Post by: ruby92 on March 24, 2010, 10:52:20 am
which of the following gives her result expressed to the appropriate no of significant figures?
a) 327.7 b)328 c)330 d)300
Post by: ruby92 on March 24, 2010, 11:03:39 am
the resistor is marked as 4.7 +/- 2% ohms
if these values were used to calculate the power dissipated what is the percentage uncertanity
a) 2% b) 4% c)6% d) 8%
Post by: A.T on March 24, 2010, 11:10:35 am
the current in a resistor is marked as (2.50+/-0.05)mA
the resistor is marked as 4.7 +/- 2% ohms
if these values were used to calculate the power dissipated what is the percentage uncertanity
a) 2% b) 4% c)6% d) 8%
For resistance the percentage uncertainity is given which is 2%
For current u can find it by 0.05/2.5 * 100 = 2%
The formula is P=I^2 * R
so we'll multiply the current's uncertainity by 2 which will give 4%
now add both % uncertainities = 4+2=6%
hence answer is C
Post by: ruby92 on March 24, 2010, 11:13:28 am
and dont u have to find the fractional uncertanity for both and add them?
Post by: A.T on March 24, 2010, 11:30:05 am
but current is given a mA we dont take the 'm' into account here?For uncertainity u dont have to count the milli but for findind the actual power u will need to count it.
and dont u have to find the fractional uncertanity for both and add them?
The simplest way to find the % uncertainity is find the % unc. of all the variables and add then so there is no neeed to find fractional unc.
Post by: ruby92 on March 24, 2010, 11:33:47 am
Post by: ruby92 on March 24, 2010, 11:58:57 am
any ideas
Post by: astarmathsandphysics on March 24, 2010, 02:08:48 pm
Post by: ruby92 on March 24, 2010, 02:15:59 pm
Post by: ruby92 on March 24, 2010, 02:43:23 pm
how long does the stone take to strike the ground?
with what velocity does the stone strike the ground
ASAP please
thnkx
Post by: A.T on March 24, 2010, 05:49:55 pm
Because 330 is in the range 327.66 -3% to 327.66+3%Arent all the values in that range?
i dont think this is the correct logic.
Post by: ruby92 on March 24, 2010, 05:54:34 pm
could someone please explain this and the above questions...urgently!!!! ???
Post by: astarmathsandphysics on March 24, 2010, 08:44:00 pm
0.196=4.9t^2 so t=0.2 seconmd
v=x/t=20/0.2 =100m/s
Post by: ruby92 on March 24, 2010, 09:48:21 pm
Post by: A.T on March 24, 2010, 10:19:59 pm
hw did u get 4.9 as the acceleration9.81/2
9.81 = g
Post by: ruby92 on March 24, 2010, 10:24:18 pm
given momentum and time how do u find force?
Post by: A.T on March 24, 2010, 10:27:52 pm
mv= momentum
change in momentum per unit time is force.
change in momentum is (mv-mu) = m(v-u)
Hence F=(m(v-u))/t
as (v-u)/t = a , we get F=ma
Dunno if this anwers your question LOL
Post by: ruby92 on March 24, 2010, 10:33:16 pm
i found out that the V total velocity is 3
and that kinetic energy lost is -60J
how do we find the exchange force..which By the way is 73N
wht exactly are the differences in momentum and kinetic energy when objects collide elastically and inelastically
???
Post by: Monica on March 25, 2010, 12:00:38 am
Post by: astarmathsandphysics on March 25, 2010, 09:43:51 am
inelastic collisions momentum conserved, energy not conserved
Post by: ruby92 on March 25, 2010, 10:38:57 am
find the velocity of the boron nucleus if the collison is perfectly elastic?
Post by: A.T on March 25, 2010, 12:58:07 pm
a nucleus of mass 1.67*10^-27 kg moving with a velocity of 2*10^4 collides head on with a boron nucleus of mass 1.7*10^-27kg orginally at rest.Data you know:
find the velocity of the boron nucleus if the collison is perfectly elastic?
m1=1.67*10^-27 kg
m2=1.7*10^-27 kg
u1=2*10^4 ms-1
u2=0
In elastic the total momentum(mv) and the energy(1/2 mv^2) is conserved.
For conserved momentum u have eq:(denote as eq 1 )
m1u1+ m2u2 = m1v1 + m2v2
And for energy conserved the eq is: (denote as eq 2)
(1/2) m1 u1² + (1/2) m2 u2² = (1/2) m1 v1² + (1/2) m2 v2²
now substitute given values in both eq (1 and 2) and solve them simultaneously and u'll get values of v1 and v2. v2 will be the velocity of the boron nucleus.
this link can help u for your question: http://answers.yahoo.com/question/index?qid=20081104122100AABK0SO
If collisions were INelastic then:(Just for ur concept)
m1=1.67*10^-27 kg
m2=1.7*10^-27 kg
u1=2*10^4 ms-1
u2=0
Using m1u1+ m2u2 = m1v1 + m2v2
as the collision is perfectly elastic the final velocity of both objects after collision would be zero,and knowing that u2=0 we get:
m1u1=V(m1+m2)
put the given values and you will get v = 9910 ms-1 (correct to 3 significant fig.)
