ALL CIE PHYSICS DOUBTS HERE !!!

[astarmathsandphysics]

a)cos it is easier to breathe in.
see attachment

[astarmathsandphysics]

a)gamma rays
b)see attachment

[ashwinkandel]

Questions from Oscillation
I have attached four questions below.

[astarmathsandphysics]

I will make some videos tomorrow for these

[astarmathsandphysics]

Here are your answers.
Will also make some videos

[ashwinkandel]

Sir(astarmathsandphysics), I think you forgot the answer to fourth question.

[astarmathsandphysics]

You are right sorry.

[astarmathsandphysics]

a)natural resonance frequency of system shown be nowhere near frequency of earthquake waves.
b)gradient is negative and constant so a=-kx
c)k=omega^2=10/0.02=500 so omega=sqrt(500)=22.4
f=omega/2pi=22.4/6.28=3.6 Hz

[silvercameron]

can someone pls explain why the answer to Q.no. 11 is A? See attachment.

[astarmathsandphysics]

Use Newtons Law of restitution
e=(separation speed)/(approach speed)=(v2-v1)/(u1+u2)
e=1 for a perfectly elastic collision so v2-v1=u1+u2
answer is D

[astarmathsandphysics]

Q10
Use Newtons Law of restitution
e=(separation speed)/(approach speed)=(v2-v1)/(u1+u2)
e=1 for a perfectly elastic collision so v2-v1=u1+u2
answer is D
q11
Applying F=ma for 2 kg mass 2g-T=2a
applying F=ma for 8kg mass T-6=8a
add these to get 2g-6=10a so a=(2g-6)/10=1.36

[Physics09]

http://www.xtremepapers.com/CIE/International%20A%20And%20AS%20Level/9702%20-%20Physics/9702_w10_qp_12.pdf
Explanation to the answer of MCQ 27, urgently please!!

[astarmathsandphysics]

see attachment

[Physics09]

Thanku so much!
Also, MCQ 34 and 27
http://www.xtremepapers.com/CIE/International%20A%20And%20AS%20Level/9702%20-%20Physics/9702_s10_qp_12.pdf
Thanks again!

[SkyPilotage]

Q27 - Electric Field direction is always from +ve to -ve plate. SO if the electric field direction is away from you means the +ve plate is closer to you than the -ve plate, therefore the electron beam goes towards the +ve plate which is towards you so the answer is D!

Q34 - Total Resistance is 1.5R ---> so I^2 = P / 1.5R = 8 A^2 ---> I = 2.83 A
Here I took the value of R as 1 ohm, you can take it as any value you want as long as you are cosistent with all the resistors P,Q and R.
So P across P = (2.83)^2 x 1 = 8 Watts
P across Q = (2.83 / 2 ) ^2 x 1 = 2 Watts
P across R = P across R2 = 2 Watts.
Answer is A

There you have it!