Post by: halosh92 on June 02, 2010, 08:20:35 am
Q4d
nov 2005
Q4ai, ii
in the ms its 42+/- 4 cm/s
dont we have to change it to meters ?
and where did they get the +/-4 from
similarly for the second part, where did they get that uncertainity from?
Post by: Meticulous on June 02, 2010, 08:42:16 am
nov 2007
Q4d
It's exactly like the steel and concrete combination. Why is concrete stronger in compression and weaker in tension? Because in tension the extensions of top and bottom layers vary, and so it shatters. The opposite goes for steel.
The exact same case here. However, the difference in extensions in the thicker rod is greater, and so it breaks easier. i.e require less bending.
Post by: Meticulous on June 02, 2010, 08:49:39 am
nov 2005
Q4ai, ii
in the ms its 42+/- 4 cm/s
dont we have to change it to meters ?
and where did they get the +/-4 from
similarly for the second part, where did they get that uncertainity from?
No you don't have to change them to meters. Actually, it's preferable to keep them as it's given to you! It's given to you in cm, then keep it in cm. BUT, if you change it to meters, it would still be right. However, the probability of making an error is increased as now you have to divide by a 100 (you could type 10 in your calculator)
And, for the second part, it's not uncertainty. It's just the range of ACCEPTABLE values by students, as your answer varies from another's. It's not for you, it's for the examiners. As long as your answer lies in the range, you're safe.
Post by: Meticulous on June 02, 2010, 08:50:20 am
Post by: halosh92 on June 02, 2010, 08:59:01 am
No you don't have to change them to meters. Actually, it's preferable to keep them as it's given to you! It's given to you in cm, then keep it in cm. BUT, if you change it to meters, it would still be right. However, the probability of making an error is increased as now you have to divide by a 100 (you could type 10 in your calculator)
And, for the second part, it's not uncertainty. It's just the range of ACCEPTABLE values by students, as your answer varies from another's. It's not for you, it's for the examiners. As long as your answer lies in the range, you're safe.
so basically the +/-4 is not the uncertainity.
and for this question the final answer MUST be in m.
so i have to divide 42 by 100 ..correct?
THX A BUNCH ;D
Post by: Meticulous on June 02, 2010, 09:07:23 am
so basically the +/-4 is not the uncertainity.
and for this question the final answer MUST be in m.
so i have to divide 42 by 100 ..correct?
THX A BUNCH ;D
Yes. +/- 4 is not part of the uncertainty. They will ask for it DIRECTLY in the exam.
And NO! You got me totally wrong! LOL I said it's preferable to keep it in CM, as it is given to you in the question, so ther's a less chance of making errors!
Was I clear, again? :P
And don't thank me, it's my job ;)
Post by: halosh92 on June 02, 2010, 09:44:08 am
Yes. +/- 4 is not part of the uncertainty. They will ask for it DIRECTLY in the exam.
And NO! You got me totally wrong! LOL I said it's preferable to keep it in CM, as it is given to you in the question, so ther's a less chance of making errors!
Was I clear, again? :P
And don't thank me, it's my job ;)
hhahah i got u yep....but if u look at the question in the space where we have to write the answer the unit theyve given is meters xD
hahah and dude i cant help NOT thanking u :P
Post by: Meticulous on June 02, 2010, 09:46:26 am
hhahah i got u yep....but if u look at the question in the space where we have to write the answer the unit theyve given is meters xD
hahah and dude i cant help NOT thanking u :P
Oooppss.. my bad ;D then yes..your answer should be in meters if it's written to you.
And I know you cannt help thanking me 8) :P LOL JK..its alright :)
Post by: halosh92 on June 02, 2010, 04:01:35 pm
june 2005
p2
Q6bi
could someone draw and explain why the directions are in such a manner???
thx
Post by: nid404 on June 02, 2010, 04:57:32 pm
;D ;D
june 2005
p2
Q6bi
could someone draw and explain why the directions are in such a manner???
thx
I know that the +ve charge will experience a force towards the right. That is because electric field is from +ve plate to the negative plate. +ve charge will experience a pull towards the negative plate. The -ve charge just the opp.
The ms says correct directions with line of action of arrows passing through charges...Now I wonder if we have to show the force that the charge experiences due to the other. (force that once charge has on the other) :-\
Post by: halosh92 on June 02, 2010, 05:22:02 pm
I know that the +ve charge will experience a force towards the right. That is because electric field is from +ve plate to the negative plate. +ve charge will experience a pull towards the negative plate. The -ve charge just the opp.
The ms says correct directions with line of action of arrows passing through charges...Now I wonder if we have to show the force that the charge experiences due to the other. (force that once charge has on the other) :-\
ahh u c the next questions are about couples, so i guess its just these 2 forces, maybe we just have to extend the lines? ???
Post by: sizbeauty on June 02, 2010, 06:25:50 pm
Post by: halosh92 on June 03, 2010, 12:59:31 pm
Q4c
Post by: Meticulous on June 03, 2010, 01:07:51 pm
Post by: halosh92 on June 03, 2010, 01:22:04 pm
Please halosh, post the q paper.
there u go
Post by: Meticulous on June 03, 2010, 01:26:49 pm
Post by: halosh92 on June 03, 2010, 01:32:52 pm
Have you calculated the momentum before and after the collision? What were your answers?
nop i calculated the change in momentum, its 0.14
Post by: Meticulous on June 03, 2010, 01:34:02 pm
its not conserved
right?
Post by: halosh92 on June 03, 2010, 01:43:52 pm
Well then..
its not conserved
right?
actually the ms says :
plate (and Earth) gain momentum
equal and opposite to the change for the ball
so momentum is conserved
and sorry that was change in energy not momentum ;D
we didnt have to calculate momentum
Post by: Meticulous on June 03, 2010, 01:52:13 pm
The thing is..mometum is like energy...always conserved.
The q here was tricky, to the ball, it LOST momentum..but still the lost momentum went to earth
so basically its conserved
Right?
Post by: halosh92 on June 03, 2010, 01:59:42 pm
LOL no prblm
The thing is..mometum is like energy...always conserved.
The q here was tricky, to the ball, it LOST momentum..but still the lost momentum went to earth
so basically its conserved
Right?
alright ;)
thx ;)
Post by: Meticulous on June 03, 2010, 02:03:01 pm
Post by: halosh92 on June 04, 2010, 04:16:31 pm
Q3ci
is it towards or away from the disc?
Q4biii
is there a specific formula here or wat?
Post by: nid404 on June 04, 2010, 04:45:52 pm
2003 nov
Q3ci
is it towards or away from the disc?
away from the disc
Post by: halosh92 on June 04, 2010, 04:48:39 pm
away from the disc
how do we know ???
and could u do the other question too?
thx :)
Post by: nid404 on June 04, 2010, 05:00:08 pm
I'll be back after dinner :)
Post by: halosh92 on June 04, 2010, 05:02:46 pm
It forms a couple. Forces will act away from each other.