Post by: ruby92 on March 25, 2010, 01:02:50 pm
its a head on collison
how would u find the momentum transferred from one block to the other. if its elastic? and if its inelastic?
tnkx :)
Post by: astarmathsandphysics on March 25, 2010, 01:49:53 pm
Post by: A.T on March 25, 2010, 02:19:46 pm
ok a mass of 8kg moving at 6ms-1 towards another block of mass 4kg at 3ms-1If inelastic: (they would become one particle(join) after collision and will have the same velocity)
its a head on collison
how would u find the momentum transferred from one block to the other. if its elastic? and if its inelastic?
tnkx :)
formula for inelastic:
m1u1 + m2u2 = V(m1+m2)
Use it to find V(the combined velocity) multiply it by masses to get the momentum.
For elastic u'll have to do all the long working shown in last last question,but as far as i know in A-levels they give u a value of one of the final velocity which makes it easier and u can simply find it using
m1u1+ m2u2 = m1v1 + m2v2
Post by: astarmathsandphysics on March 25, 2010, 02:42:49 pm
You do need the speed of one of the particles after the collision, or a condition - the become one particle so have the same speed, one has twice the speed of the other, one particle comes to rest etc so you can use the formula Impulse=m(v-u)
Post by: ruby92 on March 25, 2010, 05:36:38 pm
What is the tension of the rope?
(a) 55.0 N
(b) 92.4 N
(c) 147.4 N
(d) 176.6 N
(e) 294.6 N
explained as simple as possible :(
Post by: ruby92 on March 25, 2010, 08:49:36 pm
could someone please solve this fast :( :(
Post by: A.T on March 25, 2010, 09:10:26 pm
A 30 kg block (M1) is placed on a frictionless plane that inclines at a 30° angle with respect to the surface of Earth. This block is connected to another 20 kg block (M2) via a weightless rope over a frictionless, ideal pulley. The second block is hanging vertically.
What is the tension of the rope?
(a) 55.0 N
(b) 92.4 N
(c) 147.4 N
(d) 176.6 N
(e) 294.6 N
explained as simple as possible :(
Resolving forces on the 30 kg mass:
T-30gsin(30)=ma
T-30gsin(30)=30a ....1
Resolving forces on 20 kg mass:
W-T=ma
20g-T=20a.....2
solving 1 and 2 simulataneously:
T-30gsin(30) + 20g-T = 30a + 20a
a=0.981 ms-2
T=20g-20a from 2
T=20*9.81 - 21(.981)
T=175.599=175.6 N
Hence answer is C
A 10-kg sign is suspended by three ropes, each supporting an equal portion of the sign's weight. The two end ropes make an angle of 70º to the horizontal. What is the tension on each of the ropes?
could someone please solve this fast :( :(
According to question it seems the center rope in vertical.
IF all ropes were vertical the tension in each string would be:
3T=W
3T=10g
T=100/3 N
Hence the center rope has a tension of 100/3 N
now for the end ropes, as they are at 70 degree angle both their vertical component is:
Tsin(70)=100/3
T=35.5N
hence tension in the both end rope is 35.5N
Post by: ruby92 on March 25, 2010, 09:16:31 pm
isnt it d??
Post by: ruby92 on March 25, 2010, 09:20:26 pm
By the way Thanks a lot :D
Post by: A.T on March 25, 2010, 09:27:33 pm
c are u sure?from my concpets it IS C.
isnt it d??
Post by: A.T on March 25, 2010, 09:29:42 pm
why did u take the vertical component???
By the way Thanks a lot :D
It depends if tension was in horizontal direction you would have to resolve horizontally,but now its vertical so we will have to resolve vertically and hence find the vertical component.
Post by: ruby92 on March 25, 2010, 09:32:29 pm
thkxxx again
Post by: ruby92 on March 26, 2010, 05:19:52 pm
in a force extension graph the area under the graph represents strain energy only upto the elastic limit?
Post by: astarmathsandphysics on March 26, 2010, 05:30:42 pm
Post by: halosh92 on March 26, 2010, 08:16:33 pm
URGENT
thx
Post by: A.T on March 26, 2010, 08:51:08 pm
could someone explain briefly which formulas do we use exactly for elastic and inelastic collisionsFor elastic:
URGENT
thx
m1u1+m2u2=m1v1+m2v2 OR you can use the energy conservation formula
for inelastic
m1u1=m2u2=V(m1+m2)
Post by: ruby92 on March 26, 2010, 09:12:02 pm
1.When a battery in a toy car delivers a current of 0.75A, the p.d across its terminals is 1.3 V. when the car's motion is reversed the current delivered by the battery and the p.d. across its terminals becomes 12A and o.80v respectively.
calculate for the battery
a)the emf
b)the internal resistance
2.physics oct nov 2005 q6 part A(explained in detail)
Post by: astarmathsandphysics on March 26, 2010, 09:23:40 pm
Post by: ruby92 on March 26, 2010, 09:26:51 pm
Post by: astarmathsandphysics on March 26, 2010, 09:27:23 pm
for perfectly inelastic collsions - bodies stick together
Post by: astarmathsandphysics on March 26, 2010, 09:29:09 pm
Post by: ruby92 on March 26, 2010, 09:37:54 pm
there is no diagram or anything
ive attached the paper with the second problem.
its q6 part a
Post by: A.T on March 26, 2010, 10:28:06 pm
the first question...it isnt from a worksheet the physics teacher gave us...LOL wut? :P
And for no oct/nov 2005 Q
Similarity is that they both have same wavelengths which is 3.
difference is phase difference the one is a sine wave the other is a cosine wave and they have a phase difference of pie(180 degree)(half wave)
Post by: ruby92 on March 26, 2010, 10:32:28 pm
and its Q6 not 5 part a :)
Post by: astarmathsandphysics on March 26, 2010, 10:44:04 pm
Is this the right question? V Simple
Post by: ruby92 on March 26, 2010, 10:56:10 pm
Post by: halosh92 on March 27, 2010, 06:41:01 am
electricity question:
1.When a battery in a toy car delivers a current of 0.75A, the p.d across its terminals is 1.3 V. when the car's motion is reversed the current delivered by the battery and the p.d. across its terminals becomes 12A and o.80v respectively.
calculate for the battery
a)the emf
b)the internal resistance
2.physics oct nov 2005 q6 part A(explained in detail)
whats the answer to the first question though????????
