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AS Physics P1 MCQs Preparation Thread.

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http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s09_qp_1.pdf
Could anyone please explain no. 11?
I don't understand why P is less than R.. since the body is underwater, isn't P supposed to be greater than R??

What exactly makes you think that?

The pressure will be greater at the greater depth. Force = Pressure * Area so the force is going to be greater at the bottom too. The fluid at the top is going to push down on the particles at the bottom due to gravity so there's more pressure at the bottom.
 
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#9 & #26 M/J 2004 please
(9) Ethanoic acid is a weak acid so it wont ionise completely. Sodium hydroxide is an alkali so it doesn't produce H+. Sulphuric acid has 2 H+.
So answer is B. Nitric acid is a strong acid and produces one H+

(26) As you can see, there are two intermediates formed in the process. Hence, the energy diagram will have two humps, one for each intermediate.
The second hump should always be lower because the reaction can only proceed as long as the energy barrier for the following step is less then the preceding one.
So the correct answer is D

Hope that helped :)
 
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Th
(9) Ethanoic acid is a weak acid so it wont ionise completely. Sodium hydroxide is an alkali so it doesn't produce H+. Sulphuric acid has 2 H+.
So answer is B. Nitric acid is a strong acid and produces one H+

(26) As you can see, there are two intermediates formed in the process. Hence, the energy diagram will have two humps, one for each intermediate.
The second hump should always be lower because the reaction can only proceed as long as the energy barrier for the following step is less then the preceding one.
So the correct answer is D

Hope that helped :)
This is chemistry not physics x_x thanks anyways i got help :)
 
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I is directly propotional to a^2
I=ka^2
I=k*(2a)^2
I=4ka^2
I=4 times the previous Io


I is directly propotional to f^2

Frequency is halved since the time period is doubled..

so
I=kf^2
I=k(1/2a)^2
I=1/4kf^2
I=one fourth of the previous Io

so.. we will multiply both the coefficents of k to get the final value.

1/4 * 4 = 1
I=ka^2

therefore I remains same.
 
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It's just math. They said the intensity is proportional to the square of the amplitude as well as square of the frequency

P = Io.

Q has double the amplitude, so 4x the intensity = 4Io.
Q has half the frequency,so 1/4x the intensity = 4Io * 1/4 = Io.
 
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I is directly propotional to a^2
I=ka^2
I=k*(2a)^2
I=4ka^2
I=4 times the previous Io


I is directly propotional to f^2

Frequency is halved since the time period is doubled..

so
I=kf^2
I=k(1/2a)^2
I=1/4kf^2
I=one fourth of the previous Io

so.. we will multiply both the coefficents of k to get the final value.

1/4 * 4 = 1
I=ka^2

therefore I remains same.


thanks ... one question...
When the strain in the specimen is increased, what happens to the resistance of the wire?
as the ansers .. > It increases, because the length increases and the cross-sectional area decreases. ...but how ..? strain is inversly proportional to length
 
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thanks ... one question...
When the strain in the specimen is increased, what happens to the resistance of the wire?
as the ansers .. > It increases, because the length increases and the cross-sectional area decreases. ...but how ..? strain is inversly proportional to length

Can you link the paper.
 
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