http://www.xtremepapers.com/papers/...nd AS Level/Physics (9702)/9702_w10_qp_22.pdf
How do we do this by calculation. Help plzzz
How do we do this by calculation. Help plzzz
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http://www.xtremepapers.com/papers/...and AS Level/Physics (9702)/9702_s06_qp_2.pdf
no 1cii
no 6
no 7
Bump
For 1cii, i dont know how to justify it..
For no 6, how do i know how many nodes are there? And why is 1/2lambda is 32.4 cm?
For no.7, i dont understand (a) (b) and (c)..how do i know which lamp is faulty? dont remember being taught this..
Please help. Waves in tube and electronics are my weakest topics
Nope. it aint included.
f=mass*acceleration , f=mass*(v-u/t) so force= mv-mu/t rearranging the equation u get force*time=mv-mu and mv-mu is change in momentum and force*time is impulse so impulse=change in momentumcan any one explain me what is impulse?
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]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 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.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
From my understanding, the answer should be C.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.
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 )could somebody please help me with question 5 aii.) and iii.)
qn paper: http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w11_qp_22.pdf
ms: http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w11_ms_22.pdf
i just don't understand kirchoff's second law;(
omg tx soo much .....i actually understood now....God blessI'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 )
View attachment 10790
The mark scheme is incorrect.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 ??
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 .......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%
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.
Can someone please tell me what definitions are commonly asked in questions, so that I can prepare them? Jazak Allah.
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