Post by: astarmathsandphysics on March 27, 2010, 07:11:12 am
Post by: ruby92 on March 27, 2010, 08:51:58 am
where we all end up ???
Post by: ruby92 on March 27, 2010, 11:14:24 am
Question:
1. State kirchoff's laws. use the laws to deduce the laws of Ia and Ib as shown in the figure.
The circuit is here:
http://tinypic.com/r/t5j3iu/5 (http://tinypic.com/r/t5j3iu/5)
2.the circuit is attached.We need to find I1 I2 and I3
could someone solve these immediately
i have an exam tommorrow :( :(
Post by: halosh92 on March 27, 2010, 12:24:44 pm
I1= 0.44
I2=0.39
I3=0.834
?????
Post by: astarmathsandphysics on March 27, 2010, 01:40:15 pm
Post by: astarmathsandphysics on March 27, 2010, 02:21:08 pm
I2=2I3 whe I2 is current in 2 ohm resistor etc by symmetry
2I2+3I2=3 so I2=3/5A IA=IB=3/5A
Post by: ruby92 on March 27, 2010, 02:42:54 pm
V2+V3=3this is the answer to the first question?
I2=2I3 whe I2 is current in 2 ohm resistor etc by symmetry
2I2+3I2=3 so I2=3/5A IA=IB=3/5A
Post by: astarmathsandphysics on March 27, 2010, 02:44:36 pm
V=IR so 1=6I so I=1/6A
similarly I2=1/5 and I3=1/5+1/6 sine the currents are opposed
Post by: ruby92 on March 27, 2010, 03:22:44 pm
in the circuit a third current -Ia-Ib is there
for the first loop going in an anti clockwise direction
2Ia+3(-Ia-Ib)-2=0
2Ia-3Ia-3Ib-2=0.
-1Ia-3Ib-2=0.....(1)
in the second loop.
going anticlockwise
-2Ib-3(-Ia-Ib)+2=0
-2Ib+3Ia+3Ib+2=0
1Ib+3Ia+2=0....(2)
solving simulatneously.....
Ib as -o.5
and Ia by substitution as -0.5
what exactly have i done wrong?
Post by: Phosu on March 27, 2010, 04:30:33 pm
haha heres how i did it.
Ia + Ib = Ic , heres the basic mistake which you did, which caused the whole thing to go rong.
Ia be a, and Ib be b
2-2a-3Ic=0
2-2b-3Ic=0
any 2 above can be used as equation 1
2-2a-3Ic=2-2b-3Ic
3Ic and 2 will get cancelled
you get an equation -2a+2b=0, this will be used as equation 2
now simultaneously solve
i will use loop a, and split Ic current
2-2a-3(a+b)=0
2-2a-3a+3b=0
1)2-5a+3b=0
2)-2a+2b=0
and voila! :D
Post by: ruby92 on March 27, 2010, 04:35:24 pm
and wht have i done wrong??
Reply fast
Post by: Phosu on March 27, 2010, 04:44:15 pm
Post by: ruby92 on March 27, 2010, 04:47:18 pm
wht did i do wrong???
Post by: Phosu on March 27, 2010, 04:56:31 pm
Post by: ruby92 on March 27, 2010, 05:03:29 pm
ok soo why is Ic =Ia+Ib
???are a and b going out of the junction?
Post by: ruby92 on March 27, 2010, 05:12:27 pm
a) What is the wavelength of the sound?
0.5m
1m
1.5m
2m
b) Ignoring the reduction of amplitude with distance from each speaker, what is the amplitude of the sound;
i) halfway between the speakers?
0
1 A
2 A
3.142 A
ii) 0.5m from either speaker?
0
1 A
2 A
3.142 A
iii) 1m from either speaker?
0
1 A
2 A
3.142 A
Post by: Phosu on March 27, 2010, 05:25:34 pm
:-[ sorryif you complete the loop of both Ia and Ib you will find that they both meet up before entering the 3ohm resistor.
ok soo why is Ic =Ia+Ib
???are a and b going out of the junction?
so Ia+Ib=Ic
i guess the going out part was just to create confusion
Post by: halosh92 on March 27, 2010, 05:27:52 pm
Two loudspeakers both emit sounds with a frequency of 340Hz. Both sounds have equal amplitude and phase. The amplitude is A. The speakers are 4m apart and face each other. (The speed of sound in air is 340ms-1)
a) What is the wavelength of the sound?
0.5m
1m
1.5m
2m
b) Ignoring the reduction of amplitude with distance from each speaker, what is the amplitude of the sound;
i) halfway between the speakers?
0
1 A
2 A
3.142 A
ii) 0.5m from either speaker?
0
1 A
2 A
3.142 A
iii) 1m from either speaker?
0
1 A
2 A
3.142 A
which paper is this?
