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As physics p1 MCQS YEARLY ONLY.

Zepudee

Zepudee

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MaaAsif said:
Mass of 1 particle= 6.6x10^-27
Mass of 5x10^4 particles= 3.3x10^-22

Then, Pressure= Force/Area
Force = Rate of change of momentum. Since final velocity is zero, force=mass*initial velocity/time
Pressure=Mass*Initial velocity/AreaxTime

Convert the area of 1cm^3 into m^3 which gives us 1x10^-4 m^3. Time here is 1 second.

Hence: Pressure= 3.3x10^-22 * 1.5 × 10^7 / 1x10^-4 * 1
Pressure= 4.95x10^-11 which can be rounded off to give 5.0x10^-11 Pa.
Click to expand...

Thank you so much sir :D
How about this?
http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_w13_qp_12.pdf
no 30 , ans is A
 
Zepudee

Zepudee

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MaaAsif said:
Lol. Sir? Doesn't my profile clearly indicate that i'm a female? :O
As for Q30, we know that intensity is directly propotional to the square of amplitude. Hence the ratio can be written as:

I/ A^2 = I' / A^2' (where I' and A^2' indicate the intensity and amplitude of the emerging light)
I / A^2 = I' / (Acos60)^2'
Cross multiply and you'll get 0.25I=I'.

Hope this makes sense. I'm very bad at explaining stuff. :p
Click to expand...

WOOO, it makes sense!! Hehehehe thank youu! Oops, sorry about that anyways. And oh, yup. Always :3 I love Snape.
 
Hadi Murtaza

Hadi Murtaza

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Joe smith said:
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w13_qp_11.pdf Q11 please?? really need help in it
Click to expand...
Mass of each ɑ particle = 6.6 × 10⁻²⁷ kg
Total mass of particles colliding per second = (6.6 × 10⁻²⁷ kg) × (5.0 × 10⁴ s⁻¹) = 3.3 × 10⁻²² kg/s
Force = Total mass of particles colliding per second × Velocity = (3.3 × 10⁻²² kg/s) × (1.5 × 10⁷ m/s) ≈ 5.0 × 10⁻¹⁵ kg m/s²
Pressure = Total mass of particles colliding per second / Area = (5 × 10⁻¹⁵ kg m/s²) / (1 × 10⁻³) = 5 × 10⁻¹¹
Answer: C
 
Y

Yogita.T

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MaaAsif said:
Mass of 1 particle= 6.6x10^-27
Mass of 5x10^4 particles= 3.3x10^-22

Then, Pressure= Force/Area
Force = Rate of change of momentum. Since final velocity is zero, force=mass*initial velocity/time
Pressure=Mass*Initial velocity/AreaxTime

Convert the area of 1cm^3 into m^3 which gives us 1x10^-4 m^3. Time here is 1 second.

Hence: Pressure= 3.3x10^-22 * 1.5 × 10^7 / 1x10^-4 * 1
Pressure= 4.95x10^-11 which can be rounded off to give 5.0x10^-11 Pa.
Click to expand...

I have one question, which is:
How do we get 1cm^3 as 1*10^-4
Shouldn't it be 1*10^-6?
 
Thought blocker

Thought blocker

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Yogita.T said:
An archer draws his bowstring back to position X. The bowstring and arrow are shown. The
tension T in the string is also shown. Then he draws the bowstring back further to position Y.

If anyone could please help in solving this question, it would be much appreciated
It's question 14 of P12 W13
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w13_qp_12.pdf
Thanks so much :)
Click to expand...
Increase in force = 2(120cos(55)) – 2(100(cos(65)) = 53N
 
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