it!
Just go through all the past papers, markschemes and examiner reports. Read all the experiments described in the book
and read the examiner tips from cambridge students. if u have the physics coursebook, there's a section in the end of the book for physics paper 5 particularly
Post by: whale95 on November 06, 2012, 04:04:28 pm
Thanks for answering! :)
Post by: astarmathsandphysics on November 07, 2012, 03:22:49 pm
Post by: baditude on December 07, 2012, 03:58:00 pm
Thanks
Post by: astarmathsandphysics on December 07, 2012, 09:09:04 pm
Post by: silvercameron on January 10, 2013, 04:11:20 pm
if (dN/dt) is directly proportional to N, then when we write the equation as (dN/dt)= (-lambda)t, where lamda is the decay constant, why does the constant have the negative sign before it?
This one is from ideal gases:
how is the equation pV=nkT derived?
Post by: kruti1996 on July 04, 2013, 03:16:17 am
please help me out with June 2012, paper 11, question 12.
thank you
Post by: astarmathsandphysics on September 23, 2013, 03:20:08 pm
this one's from nuclear physics:
if (dN/dt) is directly proportional to N, then when we write the equation as (dN/dt)= (-lambda)t, where lamda is the decay constant, why does the constant have the negative sign before it?
This one is from ideal gases:
how is the equation pV=nkT derived?
N is decreasing so dN is negative. dN is the change in N in a time period dt. If N is decreasing, at the end of the time period there are less atoms than at the start.
Post by: astarmathsandphysics on September 23, 2013, 03:52:59 pm
Hi people,
please help me out with June 2012, paper 11, question 12.
thank you
Post by: astarmathsandphysics on September 23, 2013, 03:55:00 pm
here
http://www.astarmathsandphysics.com/a_level_physics_notes/thermal_physics_and_gases/a_level_physics_notes_kinetic_theory_of_gases.html
Post by: Tasrina on December 30, 2014, 08:19:35 pm
Post by: astarmathsandphysics on December 30, 2014, 11:45:32 pm
Post by: Tasrina on December 31, 2014, 12:48:14 pm
Post by: Tasrina on December 31, 2014, 02:51:38 pm
Post by: astarmathsandphysics on December 31, 2014, 07:22:00 pm
1. pV/T=constant
2. pV=nRT3.
3. KE=3/2 kT
4. Energy of planet in orbit = 1/2 GPE
5. R^3 = k T^2 for planet in orbit
6. Period of geostationary orbit
7. r=mv/Bq for circular motion in electric field
There may be others.
I will add to this list.
Post by: Tasrina on January 01, 2015, 11:24:08 am
Post by: astarmathsandphysics on January 01, 2015, 11:42:02 am
Post by: astarmathsandphysics on January 01, 2015, 11:46:43 am
Post by: astarmathsandphysics on January 01, 2015, 11:49:50 am
Post by: Tasrina on January 02, 2015, 03:11:35 pm
Thanks
Post by: astarmathsandphysics on January 02, 2015, 09:28:42 pm
satellite moves round earth's equator every 24 hours=86400 seconds to stay over same point.
