Physics: Post your doubts here!

[Wolfgangs]

http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_s10_qp_13.pdf
q35 please

 

[Thought blocker]

Heya guys! My paper was great...... but I won't get above 35
Shits
I wanna screw myself................... TIME ZYADA Q NI DETE ????????????????
I knew answers but TIME!!! HELL WITH YOU . _ .

 

[Wolfgangs]

Heya guys! My paper was great...... but I won't get above 35
Shits
I wanna screw myself................... TIME ZYADA Q NI DETE ????????????????
I knew answers but TIME!!! HELL WITH YOU . _ .

Sample 11 or 12? I hope the paper will be easier for sample 11. Anyway, I know you'll get a good grade overrall, you seemed to be skilled in Physics

 

[Thought blocker]

Sample 11 or 12? I hope the paper will be easier for sample 11. Anyway, I know you'll get a good grade overrall, you seemed to be skilled in Physics

12 Hope my guesses in 5 questions are correct........

 

[TheJDOG]

Damn, just did Paper 12 and there were few questions repeated, I'm gonna honestly say this paper wasn't smooth with me :/ It wasn't easy and wasn't hard.
Anyways, good luck all!

 

[_Ahmad]

HELP!!!!!!!!!

http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_s13_qp_11.pdf

Q24,37

 

[Wolfgangs]

Did you guys get so many circuit questions? I hope we will get at least 3 radioactive questions

 

[TheJDOG]

Did you guys get so many circuit questions? I hope we will get at least 3 radioactive questions

If I remember correctly, we got like 5 or 6 electricity questions.

 

[Wolfgangs]

If I remember correctly, we got like 5 or 6 electricity questions.

Were there repeated questions from these electricity questions?

 

[Wolfgangs]

http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_s12_qp_12.pdf
Can someone please explain this question?

 

[MiniSacBall]

HELP!!!!!!!!!

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s13_qp_11.pdf

Q24,37

Well, i figured out the answer just now,

For Q24,

Power per unit area is intensity, if you remember! [Intensity = Power/Area] ,
As Intensity (and its energy) is directly proportional to amplitude square,
So, if the wave has an amplitude of A and falls on an area of S,
So, power per unit area [intensity, P] = A^2/S
Thus when A = 2A, and S = 1/3 S
So Intensity = (2A)^2 / (1/3 S),
Thus, after solving it you will find 12A^2/S,
=12P

For Q37:
It will be D not C becoz:

If the variable resistance is zero the
current will be large and the voltmeter reading will be zero. When the variable resistance is 10 Ω the current
will be reduced, but not zero, and the voltmeter reading will be high.

 

[MiniSacBall]

http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_w13_qp_11.pdf

Q38

 

[MiniSacBall]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s12_qp_12.pdf
Can someone please explain this question?

Which one!

 

[Anonymous981]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s13_qp_11.pdf
Q8

Someone plz?

 

[Wolfgangs]

Which one!

37

 

[MiniSacBall]

Someone plz?

First,
Derive equations of both trains! using s= ut +1/2 at^2
Well first train has constant speed of 10 m/s, thus a = 0 so the equation will be s=10t
For second train it will be s= 0.25t^2 ,
When the trains will pass each other the distance will same so
10t=0.25t^2
=> 0.25t^2 - 10t = 0
solve, you will get t = 0, and t= 40,
Thus t = 40s

 

[_Ahmad]

http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_w13_qp_11.pdf

Q38

since this is a parallel circuit so the emf across 4R will be same as emf across all three remaining resistors (R,2R,3R)

lets say R=1 ohms
since they have given the V across 2R we can find the current in the down branch of the circuit
V=IR
2/2=I
I=1 A
now you can find the V across all the remaining resistors in the down branch (R,3R)

R,
V=1*1=1V

3R,
V=1*3=3V

so total V or emf =1V+2V+3V=6V

 

[_Ahmad]

Well, i figured out the answer just now,

For Q24,

Power per unit area is intensity, if you remember! [Intensity = Power/Area] ,
As Intensity (and its energy) is directly proportional to amplitude square,
So, if the wave has an amplitude of A and falls on an area of S,
So, power per unit area [intensity, P] = A^2/S
Thus when A = 2A, and S = 1/3 S
So Intensity = (2A)^2 / (1/3 S),
Thus, after solving it you will find 12A^2/S,
=12P

For Q37:
It will be D not C becoz:

If the variable resistance is zero the
current will be large and the voltmeter reading will be zero. When the variable resistance is 10 Ω the current
will be reduced, but not zero, and the voltmeter reading will be high.

Can you please explain how you derived A^2/S PLEASE

 

[Anonymous981]

First,
Derive equations of both trains! using s= ut +1/2 at^2
Well first train has constant speed of 10 m/s, thus a = 0 so the equation will be s=10t
For second train it will be s= 0.25t^2 ,
When the trains will pass each other the distance will same so
10t=0.25t^2
=> 0.25t^2 - 10t = 0
solve, you will get t = 0, and t= 40,
Thus t = 40s

Thank you sooooo much!
Can you explain q21 and 35 of the same paper?

 

[MiniSacBall]

37

For 37, the answer is D

When the switch will be connected, the current will pass through both red and green, so the they both will light dimly

I sorry in not giving an excellent explaining but i suck at electricity :/