I'll be back after dinner :)
ooh ::)
thx nid ;D
also Q5c shouldnt be a straight CONSTANT line...as velocity is dependent on voltage which is unaltered in this case ? ???
Post by: nid404 on June 04, 2010, 06:35:29 pm
a=slit width
x=1/2 L
D= distance from screen
Post by: nid404 on June 04, 2010, 06:42:24 pm
Post by: halosh92 on June 04, 2010, 06:51:18 pm
5)c) Depends on the force i guess. I'm not too sure if the graph would be a linear straight line or whether it would be constant. Cause electric field strength is constant
the ms says its a curve :S
Post by: nid404 on June 04, 2010, 06:52:27 pm
the ms says its a curve :S
-___-
Wait for A.F :P
Post by: aloha32 on June 05, 2010, 11:41:29 am
I need help with the June 09 paper 2 part 1 question 5 b.I don't understand how we get 2 minimas.Can anyone help plz?
thanks ! :)
Post by: halosh92 on June 05, 2010, 12:41:20 pm
to calculate the power dissipated, in a series circuit and in a parallel circuit what do we do
do we just add up the powers?
Post by: nid404 on June 05, 2010, 12:44:04 pm
give an example?
Post by: halosh92 on June 05, 2010, 12:54:51 pm
didn't get you.
give an example?
like if we have a circuit with two loops
first loop contains 2 resistors
and second loop contains 1 resistor
each has powers of 2W
how do we calculate the total power......for only the 1st loop
and the total power for the whole circuit.
Post by: Meticulous on June 05, 2010, 05:43:39 pm
Post by: halosh92 on June 05, 2010, 07:50:26 pm
Q3bi
i know its length= thita * radius
but why do we add SIN to the thita ???
Post by: Meticulous on June 05, 2010, 07:52:42 pm
Post by: halosh92 on June 05, 2010, 08:03:32 pm
halosh..please post the q paper. or its link. for the 17000 time i do edexcel..lolhahahahah i did no THAT
there u go :D
Post by: *Hope* on June 05, 2010, 08:12:42 pm
2003 june 2003actually if u notice that a right angle triangle can be formed from the those two lines
Q3bi
i know its length= thita * radius
but why do we add SIN to the thita ???
the radius is 3\2 which is equal to 1.5
this will be the hyp of a right angle triangle
so the angle 6.5 can be used to find the opp of the triangle which will be the extension
so sin6.5*1.5 will be the extension(opp)
Post by: halosh92 on June 05, 2010, 08:19:09 pm
actually if u notice that a right angle triangle can be formed from the those two lines
the radius is 3\2 which is equal to 1.5
this will be the hyp of a right angle triangle
so the angle 6.5 can be used to find the opp of the triangle which will be the extension
so sin6.5*1.5 will be the extension(opp)
thx... ;) my head would break if something like this comes in the exam..how the hell r we supposed to think of this >:(
Post by: zxcvbnm on June 05, 2010, 08:23:30 pm
where x is the fringe separation, y is the wavelength, D is distance between slits and screen and a is slit separation.
please explain the effects on intensity of changing y,D and a....for y and D, i know that the intensity changes, but for a, the intensity remains the same....why is that?
Post by: *Hope* on June 05, 2010, 08:28:17 pm
thx... ;) my head would break if something like this comes in the exam..how the hell r we supposed to think of this >:(haha..I know!! I got it as a quiz once!! And omg!!! I just stoppped thinking!!lool!! But insA no worries hopefully nothing will come like this in the exam :)
Post by: halosh92 on June 05, 2010, 08:53:19 pm
same paper another question
Q5ai
its not a straight line so shouldnt we draw a tangent here?
and could u explain Q4b
Post by: *Hope* on June 05, 2010, 09:13:26 pm
yes exactly nshalla we should pray ;Dq5ai: u actually find the exact point and sub in the equation R=V/I
same paper another question
Q5ai
its not a straight line so shouldnt we draw a tangent here?
and could u explain Q4b
simply at 6v, the current is 40mA but u have to convert it to Amperes so 40*10^-3
the final ans is 150
4b: for i1...it is just guessing but as the ms said: 0.3mm to 3mm
i2 the general formula is x=Dh/a
bii1: as x is the seperation. if u double the D in the eq, the x will double so the separation increases
maximum brightness will be less cuz u r increasing the distance of the light to the screen so there will be less light reaching and so less max brightness
bii2: again as h is increased so if u sub in the eq , the x will increase
the max brightness will decrese..am not sure though of the reason..but as the h increases the frequency decreases so as intensity is A^2 or f^2, the intensity will decrease
bii3:as all r kept constant the x will remain constant which is the seperation
the maximum brightness increases as the intensity increases
Hope i helped and was clear... :)
Post by: halosh92 on June 05, 2010, 09:21:29 pm
q5ai: u actually find the exact point and sub in the equation R=V/I
simply at 6v, the current is 40mA but u have to convert it to Amperes so 40*10^-3
the final ans is 150
4b: for i1...it is just guessing but as the ms said: 0.3mm to 3mm
i2 the general formula is x=Dh/a
bii1: as x is the seperation. if u double the D in the eq, the x will double so the separation increases
maximum brightness will be less cuz u r increasing the distance of the light to the screen so there will be less light reaching and so less max brightness
bii2: again as h is increased so if u sub in the eq , the x will increase
the max brightness will decrese..am not sure though of the reason..but as the h increases the frequency decreases so as intensity is A^2 or f^2, the intensity will decrease
bii3:as all r kept constant the x will remain constant which is the seperation
the maximum brightness increases as the intensity increases
Hope i helped and was clear... :)
that was very helpful thanku so much :D
Post by: *Hope* on June 05, 2010, 09:23:41 pm
that was very helpful thanku so much :Dnp..glad to help ;)
Post by: Meticulous on June 05, 2010, 09:29:42 pm
HAD NET PROBLEMS ;D
THANKS HOPE! :D +rep.!
Post by: halosh92 on June 05, 2010, 09:30:54 pm
I AM VERY VERY SORRY HALOSH ;D
HAD NET PROBLEMS ;D
THANKS HOPE! :D +rep.!
np A.F ure always helpful ;)
Post by: *Hope* on June 05, 2010, 09:44:47 pm
I AM VERY VERY SORRY HALOSH ;DThanks A.F :)
HAD NET PROBLEMS ;D
THANKS HOPE! :D +rep.!
Post by: leebux101 on June 06, 2010, 12:47:20 am
Post by: nid404 on June 06, 2010, 06:34:34 am
Post by: halosh92 on June 06, 2010, 07:21:24 am
if the final uncertainity is 0.0045 do we round it off 0.005
or do we take it only to 1 sf without rounding it off
and to how do we know to how many sf we take the uncertainity??
and in between the questions when were calculating the percentage uncertainities inbetween...do we round these numbers also?
and to how mnay sf also do we take the percentage uncertainity?
thxxx
Post by: nid404 on June 06, 2010, 07:23:04 am
Post by: halosh92 on June 06, 2010, 07:32:42 am
m(v-u)
45/1000 (-3.6-4.2)
= -0.35
but the ms says its positive and they have used 4.2 as v
and 3.6 as u
hows that possible?
