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Physics: Post your doubts here!

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I don't know a proper method for it, but there's a very good way, that I use for solving such questions. It really works :p

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Whenever you encounter such a question, place the tip of your pencil lightly on the paper. Then without moving your pencil, move the sheet of paper beneath in the direction opposite to the wave motion. Side by side as those humps arrive, move the piece of paper, up or down opposite to the position of the hump. Try it :p You'll get to know of this method. :p
 
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It still does not make sense to me. Th water is displaced not only by the water molecules but also due to the object and a denser object will displace more or whatsoever. :p
And sure.
Sorry, I think my explanations were vague. Okay let me describe the thing again:

The upthrust is the weight of water displaced. The amount of water displaced has nothing to do with the density. The aomunt of water displaced depends on the volume of the cuboid. Volume is same for all the cuboids, so they will displace equal amounts of water. So they will have the same upthrust. :p Clear now?
 
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I don't know a proper method for it, but there's a very good way, that I use for solving such questions. It really works :p

View attachment 59926
Whenever you encounter such a question, place the tip of your pencil lightly on the paper. Then without moving your pencil, move the sheet of paper beneath in the direction opposite to the wave motion. Side by side as those humps arrive, move the piece of paper, up or down opposite to the position of the hump. Try it :p You'll get to know of this method. :p
I can't visualise it lol I got the question tho. :p
 
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Sorry, I think my explanations were vague. Okay let me describe the thing again:

The upthrust is the weight of water displaced. The amount of water displaced has nothing to do with the density. The aomunt of water displaced depends on the volume of the cuboid. Volume is same for all the cuboids, so they will displace equal amounts of water. So they will have the same upthrust. :p Clear now?
YEAH THANKS REALLY.
 
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Guys does anyone have edexcel AS physics as a pdf or active book or know where to find it ??!!
 
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Can someone do part c please. Need it urgent.
Using the formula you proved above, you can find E, the Kinetic Energy required to statisfy their condition.
Next, using the relation E = VQ, you can find V. This is because the kinetic energy is being provided by the electric potential energy. You already know the charge and mass of an electron from the data provided at the beginning of paper.
 
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Paper 4 (A Level Structured Questions): The assessment of core and applications topics will be integrated. There will no longer be a section A and a section B.

What do they mean by the first statement ?
 
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Paper 4 (A Level Structured Questions): The assessment of core and applications topics will be integrated. There will no longer be a section A and a section B.

What do they mean by the first statement ?
The applications will be merged with the questions. There will not be a separate section B for applications.
 
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do we have to study the As part like waves etc for A2 ?
No this is not what it means.
You must have seen the past papers they had a section A which consisted of questions from the A2 syllabus except the applications which were in section B. Now they'll be merged together. There won't be any section B.
 
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someone help!!!!........
9702/s14/qp13/question 32

Thanks in advance.....
The electric field strength, E, is given by this formula:

E = F/Q (where F is force, Q is charge and E is electric field strength)

We know F = ma, so E becomes:
E = ma/q ----(i)

You're given that, K.E = k. Use the equation of K.E (1/2mv^2) to find the value of m and substitute this value of 'm' into (i)
m = 2K/v^2 ---- substituting in (i)

(i) will become:
E = ( (2K/v^2) * (a) )/ q ------(ii)

Now use the equation of motion: v^2 = u^2 + 2as
The charge starts moving from rest, so initial speed (u) is zero. And it covers a distance of 'd' for reaching plate Y, so s=d. So,
v^2 = 2ad ---- put this one into (ii)

E = ( (2K/2ad) * (a) )/ q

Simplify it, you'll get :

E = K/qd

So the answer is C.
 
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