Post by: ruby92 on March 27, 2010, 05:48:23 pm
http://www.acoustics.salford.ac.uk/feschools/waves/super.htm (http://www.acoustics.salford.ac.uk/feschools/waves/super.htm)
Post by: halosh92 on March 27, 2010, 06:14:30 pm
this is from this websitethe first part its 1m
http://www.acoustics.salford.ac.uk/feschools/waves/super.htm (http://www.acoustics.salford.ac.uk/feschools/waves/super.htm)
all the other parts have the same amplitude
but which amplitude i have no idea
Post by: ruby92 on March 27, 2010, 06:34:19 pm
Post by: ruby92 on April 19, 2010, 08:24:13 pm
Post by: nid404 on April 19, 2010, 08:25:57 pm
F=1800X2=3600N
yaa?
Post by: Saladin on April 19, 2010, 08:45:37 pm
F=ma
F=1800X2=3600N
yaa?
That is correct if you assume that the tension is unifor in the rod connecting the two, and it is paralell to the ground.
Post by: T.Q on April 19, 2010, 11:30:46 pm
its 2A then 0A then 2 A
the answer is 0A THEN 0A THEN 0A
Post by: nid404 on April 20, 2010, 07:09:59 am
That is correct if you assume that the tension is unifor in the rod connecting the two, and it is paralell to the ground.
I guess cuz nothing's mentioned in the question
Post by: halosh92 on April 20, 2010, 11:35:15 am
-''Show an understanding of the origin of the upthrust acting
on a body in a fluid''
what does that mean???
-''Recognise that, for a perfectly elastic collision, the relative
speed of approach is equal to the relative speed of
separation''
and could someone briefly explain this and give an example, how are the questions diiferent from the normal m1u1 +m2u2=m1v1 + m2v2
??????? :-[ :-\
Post by: astarmathsandphysics on April 20, 2010, 11:42:02 am
Post by: ruby92 on April 20, 2010, 05:39:51 pm
V of approach=V of seperation..so v1+u1=v2+u2
Post by: ruby92 on April 20, 2010, 07:32:10 pm
is this right?
Post by: Saladin on April 20, 2010, 08:08:43 pm
Post by: Amr Fouad on April 20, 2010, 09:35:14 pm
the ultimate tensile stress is a point before the fracture/breaking point.where the materia "gives" or "yields'??
is this right?
nope..only the first part is right..UTS is NOT the yield point..the yield point is the point where plastic deformation starts...UTS is simply the stress where the material breaks..
Post by: ruby92 on April 23, 2010, 05:50:32 pm
if anyone has notes on superposition and resonance particularly explaining the wavelenght and lenght if you could post it here it wld be realli helpful :)
Post by: ~ A.F ~ on April 23, 2010, 06:00:47 pm
what wld a displacement position graph for a longitudnal wave look like?
if anyone has notes on superposition and resonance particularly explaining the wavelenght and lenght if you could post it here it wld be realli helpful :)
Graph is an exact duplicate of the one for transverse wave
Notes: http://www.s-cool.co.uk/alevel/physics.html go to oscillations and waves part
Post by: ruby92 on April 24, 2010, 04:17:41 pm
O/N 2001 paper1
Q 2 and 15
Post by: astarmathsandphysics on April 24, 2010, 08:14:01 pm
Post by: ruby92 on April 24, 2010, 08:41:24 pm
i also have a doubt in M/J 2002 paper 1
2,15,26,28,35
Post by: astarmathsandphysics on April 24, 2010, 09:30:56 pm
Post by: astarmathsandphysics on April 24, 2010, 11:02:19 pm
mj 02
q2. B cos x+-y=z
15.d cos resulta of 3 and 4 is 5 to upper right
26. 1 period=4 squares =4*2ms=8ms
1/8ms=125Hz B
28. path difference =1/2 wavelength C
35.resistance in ratio of voltages accros resistance 2:1:2 C
Post by: ruby92 on April 26, 2010, 05:20:05 pm
could you please explain Q's 15 and 35 in more detail
Also o/n 2002 paper 1; 36
Post by: halosh92 on April 26, 2010, 06:18:51 pm
i dont get it , dont we have to use the fdsin thita formula here..shouldnt the distance be perpendicular to the force..
in the ms..they multiplied the distance (in the direction of X) by X
??????????????????URGENT
thx
Post by: ruby92 on April 26, 2010, 08:16:59 pm
34,40,18,30,35
Post by: ruby92 on April 26, 2010, 10:03:51 pm
12,7(why isnt it A how is it B?arent we supposed to find the gradient?)and 37,16
Post by: astarmathsandphysics on April 26, 2010, 10:09:59 pm
30.D since 2*2/3 -1/3 =1 charges must add up to 1
18.D since KE is propritional to v^2 F=Energy =braking force*distance=1/2mv^2
30.average current =(100+20)/2=60mA
charge=It=60mA*8=480mC
45. E is uniform between paralell plates
Post by: astarmathsandphysics on April 26, 2010, 10:12:56 pm
apply v=u+at vertically
12. 8*2+2*4=6v v=4
Post by: ruby92 on April 27, 2010, 08:05:10 pm
2.why not A? how do we know the direction of the resultant force?
could you please explain Q's 15 and 35 in more detail
Also o/n 2002 paper 1; 36
no 12(o/n 2003) ??? ???
Post by: astarmathsandphysics on April 27, 2010, 08:16:22 pm
Post by: ruby92 on April 27, 2010, 09:36:21 pm
Post by: ruby92 on April 27, 2010, 11:01:39 pm
8 (why D why not C)
9
31(why D and why not C)
O/N 2004 paper 1
20, 18,11,31,33,26.