Post by: nid404 on June 06, 2010, 07:53:03 am
if you use your way you get -0.35
if you use what they did you get +0.35
Since momentum is a vector quantity I would have gone with the first option. But since this is change in momentum, they have ignored the direction i guess :-\
Post by: halosh92 on June 06, 2010, 08:10:02 am
if you use your way you get -0.35
if you use what they did you get +0.35
Since momentum is a vector quantity I would have gone with the first option. But since this is change in momentum, they have ignored the direction i guess :-\
hows that possible ??? ???
Post by: leebux101 on June 06, 2010, 09:27:01 am
i am sure of tht
notes on any topic. but they should be short brief at-a-glance kinda ones...
Post by: nid404 on June 06, 2010, 10:55:14 am
uncertainties are always rounded off to one sf
i am sure of tht
notes on any topic. but they should be short brief at-a-glance kinda ones...
Nope not always. Check mcq papers...it's not always 1 dp. It's should be either equal to or less than the significant figures of your answer.
Ohkay I'll look
Post by: leebux101 on June 06, 2010, 11:00:04 am
Post by: thecandydoll on June 06, 2010, 12:33:43 pm
M/J 2007-----Physics Q4d (i) and (ii)
And M/J/2007----PHYSICS q1 the whole thing!
I dont get it =(
thank youuuu!
Post by: ruby92 on June 06, 2010, 02:02:43 pm
3 bii????
Post by: halosh92 on June 06, 2010, 02:44:20 pm
m/j 2004 paper 2
3 bii????
3bi) the gradient is d/t^2
3bii) s= ut + 1/2at^2
u=0
s=1/2at^2
s/t^2 * 2 = a (making a as the subject)
substitute the gradient instead of s/t^2 and multiply by 2 thats all
voila ! :)
Post by: Meticulous on June 06, 2010, 03:22:13 pm
3bi) the gradient is d/t^2
3bii) s= ut + 1/2at^2
u=0
s=1/2at^2
s/t^2 * 2 = a (making a as the subject)
substitute the gradient instead of s/t^2 and multiply by 2 thats all
voila ! :)
WOW! finally halosh answering and not asking :P jk :P
+rep.!
Post by: halosh92 on June 06, 2010, 03:26:12 pm
WOW! finally halosh answering and not asking :P jk :P
+rep.!
OUCH that hit me right in the heart :P :P
haha i used to help alot before the exams started now am just blank ::)
thx! ;)
Post by: Phosu on June 06, 2010, 04:03:29 pm
question 5b
Post by: aldehyde1612 on June 06, 2010, 04:26:35 pm
Anyone solve this for me ? :D
M/J 2007-----Physics Q4d (i) and (ii)
And M/J/2007----PHYSICS q1 the whole thing!
I dont get it =(
thank youuuu!
i need help with them too... can someone please answer!
and oct/nov 2007... q6. b ... how did u get the pd to be 5.0V ???
Post by: poisonedrose on June 06, 2010, 05:09:25 pm
i need help with them too... can someone please answer!The power supply gives 230 volts, but the shower unit only needs 225 volts( it says there in the question) so if you subtract 230 and 225, you get 5 volts.
and oct/nov 2007... q6. b ... how did u get the pd to be 5.0V ???
Post by: halosh92 on June 06, 2010, 05:14:36 pm
extensions are added
in parallel :
extensions divided by number is springs
in parallel:
spring constants are added
in series:
i have no idea wat happens :)
could someone check if this is correct or wrong?
Post by: nid404 on June 06, 2010, 05:14:45 pm
jun 09 var 1
question 5b
phase difference=( 2pi/ lamda) X path difference
phase difference = pi [ for destructive interference]
lambda(wavelength) = 2pi/ pi X path difference.
path difference is S2M - S1M
find S2M using Pythagoras = root of 802 + 1002
=128
path difference= 128-100=28
wavelength= 2pi/pi X 28
=56cm
At f= 1KHz wavelength= 0.33m/ 33cm
At f=4kHz wavelength= 0..0825/8.25cm
wavelength changes from 33cm to 8.25cm
for minima...phase difference should be either one of them pi, 3pi,5pi,7pi [sub them in the eqn]
minima is observed when lambda= 56cm, 18.7cm, 11.2cm, 8cm
which are the 2 values between 33 and 8.25?? 18.7 and 11.2...so only 2 minimas
Post by: poisonedrose on June 06, 2010, 05:24:03 pm
Anyone solve this for me ? :DUsing an appropriate scale, draw a horizontal line representing horizontal velocity of 18m/s.I used 2cm (10 boxes)=5 m/s. And it is a straight, horizontal line because horizontal velocity does not change,.
M/J 2007-----Physics Q4d (i) and (ii)
And M/J/2007----PHYSICS q1 the whole thing!
I dont get it =(
thank youuuu!
then calculate vertical velocity at point of impact with ground by using the equations of motion with the formula v2=u2+2as. U being 0. a being 9.8 and s being 16. The vertical velocity gottn is approximately 17.7m/s which we can draw using our scale, as a vertical straight line.
We join the two lines- making a triangle and we can find the resultant velocity by the using Pythagoras theorem .. (which would be the hypotenuse) a2+b2=c2.
Once you are drawn with the vector diagram, measure the asked angle with a protractor or with trigonometry
Post by: zxcvbnm on June 06, 2010, 06:07:24 pm
Post by: sizbeauty on June 06, 2010, 07:16:59 pm
Q4(b)(part 1)
oct/nov02
Q5(b)(part1) how do u calculate the phase difference in this one ??? ??? ??? ???
plz help me!!!!!!
Post by: nid404 on June 06, 2010, 07:17:42 pm
may 06-question 5, part (c)
check the attachment.
Post by: nid404 on June 06, 2010, 07:23:39 pm
m/june02
Q4(b)(part 1)
a) v2=u2+2as
v2=2X9.8X1.6
v=5.6m/s
b) p.e lost=k.e gained
p.e lost=mgh=73/1000 X 9.8 X 1.6=1.14
k.e gained=90% of p.e lost= 1.03J
1.03=mgh
1.03=73/1000 X 9.8 X h
h=1.44 m
Post by: nid404 on June 06, 2010, 07:27:57 pm
nov 02
Q5(b)(part1) how do u calculate the phase difference in this one ??? ??? ??? ???
plz help me!!!!!!
(http://4.bp.blogspot.com/_vbLcDpetdkY/SYUOq0sdO-I/AAAAAAAAAHI/8vUnINuy6yg/s320/fig+3.png)
Post by: zxcvbnm on June 06, 2010, 07:29:36 pm
Post by: halosh92 on June 06, 2010, 07:37:15 pm
in series:
extensions are added
in parallel :
extensions divided by number is springs
in parallel:
spring constants are added
in series:
i have no idea wat happens :)
could someone check if this is correct or wrong?
could someone just check this.
