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here you go
even i downloade this one but somehow it is for some other paper :S
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here you go
In M/J 2004 #25 ,why is point P moving downwards?
even i downloade this one but somehow it is for some other paper :S
Thanks matePressure=Density*G*depth
the more the depth the more the pressure.. pressure at same depths is constant. So answer should be A.
i did, they have different question papers :S
u wanted P1 of May june 2001 ryt??
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??
Thank you very much!in this u need to make a new wave nd u will see that point P moves downwards cz the direction of wave is to the right....nd Q barely moves hence the answer is A ..hope u got it...
(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+.#9 & #26 M/J 2004 please
This is chemistry not physics x_x thanks anyways i got help(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
Omg.. I thought I was in the chemistry thread!!Th
This is chemistry not physics x_x thanks anyways i got help
Omg.. I thought I was in the chemistry thread!!
Anyways, it helped me revise a bit of chemistry
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
Can you link the paper.
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