Physics: Post your doubts here!

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Swoorup

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@Abdulrab
The link you have given seems invalid. Could you link papers contained database of this site itself.
 
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Abdulrab

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(NOV 05 Q 8,11,34 AND 40) AND (3,7,22 AND 32 FROM JUNE 05) AND ( 9,36 AN 39 FROM jUNE 06) THANKS
 
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NOV-5 Q 8
The answer is A, as F is directly proportional to A, as F=ma.
So A against time would give you the same sloped graph as in the question.
And since A is constant all the time, speed would be raised at a constant rate, hence a straight line.

Q 11
The answer is A because only gravitational force is acting on the projectile. And G always acts downwards.
The answer cannot be C because it does not follow a circular path.

Q 34
For this question, do not consider the slope itself, because it is rather confusing if you do so.
Know that R=V/I, so calculate for each point the value of R. I will be supposing the following values:
At A, v=2, I =1 , R=2
At B, v=3, I =2, R=1.5
At C, v=4, I =3, R=1.33
At D, v=6, I =3.5, R=1.71

Hence the answer is C as it has the smallest value.

Q40
Its obviously proton , D .
Subtract the Initial equation by F, and then you would get 1 at the top and 1 at the bottom, which signifies proton.

Review what I just said and ask if you have doubt in those answer again.
Ill do the rest after some time. Then post you the reasons again.
 
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Considering 2 kg mass, we have the equation: T stands for tension, F for Friction
mg-T=ma
20-T=2a....(i)
Now for 8 kg mass,
T-F=ma
T-6=8a....(ii)
Equating both the equation we get a= 1.8 ms^-2
 
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Just use the formula: 1/R=1/r1 + 1/r2
Since there are 6 copper wire the equation would be
1/R = 6/10+1/100
R=1.6 ohm
 
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Considering 2 kg mass, we have the equation: T stands for tension, F for Friction
mg-T=ma
20-T=2a....(i)
Now for 8 kg mass,
T-F=ma
T-6=8a....(ii)
Equating both the equation we get a= 1.4 ms^-2
 
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KaiserFury

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Urgent help needed, exam tomorrow, AS Physics MCQ questions

Hello,

I have a couple of AS Physics MCQ questions, and I am in need of urgent help as my exam is tomorrow.

Question 1) A plane wave of amplitude A is incident on a surface of area S placed so that it is perpendicular
to the direction of travel of the wave. The energy per unit time reaching the surface is E.
The amplitude of the wave is increased to 2 A and the area of the surface is reduced to 0.5S.
How much energy per unit time reaches this smaller surface?

The answer is 2E.

2) Number 37 in this past paper:

http://www.xtremepapers.com/CIE/Interna ... _qp_11.pdf

The answer is A, but I don't understand why.

Can someone please tell me how to solve either of these questions?

Thanks.
 
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anahita16

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Ok thanks!!! :) :) :)

i had used sin theta but wrongly substituted the value :Search: . From your triangle, sin theta = 30/50 = 3/5
I didn't use the tan part...

Hey you unblocked me!!! thankyou thankyou! lol

Now i can fully concentrate on the other questions phew.
 
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Abdulrab

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answer is C as whenever the question asks to give anser to appropriate no. of sig figs take 2 sig figures
 
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anahita16

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@ Arshiful
Question 5
A and B were common responses. Perhaps candidates did not see the significance of the fact that 3% of
330 is 10 but just tried to give rounding to 3 or 4 significant figures.

from Examiner's Report

but i still can't figure out what it means. sorry dear.
and 3% of 330 is 9.9.
 
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