Post Your AS LEVEL PHYSICS PAPER 2 DOUBTS

[raamish]

http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s10_qp_21.pdf

In5c) The arrows will be left and right or up and down? And why didnnt we take cos to resolve.

 

[applepie1996]

i want to ask:​

if the separation of fringes increases then is it always that brightness of fringes will decrease??​

yup it is because they move farther apart so brightness decreases

 

[applepie1996]

hello everyone!!
can anybody tell me what's the value of 'g' we use in physics???
is it 10 or 9.81????

always 9.81

 

[applepie1996]

S05 : 4c,6iv,5iii & c
W05: 4
s06: 2b,5c
S07:5c
No7: 1b,4b, c & d

Any help wud be greatly appreciated. PLease post as soon as possible ty ))

s06 q2b
imagine all the forces passing through p and when u do that u'll see that now the distance of all the forces from p is zero so sum of moment about p will also be zero
Q5c
1/2 x 16 x (6^2 - 4.5^2) + 1/2 x 16 x (3^2 - 4.5^2) = 36 all the lengths were in cm so divide answer by 100 36/100= 0.36J

 

[applepie1996]

http://www.xtremepapers.com/papers/...nd AS Level/Physics (9702)/9702_w10_qp_22.pdf

Question 1 (b). Help will be highly appreciated

the downward force is 25 N so the upward force will also be 25 N
so u find are by resolving it into it's component which is cos35 x 25 = 20.5 N
now u draw a parallelogram (complete the drawing on the figure itself)
and u'll end up getting T=20.5/tan 55

 

[Silent Hunter]

i'm sorry but i'll try 2 do it and tell u as soon as possible

answer needed please

 

[applepie1996]

answer needed please

ok i'll try 2 do it now

 

[Silent Hunter]

ok i'll try 2 do it now

thank you

 

[applepie1996]

http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s10_qp_22.pdf

Q4 (c)

i think it's done like this
lowest frequecy = 1st harmonic (first harmonic is the first sound wave produced) which over here is lambda/4
so it will lambda/4=0.45m
=1.8m
f=v/lambda
=183Hz

 

[Silent Hunter]

mean wheneva they going to give us lowest frequency we use the 1/4 lambda thing? always?

 

[applepie1996]

mean wheneva they going to give us lowest frequency we use the 1/4 lambda thing? always?

nop over here the first wave produced is 1/4 so dat is y v use it lyk dat but in different cases different stuff

 

[leosco1995]

http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s05_qp_2.pdf
Q5 b (iii)

http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s06_qp_2.pdf
Q5 (c)

 

[applepie1996]

http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s05_qp_2.pdf
Q5 b (iii)

http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s06_qp_2.pdf
Q5 (b)

Q5 b (iii) in the previous part they assume that the light incidents normally on the grating whereas in the diagram in part (iii) it forms an angle so there is a path difference so formula can't b used

 

[applepie1996]

http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s05_qp_2.pdf
Q5 b (iii)

http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s06_qp_2.pdf
Q5 (b)

S06 5 b
work done = area under the graph = 1/2 * F * x
F=k*x
so substitute F in work done= 1/2 * k *x^2
substitute x from the graph
w=1/2 * k * (x2^2 - x1^2)

 

[leosco1995]

S06 5 b
work done = area under the graph = 1/2 * F * x
F=k*x
so substitute F in work done= 1/2 * k *x^2
substitute x from the graph
w=1/2 * k * (x2^2 - x1^2)

Actually meant part (c).. I made a typo there. :\ In any case, thanks for the quick reply.

 

[applepie1996]

Actually meant part (c).. I made a typo there. :\ In any case, thanks for the quick reply.

part c ok i'll do dat 2

 

[applepie1996]

Actually meant part (c).. I made a typo there. :\ In any case, thanks for the quick reply.

Q5c
1/2 x 16 x (6^2 - 4.5^2) + 1/2 x 16 x (3^2 - 4.5^2) = 36 all the lengths were in cm so divide answer by 100 36/100= 0.36J

 

[littlecloud11]

http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s05_qp_2.pdf
Q5 b (iii)

http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s06_qp_2.pdf
Q5 (b)

Q5biii) when you're using the formula dsin(theta) = n* lambda, one of the condition is that the incident light has no path difference before entering the diffraction grating. This is only possible when the light is normal to the grating surface, if the incident light enters at an angle then there is a path difference and the formula does not apply.

Q5b) work done = average force * extension
for the extension from x1 to x2 the change in force is F1 to F2
so, work done = (F1+ F2)/2 * (x2- x1)
now substitute F=kx for both F1 and F2 and you get = (kx1+kx2)/2 * (x2-x1)
rearrange to get, 1/2k * (x2+ x1) (x2-x1) [(x2+x1) (x2-x1) = x2^2 - x1^2]
so, finally- work done = 1/2 k (x2^2- x1^2)

 

[Silent Hunter]

http://www.xtremepapers.com/papers/...nd AS Level/Physics (9702)/9702_s11_qp_23.pdf

question 6 (c)(iv) and in question 5 (b)(ii) why are we considering the parallel circuit resistors? why not the series one? is the reason cuz that in parallel we have same pd across each resistor?

 

[leosco1995]

Q5c
1/2 x 16 x (6^2 - 4.5^2) + 1/2 x 16 x (3^2 - 4.5^2) = 36 all the lengths were in cm so divide answer by 100 36/100= 0.36J

How did you get the values of x2 and x1.. I don't get that part.