Post by: nid404 on June 06, 2010, 07:39:05 pm
could someone just check this.
the spring constant changes accordingly...the extension changes inversely.
I had answered a similar question...I'll look for it :)
Post by: sizbeauty on June 06, 2010, 07:44:34 pm
Post by: nid404 on June 06, 2010, 07:49:41 pm
Thank you :)
Post by: halosh92 on June 06, 2010, 07:53:12 pm
one more question when an object is rotating due to a couple its accelerating as its changing direction right?
or not because the couples are balanced forces?
Post by: nid404 on June 06, 2010, 07:57:13 pm
Post by: halosh92 on June 06, 2010, 08:52:55 pm
Yes there is acceleration....but we don't have circular motion for AS... ;) :D
;D ;D ;D *sigh*
thx and wat about the previous question u said u did it before
sorry for the trouble ;D
Post by: nid404 on June 06, 2010, 09:06:18 pm
Post by: halosh92 on June 07, 2010, 11:04:36 am
Q5c
could someone give it in more details plz? :)
Q7b
is it going to pass straight? or is it going to deviate at a smaller angle?
one more question: 4 pie L/T^2
if we found the total percentage uncertainity of L/T^2
then dont we have to multiply by 4 ???
thx
Post by: ruby92 on June 07, 2010, 11:47:00 am
3bi) the gradient is d/t^2
3bii) s= ut + 1/2at^2
u=0
s=1/2at^2
s/t^2 * 2 = a (making a as the subject)
substitute the gradient instead of s/t^2 and multiply by 2 thats all
voila ! :)
thnkx ;D
in m/j 2005 3b
according to the ms we dont change g-->kg only for this quiestion?
Post by: TJ-56 on June 07, 2010, 11:55:37 am
4 (c) (ii) 2. & 3.
Nov 08
6 (b)
7 (b) (ii)
May 08
6 (b)
Nov 07
3 (b) (ii)
4 (c)
5 (a)
May 06
2 (b)
2 (c) (iii)
4 (b) (i)
5 (b)
Nov 05
2 (b)
7 (b) (iii)
Nov 03
5 (c)
Nov 02
5 (b) (i)
Nov 01
3 (b)
ANY help would be deeply appreciated
thanx for any help in advance!
Post by: The SMA on June 07, 2010, 01:32:17 pm
nov 2007
Q4d
halosh, can u explain to me please for Q4a, how do we know from graph if brittle, ductile or polymeric?
and also Q4c.
Post by: halosh92 on June 07, 2010, 01:51:15 pm
halosh, can u explain to me please for Q4a, how do we know from graph if brittle, ductile or polymeric?
and also Q4c.
Q4a
brittle: when the ultimate tensile stress occurs at the elastic limit ( thats when it breaks)
ductile: it can stretch after the elastic limit ( it becomes plastically deformed) then it some point it breaks
plastic: they usually have a low stress value and a large area under the graph , in the begiinng they are ductile but as stress increases they become plastically deformed
Q4c: stress= force/area
first find the minimum area of the tube from the graph , the highest stress is 9.5 * 10^8
area= (1.9 × 103) / (9.5 × 108)
= 2 * 10^-6
maximum area is given in the question
so subtract the minimum area from the maximum
Post by: halosh92 on June 07, 2010, 02:02:07 pm
1st variant
Q5c
could someone give it in more details plz? :)
Q7b
is it going to pass straight? or is it going to deviate at a smaller angle?
one more question: 4 pie L/T^2
if we found the total percentage uncertainity of L/T^2
then dont we have to multiply by 4 ???
thx
could anyone solve this??? ??? ??? ??? ??? ???
Post by: Meticulous on June 07, 2010, 02:14:57 pm
could anyone solve this??? ??? ??? ??? ??? ???
when i get up halla..in 2 hours
Post by: sweetie on June 07, 2010, 02:29:41 pm
Post by: nid404 on June 07, 2010, 03:21:50 pm
cud sum1 plz solve Q5b of the 1st variant
https://studentforums.biz/index.php/topic,8636.60.html
Post by: AN10 on June 07, 2010, 03:39:23 pm
Q4c: stress= force/area
first find the minimum area of the tube from the graph , the highest stress is 9.5 * 10^8
area= (1.9 × 103) / (9.5 × 108)
= 2 * 10^-6
maximum area is given in the question
so subtract the minimum area from the maximum
i dnt get da last step...y do we subtract the areas ???
Post by: The SMA on June 07, 2010, 03:44:15 pm
Q4a
brittle: when the ultimate tensile stress occurs at the elastic limit ( thats when it breaks)
ductile: it can stretch after the elastic limit ( it becomes plastically deformed) then it some point it breaks
plastic: they usually have a low stress value and a large area under the graph , in the begiinng they are ductile but as stress increases they become plastically deformed
Q4c: stress= force/area
first find the minimum area of the tube from the graph , the highest stress is 9.5 * 10^8
area= (1.9 × 103) / (9.5 × 108)
= 2 * 10^-6
maximum area is given in the question
so subtract the minimum area from the maximum
Thanks! I got it now ^^
Post by: The SMA on June 07, 2010, 03:49:16 pm
i dnt get da last step...y do we subtract the areas ???
we need to minus bcoz we want to find the max area which is (3.2*10^-6)-(2.0*10^-6). not the min area we solved earlier.
Post by: sweetie on June 07, 2010, 04:26:47 pm
phase difference=( 2pi/ lamda) X path difference
phase difference = pi [ for destructive interference]
lambda(wavelength) = 2pi/ pi X path difference.
path difference is S2M - S1M
find S2M using Pythagoras = root of 802 + 1002
=128
path difference= 128-100=28
wavelength= 2pi/pi X 28
=56cm
At f= 1KHz wavelength= 0.33m/ 33cm
At f=4kHz wavelength= 0..0825/8.25cm
wavelength changes from 33cm to 8.25cm
for minima...phase difference should be either one of them pi, 3pi,5pi,7pi [sub them in the eqn]
minima is observed when lambda= 56cm, 18.7cm, 11.2cm, 8cm
which are the 2 values between 33 and 8.25?? 18.7 and 11.2...so only 2 minimas
Thank You nid... +rep
but is "phase diff.=2pi/lambda *path diff " a given formula or u derived it on ur own???
Post by: sweetie on June 07, 2010, 04:28:55 pm
Post by: zxcvbnm on June 07, 2010, 04:40:37 pm
Post by: sizbeauty on June 07, 2010, 05:32:21 pm
Q4(part c)
Q6(part b)don't we just add up the powers???