Post by: astarmathsandphysics on April 28, 2010, 08:54:17 am
mj 2004 p1 q8 D - as ball speeds up acc falls as air resistance increases]
9. Work vertically s=ut+1/2at^2 u=0 cos he vertically speed is zero
1.25=0+1/2*10*t^2
t=0.5
work horizontally s=ut 10=u*0.5 u=20
31. R=pl/A and A=kd^2 so double the diameter mean divide resistance by 4. V=IR and V is same for wires in parallel so current in P is 4x cuurent in Q
Post by: astarmathsandphysics on April 28, 2010, 09:07:19 am
so 2xp(P)+xp(Q)=2xp(Q) so xp(P)=2p(Q) so p(P)1/2p(Q)
18.work dine=increase in potential energy+heat generated
9000*40=20000*12+H
H=120000J A
11.B m(v-u)=m(-v-v)=-2mv
31. C use P=VI
33.R=pl/A double the length means double R and double the diameter means 1/4 R so R is halved overall
V=IR so double V and hal;ve R mean I is 4x bigger
26 !6 I=kA^2
Post by: ruby92 on April 29, 2010, 09:27:31 pm
Q3 (C) part iii 2. how is the torque of this also 45?
Q5 of the same papers part c
m/j 2004 paper 2
1 (b)
Post by: ruby92 on April 29, 2010, 11:22:47 pm
can you post question and paper and year in same post
mj 2004 p1 q8 D - as ball speeds up acc falls as air resistance increases]
9. Work vertically s=ut+1/2at^2 u=0 cos he vertically speed is zero
1.25=0+1/2*10*t^2
t=0.5
work horizontally s=ut 10=u*0.5 u=20
31. R=pl/A and A=kd^2 so double the diameter mean divide resistance by 4. V=IR and V is same for wires in parallel so current in P is 4x cuurent in Q
i dont get 31 ??? ???
Post by: ruby92 on April 30, 2010, 01:05:45 am
5 b
3 B ii
1 b
o/n 2004 paper 2
1 b ii
m/j 2005
3 b we dont convert g to kg?
Post by: astarmathsandphysics on April 30, 2010, 08:17:58 am
on2004p2 %error =2*0.02/0.50*100=8%
mj2005 you should convert to kg it is good practice but units of grams cancels cos factor of 1000 occurs on both sides
Post by: ruby92 on April 30, 2010, 05:39:44 pm
3 c
Post by: nid404 on April 30, 2010, 07:21:28 pm
horizontal comp= 6.1 sin 35= 3.5N
Post by: ruby92 on April 30, 2010, 08:08:20 pm
the horizontal component is cos rite?
Post by: nid404 on April 30, 2010, 08:10:17 pm
Post by: halosh92 on April 30, 2010, 08:40:19 pm
paper 2
Q3b
shouldnt we consider the direction of the velocity in the formula???
and how did they know thats the time??
Post by: ruby92 on April 30, 2010, 08:44:56 pm
5 c im basically stuck at the unit conversion part here ??? ???
6 C of the same paper
7 a, b
Post by: astarmathsandphysics on April 30, 2010, 10:01:10 pm
1/2*1600(0.06^2-0.045^2)+1/2*1600(0.03^2-0.045^)=1/2*0.85=v^2
Post by: astarmathsandphysics on April 30, 2010, 10:06:10 pm
Post by: astarmathsandphysics on April 30, 2010, 10:11:33 pm
A will blow if a power supply is use
Post by: ruby92 on May 01, 2010, 12:02:39 am
0.5cm above top of tuve since w/2=16.2
where did u get the 16.2 from?
o/n 2006
4 ii 1?
Post by: halosh92 on May 01, 2010, 07:26:09 am
2002 nov physics AS cie
paper 2
Q3b
shouldnt we consider the direction of the velocity in the formula???
and how did they know thats the time??
could solve this please???? ;D ;D ;D
Post by: astarmathsandphysics on May 01, 2010, 07:40:56 am
Post by: nid404 on May 01, 2010, 07:45:21 am
could solve this please???? ;D ;D ;D
Yes of course you consider the direction of motion.
check the attachment
momentum= 0.045X 4.2 - (0.045X -3.6) = 0.35 Ns
Post by: halosh92 on May 01, 2010, 08:06:51 am
Yes of course you consider the direction of motion.
check the attachment
momentum= 0.045X 4.2 - (0.045X -3.6) = 0.35 Ns
thxx aloot nid gr8 explanation!
Post by: nid404 on May 01, 2010, 08:11:15 am
Post by: ruby92 on May 02, 2010, 04:35:22 pm
i would really appreciate is someone would draw these graphs and post them here
M/j 2004 2 b
I need this ASAP
Post by: astarmathsandphysics on May 02, 2010, 07:37:59 pm
Post by: ruby92 on May 02, 2010, 08:02:52 pm
Post by: ruby92 on May 02, 2010, 08:36:40 pm
a i in the ms the take s=12.8.
but it says the speed of the car before brakes are applied ??? ???
also isnt u=v (the speed with which the car is travelling) and v=0???
for the ii part why cant we use one of the formulas for acceleration?
Post by: halosh92 on May 03, 2010, 08:02:23 am
for beta , alpha and gamma rays:
-which one deflects more in an electric field and why?
-and which one ionises more and why?
-and when we right the decay formulas, do we show the charge of the beta electron or alpha?
a-re all the 3 radiation coming from the nucleus?
thx
Post by: Nobody on May 03, 2010, 08:26:45 am
i have a couple of questions:
for beta , alpha and gamma rays:
-which one deflects more in an electric field and why?
