Physics: Post your doubts here!

[princeali97]

Can you also explain b (i) please?

Thats easy.Remember for destructive interference(i.e minima in this case) phase difference is 180 degress or pi radians.
For constructive(i.e maxima) phase difference is 0 or 360 degrees or 0 or 2pi radians.

 

[usama321]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w10_qp_21.pdf
Q6 b part help needed

Calculate the resistance of the heater first. P = V^2/R R = 52.9 ohm

i) Voltage across both the resistors would be 230. Use V^2/R = 2000W

ii)Voltage would be equally distributed across both the heaters. = 500W

iii) Calculate the parallel resistance of the circuit. 1/52.9 + 1/52.9 = 26.45. The third resistor is in series. Voltage dissipated across it would be 52.9/52.9 + 26.45 * 230 = 460/3 Volts. The rest 230/3V would be dissipated across both the other resistors.
Use V^2/R = 666.6 = 667 W

 

[randomcod]

http://papers.xtremepapers.com/CIE/...and AS Level/Physics (9702)/9702_s09_qp_2.pdf
3bi, why is the angle 90? Thanks

 

[ZaqZainab]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s09_qp_2.pdf
3bi, why is the angle 90? Thanks

http://en.wikipedia.org/wiki/Torque read about it its not related to our course so we just need to know For the minimum value of F to achieve this torque the magnitude of the angle  that should be 90

 

[_Ahmad]

http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_s13_qp_22.pdf

Can anyone please explain Q5b iii)

 

[DeViL gURl B)]

Paper

Which variant?

variant 2

 

[omaaaar]

Calculate the resistance of the heater first. P = V^2/R R = 52.9 ohm

i) Voltage across both the resistors would be 230. Use V^2/R = 2000W

ii)Voltage would be equally distributed across both the heaters. = 500W

iii) Calculate the parallel resistance of the circuit. 1/52.9 + 1/52.9 = 26.45. The third resistor is in series. Voltage dissipated across it would be 52.9/52.9 + 26.45 * 230 = 460/3 Volts. The rest 230/3V would be dissipated across both the other resistors.
Use V^2/R = 666.6 = 667 W

But in the first part when we use V^2/R we get 1000W not 2000W

 

[UXm@N]

http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_w11_qp_21.pdf
sorry this is an AS level question
just misread site

I was just revising Electricity....

 

[DeViL gURl B)]

Guys..
If anyone has an easy method .. Or the changes to X and all when the intensity increases ...
To remember or written as they are asked in the pastpapers .. Like what happens to the dark fringes etc..
Please share or if made any notes PLEASE SHARE!! I'm really bad at that thing ...


Help would be well appreciated .. Thank u

 

[Fahm Deen]

Oct/Nov 11 paper-22 q-1(b)
Please can someone explain in details and if possible provide notes for these graph type questions.

 

[_Ahmad]

http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_s13_qp_23.pdf

can anyone PLEASE explain Q5 b iii)

 

[Haya Ahmed]

http://papers.xtremepapers.com/CIE/...and AS Level/Physics (9702)/9702_w03_qp_2.pdf

Q6 (c) can someone explain please ?! Thanks ^_^

 

[Zepudee]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s13_qp_23.pdf

can anyone PLEASE explain Q5 b iii)

You have to find the path difference to X
SX = square root of 32 and 52=5
SPX=3+4=7, so path difference is 7-5=2
Wavelength = 0.80m
Path diff=n x lambda
2/ 0.80 = 2.5.
2.5 lambda is a destructive interference so a minimum is detected.

 

[usama321]

But in the first part when we use V^2/R we get 1000W not 2000W

There are two resistors in the first circuit. Both have pd of 230 across them. Thus you will multiply it by 2

 

[usama321]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s13_qp_22.pdf

Can anyone please explain Q5b iii)

The time period of the stationary wave is 20ms. After 5 ms, it would be a straight horizontal line, drawn on the dotted line. Take a look at this animation and you will understand the logic

 

[usama321]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w03_qp_2.pdf

Q6 (c) can someone explain please ?! Thanks ^_^

In the previous part, we calculated the density of the nucleus, which is very high compared to the density of the iron ball. This means that most of the mass is concentrated at the nucleus. But as the nucleus is very small, it occupies only a small volume, and the rest of the iron ball is mostly empty space, leading to decrease in density.

 

[usama321]

DeViL gURl B)
For the b part, while the velocity is positive, the ball is going upwards. When it is negative, it is coming downwards. Use area under the graph 1/2 (2.4) * 9 - 1/2 1.6 * 6

For c part, the safest to do this question would be to form an equation of the line and find the co-ordinates at time 3.5s. However, in the mark scheme, they have written (allow 4.2 +- 0.2) thus 4 would also be accepted

 

[Wolfgangs]

http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_w11_qp_21.pdf

How to solve question Q4 c ii

 

[Sam Ivashkov]

Question! In this paper (http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_s12_qp_22.pdf) Q6c (ii), the answer in the marking scheme is 490 - 500 Hz. This is as a result of multiplying the distance between two consecutive antinodes or alternate nodes (which is 0.34 m) by 2 (giving 0.68 m) and taking that value as the wavelength. My question is, why do they multiply it by 2? The stationary wave produced isn't the first harmonic, so why do we multiply it by 2?

 

[kitkat <3 :P]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w11_qp_21.pdf

How to solve question Q4 c ii

look u calculated the speed above (i got 15.49)
and the electrical energy was 25% of the kinetic energy ( thats what i understood :/)
P=E/T
P=0.25(o.5*m*v^2)/1 (it said mass per second i.e. one second)
P=0.125*m*(15.49)^2
11000/0.125*240 =m
m=1100/3 Kg s^-1
m not sure btw :/
usama321 ??