Post by: zxcvbnm on June 07, 2010, 06:36:51 pm
Post by: sweetie on June 07, 2010, 06:39:22 pm
june 09, variant 1, question 5,b
https://studentforums.biz/index.php/topic,8636.60.html
Post by: zxcvbnm on June 07, 2010, 06:45:43 pm
phase difference=( 2pi/ lamda) X path difference
phase difference = pi [ for destructive interference]
lambda(wavelength) = 2pi/ pi X path difference.
path difference is S2M - S1M
find S2M using Pythagoras = root of 802 + 1002
=128
path difference= 128-100=28
wavelength= 2pi/pi X 28
=56cm
At f= 1KHz wavelength= 0.33m/ 33cm
At f=4kHz wavelength= 0..0825/8.25cm
wavelength changes from 33cm to 8.25cm
for minima...phase difference should be either one of them pi, 3pi,5pi,7pi [sub them in the eqn]
minima is observed when lambda= 56cm, 18.7cm, 11.2cm, 8cm
which are the 2 values between 33 and 8.25?? 18.7 and 11.2...so only 2 minimas
thankyou sweetie and nid!
but where did you get the formula from? for phase difference?
Post by: zxcvbnm on June 07, 2010, 06:47:34 pm
Post by: halosh92 on June 07, 2010, 06:55:34 pm
oh i realised that we can also use (2n+1)(lambda)/2...but i've forgotten what the n stands for...please clarify this asap!
no of orders maybe?
Post by: zxcvbnm on June 07, 2010, 06:59:57 pm
Post by: zxcvbnm on June 07, 2010, 07:21:16 pm
Post by: halosh92 on June 07, 2010, 07:26:32 pm
june 09, 21, question 1,b,iii....shouldnt it be E1-E2=I1R-2I2R?
could u upload the paper plz?
Post by: zxcvbnm on June 07, 2010, 07:36:46 pm
Post by: halosh92 on June 07, 2010, 07:46:45 pm
here...i did figure it out but not sure if i'm right
Q1 ??? there is no such similar question
do u mean Q8?
Post by: zxcvbnm on June 07, 2010, 07:54:54 pm
Post by: halosh92 on June 07, 2010, 08:15:02 pm
oh sorry! i meant 7...the one where we have to use kirchoff's second law....i figured it out....current entering=current leaving...but if you could still have a look
yep its correct see if we take anticlockwise direction starting from E1
current is in the same direction so it will be
-I1R and then it faces the (+) terminal of E2 so we take it as negative so next is -E2
then we face the next resistor but here the current is moving in the opposite direction to our clockwise direction so it will be + I2R
then we face the next resistor, so the the current is the same direction to our chosen direction so it will be -I1R
and we then face the negative terminal of E1 so we take it as +E1
so we group them together
-I1R-E2+I2R-I1R+E1=0
got it?
Post by: zxcvbnm on June 07, 2010, 08:22:28 pm
Post by: halosh92 on June 07, 2010, 08:28:16 pm
could anyone solve this??? ??? ??? ??? ??? ???
[/quote/]
someone PLZ PLZ DO THIS :(
Post by: ruby92 on June 07, 2010, 08:56:16 pm
if the lamp that shorted is C
so shouldnt the resistance of one of the normal lamps be 30?
Post by: zxcvbnm on June 07, 2010, 09:50:41 pm
Post by: Chingoo on June 07, 2010, 10:23:10 pm
Please explain in DETAIL May/June 2009 Paper 21 Question 7 (in attachment)? I suck at these questions, in-depth explanation would be loved. Thanks in advance!
Post by: astarmathsandphysics on June 07, 2010, 10:25:38 pm
if the lamp that shorted is C
so shouldnt the resistance of one of the normal lamps be 30?
no is 15 cos when S_1 is closed and other two open Resistance is 15
Post by: astarmathsandphysics on June 07, 2010, 10:28:55 pm
sum of two amplitudes is A/3
square to get I/9
Post by: astarmathsandphysics on June 07, 2010, 10:42:24 pm
7ai)R
ii)R/2 since BY is in paralell with CX
iii)R+R/2+R since AB is in series with YZ and the paralell circuit between Band Y
bi)I_1+I_2=I_3
ii)E_=I_1 R+I_2 R by applying V=IR to the two resistances
iii)E_1-E_2 =I_1 R -I_2 R +I_1 R
net voltage is E_1 -E_2 and apply V=IR to each (remeber I_2 is in opposite direction so negative)
Post by: Chingoo on June 07, 2010, 11:02:36 pm
Post by: hesho21 on June 07, 2010, 11:51:26 pm
Post by: astarmathsandphysics on June 07, 2010, 11:59:32 pm
3.2*10^-6-2*10^-6=1.2*10^-6=cross section of bubble
Post by: hesho21 on June 08, 2010, 12:10:48 am
breaking stress=9.5*10^7=F/A=1900/A so A=1900/9.5*10^9=2*10^-6 =actuall cross section of metal
3.2*10^-6-2*10^-6=1.2*10^-6=cross section of bubble
can i ask why did u subtract at the end the areas?
Post by: sweetie on June 08, 2010, 12:11:55 am
m/j 2006 7 a b c
if the lamp that shorted is C
so shouldnt the resistance of one of the normal lamps be 30?
no is 15 cos when S_1 is closed and other two open Resistance is 15
i still dont get it
cud u plz xplain it again in more detail?
Thank You soooo much :)
Post by: thecandydoll on June 08, 2010, 04:13:32 am
Oct/Nov/2009 Variant 22.
Q3--alll =(
Post by: The SMA on June 08, 2010, 04:29:36 am
Post by: cashem'up on June 08, 2010, 07:39:58 am
can somebody help me on M/J 2004 p2 Q8abc please
theres a formula u should be aware with chek pg 188 of chris mee physics textbook
Vout= VR1 / (R1 + R2)
ok now lets use it here
a) 1.0= 1.5 R1/ (R1 + 3900)
1.0R1+3900=1.5R1
0.5R1=3900
R1=7800
b)Vout= 1.5x7800 / 1250+7800 = 1.293 V
c)the voltmeter has 7800 V so parallel voltage is 7800/2 = 3900
now Vout = 1.5x3900 / 3900+3900 = 0.75V
Post by: sabrina on June 08, 2010, 09:00:03 am
physics tomorrow :'( and im going crazy. someone plz explain m/j2009 var1 q5a. shouldn't it be inphase?
Post by: sabrina on June 08, 2010, 09:02:17 am
Post by: halosh92 on June 08, 2010, 09:03:42 am
n the next too
post the paper plz
Post by: sabrina on June 08, 2010, 09:07:04 am
Post by: halosh92 on June 08, 2010, 09:09:32 am
here it is
for part a) this is a repeated question and the answer is always the same , its a fact just learn wats in the ms.
and for the next question, it solved somewhere in this thread by "nid" just look for it in the previous pages.
Post by: sabrina on June 08, 2010, 09:11:49 am
Post by: The SMA on June 08, 2010, 09:32:12 am
Post by: astarmathsandphysics on June 08, 2010, 10:07:30 am
Post by: sizbeauty on June 08, 2010, 10:34:30 am
frm where did v get 0.0362????