-and which one ionises more and why?
-and when we right the decay formulas, do we show the charge of the beta electron or alpha?
a-re all the 3 radiation coming from the nucleus?
thx
1) A beta particle is deflected more that an alpha particle in an electric field. This is because a beta particle ( a high energy, fast moving electrons) is much smaller and lighter than an alpha particle( a helium atom).
2) Alpha particles are strongly ionising but can be stopped by paper or skin.Because they have a strong positive charge (+2) and a mass of 4 (i.e. 4 times the mass of a proton)
and yes....all three radiations are originated from nucleus of the atom.
Post by: halosh92 on May 03, 2010, 09:09:48 am
1) A beta particle is deflected more that an alpha particle in an electric field. This is because a beta particle ( a high energy, fast moving electrons) is much smaller and lighter than an alpha particle( a helium atom).
2) Alpha particles are strongly ionising but can be stopped by paper or skin.Because they have a strong positive charge (+2) and a mass of 4 (i.e. 4 times the mass of a proton)
and yes....all three radiations are originated from nucleus of the atom.
thankyou ;D
Post by: ruby92 on May 03, 2010, 09:57:13 am
o/n 2008 paper 2 physics
a i in the ms the take s=12.8.
but it says the speed of the car before brakes are applied ??? ???
also isnt u=v (the speed with which the car is travelling) and v=0???
for the ii part why cant we use one of the formulas for acceleration?
this is from paper 2 CIE physics
i would really appreciate is someone would draw this graphs and post them here
M/j 2004 2 b
I need this ASAP
Post by: AN10 on May 03, 2010, 12:50:39 pm
Post by: Nobody on May 04, 2010, 06:15:43 am
plz explain mcq 27 of the paper attached.
answer is c
Post by: AN10 on May 04, 2010, 09:00:15 am
Post by: astarmathsandphysics on May 04, 2010, 10:50:02 am
E=F/q
Post by: nid404 on May 07, 2010, 06:40:50 am
I skipped this one...now im confused
Post by: cashem'up on May 07, 2010, 10:09:03 am
Post by: nid404 on May 07, 2010, 10:18:44 am
then it's ok :D :P
Post by: cashem'up on May 07, 2010, 11:14:25 am
u sure?
then it's ok :D :P
yep m 100% sure checked the ms and ER......no doubt abt..lolzz........by the way nid404 which subs have u taken for as levels
Post by: nid404 on May 07, 2010, 11:15:32 am
You?
Post by: cashem'up on May 07, 2010, 11:17:57 am
Post by: nid404 on May 07, 2010, 11:20:36 am
I give eng in nov ;D
Post by: cashem'up on May 07, 2010, 11:26:17 am
Post by: holtadit on May 07, 2010, 11:27:12 am
Yes it is :P
I give eng in nov ;D
WHy you taking english ?
Post by: nid404 on May 07, 2010, 11:28:21 am
WHy you taking english ?
Am I not supposed to ?
Post by: cashem'up on May 07, 2010, 11:37:15 am
Am I not supposed to ?haha.......funny....
have you already given eng or ur gonna give this nov.......
Post by: nid404 on May 07, 2010, 12:36:56 pm
haha.......funny....
have you already given eng or ur gonna give this nov.......
ohk....that wasn't meant to be funny ::)
This nov
Post by: cashem'up on May 07, 2010, 12:40:08 pm
funny in the sense.... a smart reply..........
Post by: nid404 on May 07, 2010, 12:42:23 pm
Not really...I would've preferred to have done with it this session. A2 isn't child's play...not for the faint hearted
Post by: halosh92 on May 07, 2010, 02:43:24 pm
THX ALOT
Post by: cashem'up on May 07, 2010, 03:15:52 pm
lol...thank you thank you :D
Not really...I would've preferred to have done with it this session. A2 isn't child's play...not for the faint hearted
fingers crossed...... i got a lot of A2 to cover and m plannin to get over it by nov......god knows wat will hapen
Post by: cashem'up on May 07, 2010, 03:17:15 pm
does anyone have paper 1 for 2001 november ????PLZ RILI URGENT
THX ALOT
same prob i had i have the other papers but not p1 for 2001 nov
Post by: Phosu on May 07, 2010, 04:35:50 pm
What is the resistance of this lamp at room temperature?
A)36? B)580? C)1.5k? D)9.2k?
Post by: T.Q on May 07, 2010, 04:42:53 pm
The filament of a 240 V, 100W electric lamp heats up from room temperature to its operating temperature. As it heats up, its resistance increases by a factor of 16.
What is the resistance of this lamp at room temperature?
A)36? B)580? C)1.5k? D)9.2k?
D IN THE ANSWER
THEN R x 16
Post by: Phosu on May 07, 2010, 04:52:33 pm
Post by: Phosu on May 07, 2010, 05:03:25 pm
D IN THE ANSWERrong ans is A
THEN R x 16
Post by: Phosu on May 07, 2010, 05:11:03 pm
2?. No change of the total volume occurs.
What is the density of the liquid mixture?
A)4/3? B)3/2? C)5/3? D)3?
Post by: halosh92 on May 07, 2010, 05:14:13 pm
how do we find the resistances?
see
the voltage at R3 is 2V
at R2 its 1V
at R1 its 2V (because all add up to 5V)
therefore R3 and R1 have same voltage therefore same resistance so its C
Post by: halosh92 on May 07, 2010, 05:16:48 pm
The filament of a 240 V, 100W electric lamp heats up from room temperature to its operating temperature. As it heats up, its resistance increases by a factor of 16.
