Physics: Post your doubts here!

[Hijab]

Wall area = A m²
Volume of air colliding with wall per sec .. V/s(m³/s) = u(m/s) x A(m²) = uA
Mass colliding per sec .. m/t (kg/s) = V/s(m³/s) x D(kg/m³) = uA D

Force = rate of change of momentum ..
F = (mv - mu) / t = m/t (v-u) .. .. assuming the air comes to rest, then v = 0 and ..
F = -(m/t)u .. .. (against the motion)

F = (uAD)u .. .. u²AD .. .. ►F = 33² x A x 1.20 .. (N)

Thankyou

 

[Hijab]

Learn the formula as it is used in some other mcqs too. Rate of flow Kg/s = density * area* velocity...
Thus we get 33 * 1.2 * 12 = 475.2 kg/s
Thought blocker has explained it quite well

Thankyou

 

[biscuitbiscuit]

http://papers.xtremepapers.com/CIE/...and AS Level/Physics (9702)/9702_w04_qp_2.pdf
"Q6(a)(i)", I didn't get the marking scheme on this one but is the answer something like this that resistance decreases with increasing p.d. since its an increasing curve (current is increasing)

 

[Hijab]

Wall area = A m²
Volume of air colliding with wall per sec .. V/s(m³/s) = u(m/s) x A(m²) = uA
Mass colliding per sec .. m/t (kg/s) = V/s(m³/s) x D(kg/m³) = uA D

Force = rate of change of momentum ..
F = (mv - mu) / t = m/t (v-u) .. .. assuming the air comes to rest, then v = 0 and ..
F = -(m/t)u .. .. (against the motion)

F = (uAD)u .. .. u²AD .. .. ►F = 33² x A x 1.20 .. (N)

Can u solve this question as well

 

[usama321]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w04_qp_2.pdf
"Q6(a)(i)", I didn't get the marking scheme on this one but is the answer something like this that resistance decreases with increasing p.d. since its an increasing curve (current is increasing)

It can be seen from the upward bending curve that the current does not increase proportionally to the voltage. It increases more than the voltage, which hints that with increasing voltage, the resistnace of C decreases.

 

[usama321]

Can u solve this question as well

Not sure about this one. Here's what the examiner report says:
The answer to this question cannot reliably be obtained by guesswork. Eliminating the pressure at P gives
(h1–h2)ρg = 8000 Pa, so (h1–h2) = 0.060 m. D is the only answer that fits.

 

[Mubbasher Iqbal]

There was this question which asked:
Making reference to the laws of electromagnetic induction, suggest why:
1) an emf is induced in the wire,
2) the emf is alternating.
For the first part obviously we'll use the faraday's law to answer.
But can any one answer the second part WITH REFRENCE TO A LAW?
I mean obviously when a wire is moving up and down, its direction changes and thus (by right hand rule) we know that the direction of induce current will change as well. but how to answer in terms of a law? please help.

 

[Hijab]

Not sure about this one. Here's what the examiner report says:
The answer to this question cannot reliably be obtained by guesswork. Eliminating the pressure at P gives
(h1–h2)ρg = 8000 Pa, so (h1–h2) = 0.060 m. D is the only answer that fits.

Yes D is the answer.... Thankyou very much for your help

 

[biscuitbiscuit]

http://papers.xtremepapers.com/CIE/...and AS Level/Physics (9702)/9702_w04_qp_2.pdf
"Q6(c), i don't get it how p.d. across C would be larger than R?

 

[ashcull14]

ashcull14

thnk u so much

 

[ashcull14]

The calculation shows that it takes 2 ms for one wave to complete its cycle. Now each square on the horizontal axis respresents 0.5ms. Thus dividing 2/0.5, it must take 4 complete squares for the wave to complete on cycle. That happens only in B

thnk u so much

 

[ashcull14]

Please someone tell how to calculate the percentage uncertainty.
Eg Measuring from vernier caliper, my reading is 3.52. How we will calculate the % uncertainty?

your error in 3.52 eg (3.52+-x)..so x/3.52*100=%uncertainity

 

[sara kamal]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s11_qp_41.pdf

please explain Q10 c 2
and Q9 b2

 

[biscuitbiscuit]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w04_qp_2.pdf
"Q6(c), i don't get it how p.d. across C would be larger than R?

This question someone!

 

[Mimick]

i can make private videos for you on waves and send it to you. you can ask me questions and doubts.

Thats nice however do u know anyone who teaches in person..I would prefer ur videos as well..Thanks

 

[chocolate_girl]

Mcq 16 and 26
http://papers.xtremepapers.com/CIE/...and AS Level/Physics (9702)/9702_s04_qp_1.pdf

 

[Thought blocker]

Mcq 16 and 26
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s04_qp_1.pdf

16:
B is incorrect because the velocity does not decrease uniformly (it's roughly parabolic), which means height of the ball does not increase uniformly, which means the potential energy of the ball does not increase uniformly (since Ep = mgh and rate of h is not constant).
26:
HOPE IT HELPS

 

[Mohammed salik]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w12_qp_13.pdf
Q11 and 26 Plz Thanks
Thought blocker

 

[sara kamal]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w12_qp_13.pdf
Q11 and 26 Plz Thanks
Thought blocker

for 26 answer is B
since 2 x distance between 2 nodes=wavelength
so as in 1m 3nodes are there ,the distance between 2 nodes =1/3=o.333
hence wavelength=2 x o.333=0.666m
so f=330/o.666=495Hz

 

[usama321]

This question someone!

For the same current, there is higher V across C than in the resistor. At 5v or above, even though the current in the resistor will be more than in C, the Voltage across C would be 5v out of the total of 7, thus again making the power dissipated across it higher.