2 ?ES = ½ × 1.8 × 10-2 ( 0.0362 – 0.0212)
= 0.077 J
plz sm1 help me!!!! ??? ??? ???
Post by: sabrina on June 08, 2010, 10:37:57 am
i have doubt in O/N2009 var2
q2bi
shouldn't it be like
for 10.7= 6*10^23
1 = x
n then cross multiply?
i have attached the paper
Post by: sabrina on June 08, 2010, 10:40:17 am
Post by: sizbeauty on June 08, 2010, 10:50:44 am
nopz u did it wrong way,
it shuld b lik if for 6*10^23 atoms,u hav 10.7cm^3 vol
so for 1 atom,it shuld b x cm^3 vol..i.e
6*10^23=10.7
1=x
and thn u can get ur ans by cross multipling n for separation,whnver v r givn the vol,v take the cube root of it so it would b (1.78 x 10^-23)1/3 =2.61 x10^-8
hope it helped :)
cn u plz luk at ma ques???
Post by: ruby92 on June 08, 2010, 11:17:33 am
2 d
Post by: The SMA on June 08, 2010, 11:18:46 am
and for part a) how to draw the waves?
Post by: sabrina on June 08, 2010, 11:31:00 am
@ sabrina,thanks a lot now i got it
nopz u did it wrong way,
it shuld b lik if for 6*10^23 atoms,u hav 10.7cm^3 vol
so for 1 atom,it shuld b x cm^3 vol..i.e
6*10^23=10.7
1=x
and thn u can get ur ans by cross multipling n for separation,whnver v r givn the vol,v take the cube root of it so it would b (1.78 x 10^-23)1/3 =2.61 x10^-8
hope it helped :)
cn u plz luk at ma ques???
Post by: halosh92 on June 08, 2010, 11:33:23 am
m/j 2007
2 d
this is sort of like the projectile
now the electron is exactly in the middle of the plates initally , vertical distance= 1.5cm
so if it has to travel upwards it will travel 1.5/2
right?
which is 0.75 cm basically so....
horizontal velocity is given in the question BUT vertical velocity in projectile cases ( why projectile? because its travelling in a curved path towards the upper plate) the vertical velocity is = 0
so s=ut + 1/2at^2
u=0 as i mentioned
s= 1/2 * (ure "a" value) *( ure "t" value) ^2
u will get an answer which is less than 0.75cm so obviouslly it will hit the plate
got it?
remember to change cms to ms
Post by: halosh92 on June 08, 2010, 11:38:56 am
m/j 2006 p2 Q6b), why do we need to times 2 the wavelength, using v=f x lambda
and for part a) how to draw the waves?
ok the wavelength they gave is basically the length between one antinode and another successive antinode ok?
(two positions loud voice as they mentioned)
so...
if u draw the wave u will first have an antonide then node then antinode the normal wave
so the distance from one antinode to the node is 32.4/2 ok?
always remember distance from node to node = half a wavelength
and wavelength from node to antinode= quarter a wavelength
so wat we found is quarter a wavelength now multiply by 4
got it?
Post by: halosh92 on June 08, 2010, 12:06:35 pm
Post by: Chingoo on June 08, 2010, 12:10:33 pm
ok the wavelength they gave is basically the length between one antinode and another successive antinode ok?
(two positions loud voice as they mentioned)
so...
if u draw the wave u will first have an antonide then node then antinode the normal wave
so the distance from one antinode to the node is 32.4/2 ok?
always remember distance from node to node = half a wavelength
and wavelength from node to antinode= quarter a wavelength
so wat we found is quarter a wavelength now multiply by 4
got it?
I think you kinda lengthened the whole story...just to add my input (in case it helps), the distance between two consecutive nodes or two consecutive antinodes is lamda/2. See this maxima and minima as a waveform with displacement (y-axis) against distance (x-axis); the minima is the node and maxima the antinode. The antinode can be on either side of the x-axis, because if you remember displacement is taken from a reference point. In case of waves the reference point is the mean position (the node) and the maximum displacement from this mean position is an antinode. If the antinode is a point 2 cm from the mean position, the least displacement is NOT -2 but 0. Hence, like halosh said, simply multiply 32.4 by 2 and you have the lamda.
In times of a sine wave, a node and antinode are 90 degrees out of phase with each other (1/4 of the sine wave graph) and an antinode and another are (270-90=)180 degrees out of phase with each other (1.2 of the sine wave graph).
Post by: halosh92 on June 08, 2010, 12:12:07 pm
I think you kinda lengthened the whole story...just to add my input (in case it helps), the distance between two consecutive nodes or two consecutive antinodes is lamda/2. See this maxima and minima as a waveform with displacement (y-axis) against distance (x-axis); the minima is the node and maxima the antinode. The antinode can be on either side of the x-axis, because if you remember displacement is taken from a reference point. In case of waves the reference point is the mean position (the node) and the maximum displacement from this mean position is an antinode. If the antinode is a point 2 cm from the mean position, the least displacement is NOT -2 but 0. Hence, like halosh said, simply multiply 32.4 by 2 and you have the lamda.
In times of a sine wave, a node and antinode are 90 degrees out of phase with each other (1/4 of the sine wave graph) and an antinode and another are (270-90=)180 degrees out of phase with each other (1.2 of the sine wave graph).
hahah yes my mother always said i talk alot ;D
Post by: cashem'up on June 08, 2010, 12:14:58 pm
Post by: sabrina on June 08, 2010, 12:16:57 pm
random errors r errors which r affected to different extend. its experimenters fault. it can be minimize by repeating the experiment several times n finding out the average. in graphs we can find out if the error is random by scattering of lines
hope its clear :)
Post by: halosh92 on June 08, 2010, 12:20:07 pm
hey guys one question i had.... if i kept a filament in constant temperature lets say in Ice then would its IV graph be a straight line through originhow is it possible to that...the filament itself will heat up because the current passing through it will heat it up i guess.
but maybe if u keep adding ice....and keep it constnt temp. then yea it will i guess. according to ohms law ???
could someone answer my question plz :(
Post by: Chingoo on June 08, 2010, 12:22:50 pm
4.10 - 2.00 = 2.10.
Even if the systematic error was absent, the DIFFERENCE would be the same i.e. 4.00 - 1.90 = 2.10 s.
Systematic errors do not affect the precision of a measurement, but they affect the accuracy.
By definition, random error is an error having a varying direction and magnitude in all readings. This is quite simple, for instance, given the same example as above--it is not possible for us to start the stopwatch at the exact moment the ball crosses X, as there is a human reaction error of 0.1 to 0.4 seconds. This error cannot be eliminated with a gradient or difference, because it is not constant. A random error does affect the precision of a measurement, but not its accuracy.
Post by: Chingoo on June 08, 2010, 12:25:41 pm
hey guys one question i had.... if i kept a filament in constant temperature lets say in Ice then would its IV graph be a straight line through origin
Like halosh said, if you keep the temperature of the filament constant by adding ice frequently, it will be a straight line through origin. Follow the rules of the Ohm's Law:
The current through a resistor is directly proportional to the potential difference across it if external conditions such as temperature and pressure remain constant.