What is the resistance of this lamp at room temperature?
A)36? B)580? C)1.5k? D)9.2k?
P=IV
100/240= I (its in decimals so i wont write it)
R=V/I
R=240/I =576
576/16= 36
Post by: halosh92 on May 07, 2010, 05:18:55 pm
A mass of a liquid of density ? is thoroughly mixed with an equal mass of another liquid of density
2?. No change of the total volume occurs.
What is the density of the liquid mixture?
A)4/3? B)3/2? C)5/3? D)3?
whats the "?" in ure doubt..retype ure question plz
Post by: Phosu on May 07, 2010, 05:46:56 pm
A mass of a liquid of density "p" is thoroughly mixed with an equal mass of another liquid of density
2?. No change of the total volume occurs.
What is the density of the liquid mixture?
A)4/3p B)3/2p C)5/3p D)3p
Post by: Phosu on May 07, 2010, 05:48:59 pm
see
the voltage at R3 is 2V
at R2 its 1V
at R1 its 2V (because all add up to 5V)
therefore R3 and R1 have same voltage therefore same resistance so its C
how did u find these voltages?
Post by: halosh92 on May 07, 2010, 05:58:53 pm
how did u find these voltages?
look at the diagram and treat each resistor as if its alone like first resistor its between the 0V and 2V
and second resistor its between 2V and 3V ...basically 3-2=1V
and the 3rd ...we know that total voltage is 5V its between 3V and 5V therefore 5-3=2V
got it?
Post by: 33oomar on May 07, 2010, 06:21:37 pm
Post by: Saladin on May 07, 2010, 06:25:03 pm
i got a proplem in IGCSE physics paper 6 May/june 08 Q4 C
What do you not understand about this?
Post by: 33oomar on May 07, 2010, 06:26:48 pm
please help me
Post by: Phosu on May 07, 2010, 06:36:16 pm
more questions!
Nov 02 pp1
9, 11, 12, 14, 17, 28, 36
Post by: 33oomar on May 07, 2010, 06:37:35 pm
i was wondering can i plot a broken x axis on a graph or not
Post by: halosh92 on May 07, 2010, 06:51:45 pm
got it Thanks!q9) s=ut+ 1/2at^2
more questions!
Nov 02 pp1
9, 11, 12, 14, 17, 28, 36
s=1/2at^2
2h=at^2
2h= a (t2 ^2 - t1^2)
2h/(t2 ^2 - t1^2) =a
11) speed of approach=speed of separation
uA - uB = vA - vB
different directions lets take this direction -----------> as (+)
so
uA-(-uB)=vA + vB
12) ------------> take as (+)
m1u1 + m2u2= m1v1 +m2v2
m1 *60 + (m2 * -30) = m1v +m2v (v is same)
cancel out the "m"
60-30 = 2v
30/2=15
Post by: nid404 on May 07, 2010, 07:00:09 pm
got it Thanks!
more questions!
Nov 02 pp1
9, 11, 12, 14, 17, 28, 36
Q12) 60m-30m= 2mv
30m=2mv
v=15cm/s
Q14) 50x +100X10= 20X60
50x=200
x=4cm
+4cm from the pivot=44cm
Q17) no change in k.e because velocity is constant
Q28 and 36 I've also explained this elsewhere...will look
Post by: halosh92 on May 07, 2010, 07:08:51 pm
got it Thanks!
more questions!
Nov 02 pp1
9, 11, 12, 14, 17, 28, 36
14) 50g is on the other side
therefore clockwise = anticlockwise let z be the unknown distance so:
(20g * 60/100) + ( 100g * 10/100) = (50g * z)
g=9.8 and convert cm to m
so z=44 cm
17) constant speed , kinetic energy= 1/2mv^2
no change in velocity so no chang ein kiinetic energy therfore change=0
28) lamda = ax/D
increase lambda , increase x
36) find current first so:
first loop and 2nd loop total resistance (15*15/15+15) =7.5
so total resistance = 7.5
total Voltage = 2V
R=V/I
I=2/7.5 =0.2666666
THATS TOTAL CURRENT.......current in one loop divide by 2 = 0.133333
so IR=V
0.1333 * 5 =0.6666
and check the choices 2/3= 0.666666
so its A
Post by: halosh92 on May 07, 2010, 07:09:36 pm
Q12) 60m-30m= 2mv
30m=2mv
v=15cm/s
Q14) 50x +100X10= 20X60
50x=200
x=4cm
+4cm from the pivot=44cm
Q17) no change in k.e because velocity is constant
Q28 and 36 I've also explained this elsewhere...will look
i already did nid :)
Post by: nid404 on May 07, 2010, 07:13:25 pm
Post by: Phosu on May 07, 2010, 08:36:28 pm
11) speed of approach=speed of separation
uA - uB = vA - vB
different directions lets take this direction -----------> as (+)
so
uA-(-uB)=vA + vB
could you plz explain the difference between speed of approach and speed of seperation. how do we know that they are same?
wen and how can we use this formula? and why cant we just use m1u1+m2u2=m1v1+m2v2
plus wen u solve the ans u've given, it gives uA+uB=vA+vB but the mark scheme says its uA+uB=vA-vB
36) find current first so:
first loop and 2nd loop total resistance (15*15/15+15) =7.5
so total resistance = 7.5
total Voltage = 2V
R=V/I
I=2/7.5 =0.2666666
THATS TOTAL CURRENT.......current in one loop divide by 2 = 0.133333
so IR=V
0.1333 * 5 =0.6666
and check the choices 2/3= 0.666666
so its A
i got till the total current part, but why did u take the resistance as 5 in the next step?
