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Physics: Post your doubts here!

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Can anyone please give me simple notes for HALL PROBE and how the HALL VOLTAGE expression is derived.
 
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Did you get an ebook? :3
I got one :3
9781471809217_grande.jpeg
 
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I think CIE is going to sue us for all the ebook priating! The publishers won't be happy and it means a lot of loss for them
 
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Is thhis for the 2016 syllabus?
Did you put it up on gceguide yet?
http://gceguide.com/Books/Cambridge International AS and A Level Physics 2nd ed.pdf

It is for 2016 syllabus :p

Not yet. Too lazy :p
you know that strange website doesn't open in Pakistan! :p
O.O what're you talking about?
I think CIE is going to sue us for all the ebook priating! The publishers won't be happy and it means a lot of loss for them
:confused: I didn't scan them. I'm just sharing the ones i downloaded from other websites. :cry:
 
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Screenshot_2016-03-25-16-49-34.png Screenshot_2016-03-25-16-49-47.png
I have posted the complete question for reference. But what I'm confused about is (b) (ii). According to the ms the new line will be below the original line reaching terminal velocity in shorter time. But if the sphere has a smaller radius thus a smaller surface area, the effect of air resistance will be less so it will take more time to reach terminal velocity.
 
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View attachment 59797 View attachment 59798
I have posted the complete question for reference. But what I'm confused about is (b) (ii). According to the ms the new line will be below the original line reaching terminal velocity in shorter time. But if the sphere has a smaller radius thus a smaller surface area, the effect of air resistance will be less so it will take more time to reach terminal velocity.
If a raindrop is falling through air, we can think of two forces acting on it:
  1. W, the weight. This makes the raindrop fall down. W is directly proportional to mass of raindrop. The mass of raindrop is directly proportional to its volume. And the volume is directly proportianal to its radius CUBED. Therefore, W is proportional to raindrop's radius CUBED.
  2. F, the frictional force. They told us in the question (in the form of a formula) that this force is directly proportianal to radius. (To power one)
So when radius is reduced, both W and F will be reduced. However, W will be reduced MORE.

When the raindrop falls, its velocity increases so F increases and, at one point, W is completely countered and so the drop is in terminal velocity.
If W is very little (like in our case) it takes a shorter time for F to counter balance the W.
 
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If a raindrop is falling through air, we can think of two forces acting on it:
  1. W, the weight. This makes the raindrop fall down. W is directly proportional to mass of raindrop. The mass of raindrop is directly proportional to its volume. And the volume is directly proportianal to its radius CUBED. Therefore, W is proportional to raindrop's radius CUBED.
  2. F, the frictional force. They told us in the question (in the form of a formula) that this force is directly proportianal to radius. (To power one)
So when radius is reduced, both W and F will be reduced. However, W will be reduced MORE.

When the raindrop falls, its velocity increases so F increases and, at one point, W is completely countered and so the drop is in terminal velocity.
If W is very little (like in our case) it takes a shorter time for F to counter balance the W.
So you're saying if I have 2 steel balls one the size of a football and other the size of a tennis ball, the tennis ball will reach terminal velocity first?
 
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Can someone explain to me how the 240 is the root mean square speed ? And why we have to multiply it by root of 2 ?
 
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