Simple ;)
Post by: sabrina on June 08, 2010, 12:26:34 pm
By definition, systematic error is an error having constant direction and magnitude in all readings. Consider a situation where your stopwatch starts from 0:00:10 s only; all your readings will have an additive 0.01 s. It will not vary with situations. Such an error can be eliminated by taking a difference/gradient. Consider you had to measure the time interval for a ball to fall between a distance XY; the first reading (when it crossed X) was 0:02:00 s, and the second reading (when it crossed Y) was 0:04:10 s. The time interval is the difference of the two:::) ::) wow good explanation.
4.10 - 2.00 = 2.10.
Even if the systematic error was absent, the DIFFERENCE would be the same i.e. 4.00 - 2.10 = 2.10 s.
Systematic errors do not affect the precision of a measurement, but they affect the accuracy.
By definition, random error is an error having a varying direction and magnitude in all readings. This is quite simple, for instance, given the same example as above--it is not possible for us to start the stopwatch at the exact moment the ball crosses X, as there is a human reaction error of 0.1 to 0.4 seconds. This error cannot be eliminated with a gradient or difference, because it is not constant. A random error does affect the precision of a measurement, but not its accuracy.
Post by: halosh92 on June 08, 2010, 12:30:17 pm
Post by: halosh92 on June 08, 2010, 12:34:37 pm
diffraction
interference
path difference
phase differenec
the exact cie meaning plz
Post by: Asamy111 on June 08, 2010, 12:35:53 pm
Post by: Meticulous on June 08, 2010, 12:42:12 pm
one more question wat exactly is
diffraction
interference
path difference
phase differenec
the exact cie meaning plz
Diffraction-It is the word used to describe the way waves spread out as they bend through and around obstacles and gaps. Diffraction also describes the interaction between waves and solid objects.
Interference-It occurs when waves overlap each other to produce a pattern where the waves reinforce each other in some places & cancel each other out in others. This also could be the definition for superposition.
Phase Difference-The phase difference between 2 particles on a wave is the fraction of a cycle by which one moves behind the other. Measured in multiples of "pie"
Path Difference-Same as phase difference but in multiples of lambda
Post by: halosh92 on June 08, 2010, 12:42:41 pm
m/j 09 .. q5 varient 1
the answer is here somewhere in this thread by "nid"
check the previous pages plz
Post by: Chingoo on June 08, 2010, 12:43:48 pm
one more question wat exactly isDIFFRACTION The spreading of waves about an obstacle or through a gap/aperture.
diffraction
interference
path difference
phase differenec
the exact cie meaning plz
INTERFERENCE The superposition of waves which support each other at a point and cancel out each other's effect at another point.
The last two I'm not sure about.
Post by: halosh92 on June 08, 2010, 01:00:13 pm
Post by: sabrina on June 08, 2010, 01:08:07 pm
what are the conditions for interfernce?1. should meet at a point
2. same type(means both should be either transverse or longitudinal)
3. same frequency, wavelength
4. if it is constructive interference so it should be in phase (path difference=lambda, phase difference=2pie)
5. if it is destructive interference so it should be out of phase (path difference=(2n+1)lambda/2, phase difference=(2n+1)pie)
hope its clear
Post by: sabrina on June 08, 2010, 01:10:17 pm
Post by: halosh92 on June 08, 2010, 01:15:13 pm
1. should meet at a point
2. same type(means both should be either transverse or longitudinal)
3. same frequency, wavelength
4. if it is constructive interference so it should be in phase (path difference=lambda, phase difference=2pie)
5. if it is destructive interference so it should be out of phase (path difference=(2n+1)lambda/2, phase difference=(2n+1)pie)
hope its clear
shouldnt we mention something about coherence??
and r u sure about the formulas isnt it for constructive:
path difference: n lamda
and destructive:
path difference= (n + 1/2) lambda and phase differnce = pie
Post by: Meticulous on June 08, 2010, 01:18:55 pm
shouldnt we mention something about coherence??
and r u sure about the formulas isnt it for constructive:
path difference: n lamda
and destructive:
path difference= (n + 1/2) lambda and phase differnce = pie
coherent means same frequency
Post by: poisonedrose on June 08, 2010, 01:19:54 pm
guys how to do o/n2008 q7biiWe know that the 2000 resistor has 3.6volts, so we subtract 6 from 3.6V giving us 2.4V
As the thermistor and 5000R are in parallel, they have the same voltage of 2.4V
Using the ratio method, we can do
3.6 : 2000
2.4 : x
Finding x- which is 1,333 (the resistance of both of the devices)
From that, we use the combination resistance formula
5000*R2/5000+R2=13333
And with that, we can simply and find out R2- which is the resistance of the thermistor.
Post by: sabrina on June 08, 2010, 01:24:41 pm
shouldnt we mention something about coherence??i m sorry abt the constructive interference.forgot to put "n".
and r u sure about the formulas isnt it for constructive:
path difference: n lamda
and destructive:
path difference= (n + 1/2) lambda and phase differnce = pie
abt destructive the formula of my best of knowledge is correct
abt coherent sources = the interference should have constant phase difference to have observable interference. i think that's it.
Post by: sabrina on June 08, 2010, 01:26:01 pm
We know that the 2000 resistor has 3.6volts, so we subtract 6 from 3.6V giving us 2.4V
As the thermistor and 5000R are in parallel, they have the same voltage of 2.4V
Using the ratio method, we can do
3.6 : 2000
2.4 : x
Finding x- which is 1,333 (the resistance of both of the devices)
From that, we use the combination resistance formula
5000*R2/5000+R2=13333
And with that, we can simply and find out R2- which is the resistance of the thermistor.
thanks alot :)
Post by: halosh92 on June 08, 2010, 01:29:09 pm
i m sorry abt the constructive interference.forgot to put "n".
abt destructive the formula of my best of knowledge is correct
abt coherent sources = the interference should have constant phase difference to have observable interference. i think that's it.
alright thx :)
Post by: neno on June 08, 2010, 01:38:35 pm
#may june 2008:Q6b
#mayjune 2009(21):Q5b
Post by: scorpion9500 on June 08, 2010, 02:27:11 pm
guys mj 2008 same q
6b
zzzz
Post by: ruby92 on June 08, 2010, 02:54:31 pm
Post by: highly_ambitious on June 08, 2010, 02:55:12 pm
Post by: The SMA on June 08, 2010, 03:23:50 pm
ok the wavelength they gave is basically the length between one antinode and another successive antinode ok?
