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Physics: Post your doubts here!

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Jaf

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can someone please explain to me the question 6c?? The mark scheme has additional values I don't know where the got them

http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w11_qp_23.pdf
http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w11_ms_23.pdf

for all I know
charge of alpha particle = 2
mass of alpha particle =4

charge of beta particle 1.6x10^-19
mass of beta particle 9.11x10^-31
The charge of the alpha particle is 2x the elementary charge (charge of a single electron/beta particle). [obviously the sign changes but that doesn't matter since it's only a ratio]
The mass of the alpha particle is 4x the rest mass of a proton (1.67 × 10^–27 kg).

Both these^ values are given in the beginning of the question paper.
 

Jaf

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http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w05_qp_2.pdfQ5 c 1 annd 2 how to show that intensity of B at pont p ( and whaere is poimt p) is 4/9 I ...and the next part also plzz help
The wave is passing through a point, and the displacement-time graph is drawn for that. P is not a point on the wave, rather P is the point the wave is passing through (think about it, there's a difference). Any how, even if you don't get it, it makes no difference to your answers. The intensity is common to all points ON a wave and so is the phase difference between the two waves.

c)i)
I/a^2 = constant (where I is intensity and a is amplitude)
Let's take the intensity of wave xB as x.

So I/3^2 = x/2^2
x = 4I/9

ii) The resultant amplitude when xA and xB meet is 1. Let the new amplitude be y.
So I/3^2 = y/1^2
y = I/9.
 

Jaf

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An athlete bends down and then jumps straight up into the air. At the instant he starts his jump
three forces act on him – his weight, an upthrust from the air and the contact force with the
ground. What is the correct order in increasing magnitude of these three forces
A. contact force weight upthrust

B. upthrust contact force weight

C. upthrust weight contact force
D. weight upthrust contact force

Provide an explanation if you can answer. It's not from past papers.
From my understanding, the answer should be C.
So first, know that even though the force is called 'up'thrust, it acts against the motion of the body and hence downwards.
The only force from the three forces that IS acting upwards is the contact force. The weight also, obviously, acts downwards.

So since the athelete accelerates (upwards; for a few fractions of a second atleast) the net force acting on the athlete must be upwards. Since there's only one upward force acting on the athlete, this must be the contact force. So we can conclude the contact force is the greatest force acting. A and B can definitely be eliminated.

Now I'm unable to explain why the upthrust should or should not be greater than the weight because of lack of information. I mean there are way to increase/decrease the upthrust. But I'm guessing since this force depends on the velocity, and the initial velocity can't be so high so as to produce a force greater than the weight, the upthrust should have a smaller magnitude than the weight.
 

Jaf

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I've redrawn the circuit to make it easier to apply Kirchoff's law. Try now (kirchoff's 2nd law is kind of hard to explain on paper :( )
photo.JPG
 
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could someone please explain Q1 PART b and Q4 PART c
WHY WON'T ANYONE ANSWER THIS!!!!
 

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I have a serious problem with uncertainties, almost all the sums i solve are wrong! For instance in question 1b(ii) 21/M/J/11 http://www.xtremepapers.com/papers/...nd AS Level/Physics (9702)/9702_s11_qp_21.pdf
I do not get how the uncertainty in A = 2 × (0.01 / 0.38) × 100 = 5.3 % I guess we multiply by 2 because of pie r square.
But because it is the radius why don't we halve the diameter's percentage uncertainty?
Also how do i decide between which significant figures, if nothing is mentioned, for instance in the question where the percentage uncertainties are .17% and .27, why is it not .17% and .267% or .20% and .27%
 

Jaf

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I'll post the answers. You'll be surprised just as I was.

a. horizontal velocity= s/t = 2.5/0.6= 4.17m/s

b. Vertically final velocity is zero as ball passes over cross-bar.
v = u + at
0 = u - (9.81*0.6)
u = 5.89m/s (ans)

c. u = root over(4.17^2 + 5.89^2)
= 7.21m/s

d. angle = tan^-1 (5.89/4.17)
= 55degrees

The mark scheme totally trolled me!!! I don't get it. How can horizontal velocity be 2.5/0.6 ??
The mark scheme is incorrect.
Put in the values in the range formula, and you don't get your original question.
range = [v^2 * sin2(theta)]/g
You get 4.98 m when you put in the values from the answers when it's really supposed to be 40 m.
 
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Can someone please tell me what definitions are commonly asked in questions, so that I can prepare them? Jazak Allah.
 
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I have a serious problem with uncertainties, almost all the sums i solve are wrong! For instance in question 1b(ii) 21/M/J/11 http://www.xtremepapers.com/papers/...nd AS Level/Physics (9702)/9702_s11_qp_21.pdf
I do not get how the uncertainty in A = 2 × (0.01 / 0.38) × 100 = 5.3 % I guess we multiply by 2 because of pie r square.
But because it is the radius why don't we halve the diameter's percentage uncertainty?
Also how do i decide between which significant figures, if nothing is mentioned, for instance in the question where the percentage uncertainties are .17% and .27, why is it not .17% and .267% or .20% and .27%
Always use the formula Pi (D)^2/4 donot use Pi R^2 it adds complicatons i have been through this problem many times and my teacher reccomended to use the other formula .......
 
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The charge of the alpha particle is 2x the elementary charge (charge of a single electron/beta particle). [obviously the sign changes but that doesn't matter since it's only a ratio]
The mass of the alpha particle is 4x the rest mass of a proton (1.67 × 10^–27 kg).

Both these^ values are given in the beginning of the question paper.

ohh thanks a lot you saved me ♥
 
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