Thanks alot for solving :)
Post by: halosh92 on May 07, 2010, 08:46:29 pm
could you plz explain the difference between speed of approach and speed of seperation. how do we know that they are same?
wen and how can we use this formula? and why cant we just use m1u1+m2u2=m1v1+m2v2
plus wen u solve the ans u've given, it gives uA+uB=vA+vB but the mark scheme says its uA+uB=vA-vB
i got till the total current part, but why did u take the resistance as 5 in the next step?
Thanks alot for solving :)
because the resistor is between the terminal look at the second loo[, and view the resistor as a single component in a battery..........nid will explain q11 better than me
np POSH ;D ;D ;D
Post by: nid404 on May 07, 2010, 09:24:38 pm
In an elastic collision...momentum as well as kinetic energy is conserved,,
conservation of momentum
total momentum before collision=total momentum after collision.
mu1+mu2=mv1+mv2
conservation of k.e
1/2mu12+ 1/2mu22= 1/2mv12+1/2mv22
cancel all the halfs first, then cancel all the 'm' terms.
in steps...here's how u do it
m(u1-v1)=m(v2-u2) 1)
m(u12-v12)= m( v22-u22) 2)
since (u12-v12)= (u1-v1)(u1+v1)
& ( v22-u22)=(v2-u2)(v2+u2)
divide the 2nd eqn by the first
m(u12-v12) m( v22-u22)
---------------------------------= ------------------------------------
m(u1-v1) m(v2-u2)
m [(u1-v1)(u1+v1)]=[m(v2-u2)(v2+u2)]
------------------- ------------------
m(u1-v1) m(v2-u2)
and then cancelling gives
(u1+v1)=(v2+u2)
or
u1-u2=v2-v1
now coming back to the question.
u1-u2 in this case will be positive because u for the other body is in the opp direction
u1-(-u2)=u1+u2
v2-v1 will be as it...both move in the same direction after collision
so u1+u2=v2-v1
that was tiring....i really hope you got it
Post by: Phosu on May 07, 2010, 09:50:27 pm
it was really helpful! :)
Post by: astarmathsandphysics on May 07, 2010, 09:59:24 pm
Post by: nid404 on May 08, 2010, 09:12:27 am
Post by: Phosu on May 08, 2010, 11:51:35 am
5, 35 and 36
Post by: halosh92 on May 08, 2010, 12:24:37 pm
Jun 03 pp1
5, 35 and 36
35) C
because electric field strength is the same at any point between the plates, so its constant
36) D
E=V/d
E=700/(5/1000)
mm to m
therefore its D and direction is always from (+) to (-)
Post by: neno on May 09, 2010, 02:05:19 pm
i have the answers but i dont understand them
#Q9-mayjune 2004
#Q25,Q20,Q11-octnov 2004
#Q7-mayjune 2005
#Q34,Q33-octnov 2005
Post by: nid404 on May 09, 2010, 05:25:34 pm
Q9) s= v2 sin 2theta / g tan theta=1.25/10
10= v2 sin (2X7.125)/ 9.8
v2= 10/ 0.025=398.125
v = root of 398.125= 20
Nov 04
Q25) wavelength= 2 times distance between corresponding antinodes. Hence C
Q11) mv-(-mv)=2mv to the left. opp direction
Q20) A density of P is half that of Q...It rises twice that of Q
Post by: nid404 on May 09, 2010, 05:33:36 pm
---> I have no idea :-[
Nov 05
Q33) D What do you not understand here?
Q34) R=v/I
the gradient at point C is least steep. indicates V/I is least...resistance is least
Post by: neno on May 09, 2010, 06:14:08 pm
r u using the formula 2as=v2-u2 do u take u2 at 0
2as=v sin theta
but y do u take 2 theata??
Post by: nid404 on May 09, 2010, 06:20:14 pm
s= v2 sin 2
Post by: neno on May 09, 2010, 06:28:38 pm
lol i got Q33 when i read it 2wice
Thanks alot By the way
Post by: neno on May 09, 2010, 06:32:07 pm
v nvr used this formula so wat are the conditions tht u use it under
when thrs in angle or it requires more
is it used in projectiles also when u wana find the velocity ??
Post by: halosh92 on May 09, 2010, 08:01:09 pm
in octnov 2004 Q25 y cant it b the frequency cuz then v already noe the speed of sound n v cud jst simply devide..
lol i got Q33 when i read it 2wice
Thanks alot By the way
i believe thats because its a stationary wave
Post by: neno on May 10, 2010, 05:15:56 pm
Post by: Phosu on May 11, 2010, 12:21:04 pm
What is the angle between the first and second order diffraction maxima?
A 22.0° B 26.6° C 44.0° D 48.6°
Post by: astarmathsandphysics on May 11, 2010, 12:29:12 pm
28 microwaves?
Where is the question from
Post by: Phosu on May 11, 2010, 12:34:15 pm
i also have a few more doubts
4, 6, 14, 16,
Post by: vanibharutham on May 11, 2010, 12:34:23 pm
for the first order maxima, i.e. the central maxima
8 sin (theta) = 3 x 1
Theta = 22.04
confirmed :)
For the second order maxima,
8 sin (theta) = 2 x 3
theta = 48.6
therefore, the angle between them is
48.6 - 22 = 26.6
Therefore B