(two positions loud voice as they mentioned)
so...
if u draw the wave u will first have an antonide then node then antinode the normal wave
so the distance from one antinode to the node is 32.4/2 ok?
always remember distance from node to node = half a wavelength
and wavelength from node to antinode= quarter a wavelength
so wat we found is quarter a wavelength now multiply by 4
got it?
thanks halosh and chingoo, i think i finally got it. :)
Post by: scorpion9500 on June 08, 2010, 03:37:26 pm
i got it
i will just give u the hint and u figure it
if s2 is closed then electricity will pass through it and wont pass through B as there is less resistance in that route
so imagine that there is no B when s2 is closed
Post by: zxcvbnm on June 08, 2010, 03:46:46 pm
Post by: zxcvbnm on June 08, 2010, 03:52:04 pm
Post by: zxcvbnm on June 08, 2010, 03:59:42 pm
here
Q4, c
Post by: Chingoo on June 08, 2010, 04:18:43 pm
Post by: zxcvbnm on June 08, 2010, 04:26:48 pm
Post by: hesho21 on June 08, 2010, 04:37:23 pm
Post by: zxcvbnm on June 08, 2010, 04:41:45 pm
Post by: halosh92 on June 08, 2010, 05:04:24 pm
what are the conditions for observing two source intrference?
sources need to be coherent
Post by: Chingoo on June 08, 2010, 06:38:03 pm
When looking at brownian motion and comparing how larger smoke particles would be seen , what do they mean by randomness of collisions would be ‘averaged out’ ?? ???
Imagine a small ball being hit by 'balls' of almost the same size; if you ever played with marbles, you'd see that hitting a marble with another of the same size compels the stationary marble to move at the same speed as the previous one, which stops. The same rule applies to snooker. Basically, this is a case of elastic collision; when two bodies with the same mass m collide, they interchange their velocities. So:
m1u1 + m2u2 = m1v1 + m2v2
mu1 + mu2 = mu2 + mu1
as v1=u2, v2=u1
This means that small smoke particles would move randomly at the same pattern and speed as the air molecules.
Now consider a large ball being hit by much smaller balls. Ever tried punching the big fat bully in your school? I bet you got a black eye ;D Basically, to conserve momentum a body with a higher mass will have a smaller velocity in contrast with a lighter body--this is just a general statement and is often contradicted, but let's keep it in mind to solve this. The bully won't budge easily if a skinny nerd punched him because his body is reluctant to change it's state of rest--inertia. Hence, smoke particles will not be easily moved by small balls hitting them.
However, this is still a shallow explanation so let's add to it with the simple concept of surface area. A smaller ball has a small surface area, hence not many balls can collide with it at the same time. A larger ball will have a much larger surface area, and many balls will hit it from all sides at the same time. Imagine all babies hitting Barney! Barney would be stagnant, unable to move. See, a smaller body would have less inertia and would react quickly and with a higher speed (than a massive body) when hit and would not allow another ball (body) to hit it in an opposite direction and cancel the forces. A larger body is reluctant to move, and hence allows many balls to hit it at once, many of which will act in opposite directions and the forces will largely cancel out. This is what is meant by 'randomness of collisions' being 'averaged out'.
Post by: halosh92 on June 08, 2010, 06:49:29 pm
Imagine a small ball being hit by 'balls' of almost the same size; if you ever played with marbles, you'd see that hitting a marble with another of the same size compels the stationary marble to move at the same speed as the previous one, which stops. The same rule applies to snooker. Basically, this is a case of elastic collision; when two bodies with the same mass m collide, they interchange their velocities. So:
m1u1 + m2u2 = m1v1 + m2v2
mu1 + mu2 = mu2 + mu1
as v1=u2, v2=u1
This means that small smoke particles would move randomly at the same pattern and speed as the air molecules.
Now consider a large ball being hit by much smaller balls. Ever tried punching the big fat bully in your school? I bet you got a black eye ;D Basically, to conserve momentum a body with a higher mass will have a smaller velocity in contrast with a lighter body--this is just a general statement and is often contradicted, but let's keep it in mind to solve this. The bully won't budge easily if a skinny nerd punched him because his body is reluctant to change it's state of rest--inertia. Hence, smoke particles will not be easily moved by small balls hitting them.
However, this is still a shallow explanation so let's add to it with the simple concept of surface area. A smaller ball has a small surface area, hence not many balls can collide with it at the same time. A larger ball will have a much larger surface area, and many balls will hit it from all sides at the same time. Imagine all babies hitting Barney! Barney would be stagnant, unable to move. See, a smaller body would have less inertia and would react quickly and with a higher speed (than a massive body) when hit and would not allow another ball (body) to hit it in an opposite direction and cancel the forces. A larger body is reluctant to move, and hence allows many balls to hit it at once, many of which will act in opposite directions and the forces will largely cancel out. This is what is meant by 'randomness of collisions' being 'averaged out'.
very interesting explanation truly :)
+ rep
Post by: Joseph_COOL on June 08, 2010, 07:02:08 pm
thanks in advance :D
Post by: ruby92 on June 08, 2010, 07:08:23 pm
2bii how do they 0.08 as time?
Post by: hesho21 on June 08, 2010, 07:24:19 pm
Imagine a small ball being hit by 'balls' of almost the same size; if you ever played with marbles, you'd see that hitting a marble with another of the same size compels the stationary marble to move at the same speed as the previous one, which stops. The same rule applies to snooker. Basically, this is a case of elastic collision; when two bodies with the same mass m collide, they interchange their velocities. So:
m1u1 + m2u2 = m1v1 + m2v2
mu1 + mu2 = mu2 + mu1
as v1=u2, v2=u1
This means that small smoke particles would move randomly at the same pattern and speed as the air molecules.
Now consider a large ball being hit by much smaller balls. Ever tried punching the big fat bully in your school? I bet you got a black eye ;D Basically, to conserve momentum a body with a higher mass will have a smaller velocity in contrast with a lighter body--this is just a general statement and is often contradicted, but let's keep it in mind to solve this. The bully won't budge easily if a skinny nerd punched him because his body is reluctant to change it's state of rest--inertia. Hence, smoke particles will not be easily moved by small balls hitting them.
However, this is still a shallow explanation so let's add to it with the simple concept of surface area. A smaller ball has a small surface area, hence not many balls can collide with it at the same time. A larger ball will have a much larger surface area, and many balls will hit it from all sides at the same time. Imagine all babies hitting Barney! Barney would be stagnant, unable to move. See, a smaller body would have less inertia and would react quickly and with a higher speed (than a massive body) when hit and would not allow another ball (body) to hit it in an opposite direction and cancel the forces. A larger body is reluctant to move, and hence allows many balls to hit it at once, many of which will act in opposite directions and the forces will largely cancel out. This is what is meant by 'randomness of collisions' being 'averaged out'.
Man that was great , and By the way I'm the big bully in my school :P :P Thx very much that just makes sense , +Rep :P
Post by: Joseph_COOL on June 08, 2010, 07:26:55 pm
Post by: Phosu on June 08, 2010, 08:29:37 pm
m/j 2009 variant 1
2bii how do they 0.08 as time?
take the duration of the slope.
2.6-1.8=0.08
Post by: ruby92 on June 08, 2010, 09:02:18 pm
not 0.08