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Physics: Post your doubts here!

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any good books that can ultimately make a difference if consulted?
Pacific phy 1 and 2 ...for AS these books are v.v.good and you can also try long man if u want to ...
phy course book is not enough as alot of topics are either missing or not explained properly so you should definitely try Pacific phy.
 
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Pacific phy 1 and 2 ...for AS these books are v.v.good and you can also try long man if u want to ...
phy course book is not enough as alot of topics are either missing or not explained properly so you should definitely try Pacific phy.
the coursebook
ZaqZainab it's actually a book for 4k. I don't wonna waste that. Do you second what Hassan Ali Abid said?
 
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4k? you can get a photocopy and its worth it dude
whose is hassan Ali abid?
even I don't know him. he's a member who replied to my query. i quoted his reply.
Pacific phy 1 and 2 ...for AS these books are v.v.good and you can also try long man if u want to ...
phy course book is not enough as alot of topics are either missing or not explained properly so you should definitely try Pacific phy.
and i asked for your opinion on this :p
 
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While checking for an elastic collision, we see whether the statement "relative speed of approach was equal to relative speed of separation" is satisfied
Now the key term is 'relative'. For two bodies moving in opposite direction to each other, let's say, you are driving at 50 km/h, and there is another car coming towards you at 50 km/h. Thus, the relative speed of the incoming car would be 100km/h, as you are going in the opposite direction.

Now let's say you are overtaking a car which is driving at around 50 km/h, while you are driving at 60 km/h. Your relative speed would thus be 10 km/h only.

Now, let's look at the question. (NOTE: The question is talking about speed, not velocity)

Ux and Uy are moving in opposite directions, thus their speeds are added as we are finding about the relative speed.
Vx and Vy are moving in opposite directions, thus again their speeds are added.
 
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Stationary wave is produced inside the tube ...it means it has nodes and antinodes...and max displacement is at antinodes and min at nodes so dust heaps are formed on nodes (as the displacement at nodes is zero).
but whare heaps are formed when water is drained off.
 
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Salam.

Can anyone help me with this A2 Paper 4 Physics question on electromagnetism and nuclear physics?

Please provide with as much explanation as possible...

Thank you.

May Allah repay your kindness.

Amin.


Electromagnetism:

Question 6. (c)

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s09_qp_4.pdf


Nuclear Physics:

Question 9 (b)

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s09_qp_4.pdf


http://papers.xtremepapers.com/CIE/...and AS Level/Physics (9702)/9702_s09_qp_4.pdf
 
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Help plx wid full a-z explanation
the answer would be B write the units of what they have given us
so we know distance is L and time is T so L is meter and time is second and a is radius which is in meters
the equation L(3+(a^2/P)=QT^2 sin teta
3,and sin teta are constants we can remove them

using units -> m(m^2/P)=Q*s^2

we have to have LHS=RHS
there is no seconds on the left hand side so we have to remove seconds we will know that one of the value of Q is s^-2 that's the only way to remove Q(xs^-2) we don't know x yet
we will have
m(m^2/p)=xs^-2*s^2 => m(m^2/p)=x As its only x and not squared or raised to anything m=x and as there is only m on the left hand side
we have to still find p we know x=m so m(m^2/p)=m
p=m^2
and so q=ms^-2
 
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