Post Your AS LEVEL PHYSICS PAPER 2 DOUBTS

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s09 qs 1 need help in calculation of the uncertainty?
d=m/v
d=m/length^3
d=580/6.0^3
d=2.69 g cm^-3

for uncertainty:
d=m/length^3
δ d/d = δ m/m + 3 δ l/l
= 10/580 + 3 × 0.1/6.0
= 0.0172 + 0.05
δ d/d = 0.0672
δ d = 0.0672 × 2.69
= 0.181
so your answer will be 2.69 + or - 0.18 g cm^-3
I HOPE U GET IT!!! :)
AND IF U HAV ANY ODER PROBLEMS U R FREE 2 ASK
 
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Can you help me with these past papers doubts

M/J 02- 7b
O/N 02- 9b, 2c
O/N 03- 3 c
M/J 04- 1b
O/N 04- 1b (ii)
M/J 05- 5c, 6b(iii)
O/N 05- 3c
 
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Can you help me with these past papers doubts

M/J 02- 7b
O/N 02- 9b, 2c
O/N 03- 3 c
M/J 04- 1b
O/N 04- 1b (ii)
M/J 05- 5c, 6b(iii)
O/N 05- 3c
yup except the 2002 papers i'll check out the others cuz i dun't hav 20002
 
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Can you help me with these past papers doubts

M/J 02- 7b
O/N 02- 9b, 2c
O/N 03- 3 c
M/J 04- 1b
O/N 04- 1b (ii)
M/J 05- 5c, 6b(iii)
O/N 05- 3c
oct/nov 2003
c)i)in string S2 force moves to the left ans in string S1 force moves to the right (i can't help u here u hav to imagine it)
ii)1)the centre is the pivot so you have to find moment from the pivot =F x s = 150 x 0.3=45Nm
ii)2)the torque is one of the two forces on the parallel sides which is 150 x 0.3 the distance =45Nm
ii)3)45=F x 0.12 , F=375N
 
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d=m/v
d=m/length^3
d=580/6.0^3
d=2.69 g cm^-3

for uncertainty:
d=m/length^3
δ d/d = δ m/m + 3 δ l/l
= 10/580 + 3 × 0.1/6.0
= 0.0172 + 0.05
δ d/d = 0.0672
δ d = 0.0672 × 2.69
= 0.181
so your answer will be 2.69 + or - 0.18 g cm^-3
I HOPE U GET IT!!! :)
AND IF U HAV ANY ODER PROBLEMS U R FREE 2 ASK
i had done all the steps i just dont know why u calculated this : δ d = 0.0672 × 2.69
 
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i had done all the steps i just dont know why u calculated this : δ d = 0.0672 × 2.69
when u add 0.0172 and 0.05 u get 0.067 which is not the uncertainty it is wat u actually get after calculating the uncertainty of the density which is 0.18/2.69=0.067 so lets take 0.18 as w so w=2.69 x 0.0067=0.18
 
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when u add 0.0172 and 0.05 u get 0.067 which is not the uncertainty it is wat u actually get after calculating the uncertainty of the density which is 0.18/2.69=0.067 so lets take 0.18 as w so w=2.69 x 0.0067=0.18
yeh i get dat.. But when we used the formula δ d/d = δ m/m + 3 δ l/l didnt we take the uncertainty out of density. Or did we not?
 
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Can you help me with these past papers doubts

M/J 02- 7b
O/N 02- 9b, 2c
O/N 03- 3 c
M/J 04- 1b
O/N 04- 1b (ii)
M/J 05- 5c, 6b(iii)
O/N 05- 3c
for oct/04
you just add 4+4 percentage uncertainty of diameter cuz u square the diameter in finding the area
 
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post your and i'll try and clear them
Can someone please help me with this question. It's a very small question. Question 5c(i) they haven't told us the directions?
http://www.xtremepapers.com/papers/...nd AS Level/Physics (9702)/9702_s10_qp_21.pdf
http://www.xtremepapers.com/papers/...nd AS Level/Physics (9702)/9702_s10_ms_21.pdf

Thank you very much! :)
and

Please help me with this question In physics paper 23, question 2 part (c). I have no idea about vector triangles.
http://www.xtremepapers.com/papers/...nd AS Level/Physics (9702)/9702_s10_qp_23.pdf
 
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1,601
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Can you help me with these past papers doubts

M/J 02- 7b
O/N 02- 9b, 2c
O/N 03- 3 c
M/J 04- 1b
O/N 04- 1b (ii)
M/J 05- 5c, 6b(iii)
O/N 05- 3c
M/J 05- 5c, 6b(iii)
5c)as the lines spread they keep going far apart and the farther apart they go the less bright they become
6b(iii)couple is force x distance so now the force is at an angle of 35 so we resolve it so it will be 2.4x10^-12 * sin35 * 2.5x10^-3
 
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Can someone please help me with this question. It's a very small question. Question 5c(i) they haven't told us the directions?
http://www.xtremepapers.com/papers/...nd AS Level/Physics (9702)/9702_s10_qp_21.pdf
http://www.xtremepapers.com/papers/...nd AS Level/Physics (9702)/9702_s10_ms_21.pdf

Thank you very much! :)
and

Please help me with this question In physics paper 23, question 2 part (c). I have no idea about vector triangles.
http://www.xtremepapers.com/papers/...nd AS Level/Physics (9702)/9702_s10_qp_23.pdf
the left circle will have the positive sign and the right circle will have the negative sign as it always moves from higher potential to lower potential
 
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