see attachment
Post by: Tasrina on January 09, 2015, 03:14:57 pm
Post by: astarmathsandphysics on January 09, 2015, 07:20:59 pm
Post by: astarmathsandphysics on January 10, 2015, 07:10:23 pm
Post by: jiuer7845 on February 12, 2022, 10:57:01 am
Pandora Charms (https://www.pandoracharmss.us/)
Pandora Charms (https://www.charms-pandora.com/)
Travis Scott Jordan 1 (https://www.travisscott-jordan1.com/)
Air Jordans (https://www.airsjordans.com/)
Jordan 11s (https://www.jordan-11s.com/)
Jordan 11 (https://www.jordans-11.com/)
Jordans Shoes (https://www.jordansshoes.org/)
Moncler Outlet (https://www.moncler-outlets.com/)
Off-White (https://www.off-white.us.org/)
Yeezy 450 (https://www.yeezy-450.com/)
Yeezy 500 (https://www.yeezys500.com/)
Nike Outlet (https://www.nikeoutlet-factory.com/)
Jordans 4 (https://www.jordans4s.com/)
Jordan Retro 4 (https://www.jordanretro4.com/)
Jordan Shoes (https://www.jordan-shoes.us.com/)
Yeezy (https://www.yeezyyeezy.com/)
Yeezy 700 (https://www.yeezys-700.com/)
Yeezy Supply (https://www.yeezys-supply.com/)
Off White Shoes (https://www.offwhiteshoess.com/)
NFL Jerseys (https://www.nflsjerseys.us.com/)
Retro Jordans (https://www.retro-jordan.com/)
Nike Air Jordan (https://www.nikeair-jordan.com/)
Jordan Shoes (https://www.jordanshoes.org/)
Jordans Shoes (https://www.jordans-shoes.com/)
Yeezy 350 V2 (https://www.yeezy350-v2.com/)
Adidas Yeezy (https://www.adidasyeezys.com/)
Yeezy (https://www.yeezyoutlet.us.com/)
Yeezy 700 (https://www.yeezy-700.us.com/)
Nike Outlet (https://www.nike-outlets.com/)
Yeezy Shoes (https://www.yeezy-shoes.us.com/)
Yeezy Foam Runner (https://www.yeezyfoam-runner.com/)
Nike Outlet (https://www.nikestoreoutlet.us.com/)
Nike Outlet (https://www.nikeoutletfactory.us/)
AJ1 (https://www.aj1.us.com/)
UNC Jordan 1 (https://www.uncjordan1.us/)
Jordan 13 (https://www.jordan-13.us/)
Jordan AJ 1 (https://www.jordanaj1.com/)
Yeezy Supply (https://www.yeezy-supply.com/)
Yeezy Zebra (https://www.yeezy-zebra.com/)
Jordan 5 (https://www.jordan-5.us/)
Jordan 1 Low (https://www.jordan1low.com/)
Air Jordans (https://www.air-jordans.us.org/)
Pandora Charms (https://www.pandoracharms.uk.com/)
Adidas UK (https://www.adidasuk.uk.com/)
Nike Store (https://www.nikestoreoutlet.us.com/)
Adidas Yeezy Official Website (https://www.adidasyeezyofficialwebsite.com/)
Yeezy 350 (https://www.yeezy350.us.com/)
Jordan 1 (https://www.jordan1.uk.com/)
Nike Outlet (https://www.nikesoutlet.us.com/)
YEEZY SUPPLY (https://www.supplyyeezys.us/)
Pandora Charms (https://www.pandoracharms.cc/)
Nike Shoes (https://www.nikeshoes.cc/)
Nike Outlet (https://www.nikeoutlet.uk.com/)
Pandora Jewelry (https://www.pandoras-jewelry.com/)
Pandora Outlet (https://www.pandoraoutlet.org/)
Jordan Shoes (https://www.jordanshoess.com/)
Air Jordan 4 (https://www.air-jordan4.com/)
Pandora Charms (https://www.charms-pandora.com/)
Adidas Yeezy (https://www.adidas-yeezy.org/)
Air Jordan 11 (https://www.air-jordan11.com/)
Air Jordan 1 (https://www.air-jordan1.com/)
Nike Jordans (https://www.nike-jordans.com/)
Jordan 1s (https://www.jordan-1s.com/)
Pandora UK (https://www.pandorauk.uk.com/)
Nike Jordan 1 (https://www.nikejordan1.com/)
Jordan 1 (https://www.jordan-1.org/)
Yeezy Slides (https://www.yeezyslides.us.com/)
Nike Air VaporMax (https://www.nikeairvapormax.us/)
Nike Vapormax Flyknit (https://www.nikevapormaxflyknit.com/)
Air Jordan 1 Mid (https://www.airjordan1-mid.com/)
Adidas yeezy (https://www.yeezyadidas.de/)
Adidas Yeezy (https://www.adidasyeezy.me.uk/)
Yeezy 350 (https://www.yeezy350.de/)
Nike Shoes (https://www.nikes.us.com/)
Nike Outlet (https://www.nikeoutletstoreonlineshopping.us/)
Yeezy (https://www.yeezystore.us.com/)
NFL Shop Official Online Store (https://www.nflshopofficialonlinestore.com/)
Nike UK (https://www.nikeuk.uk.com/)
Yeezy (https://www.yeezy.uk.com/)
Yeezy 350 (https://www.yeezy350.uk.com/)