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Physics MCQs thread.

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Hateexams93 said:
Zishi said:
No, they won't exchange their speeds unless they have same masses, and final velocity may not be equal to initial. They'll be the velocities which would add up to give the expression of U-V.
can u plz just solve the question which i posted on the previous page .. using the speed of approach and separation

That can be solved using a simple rule, when two same masses undergo elastic collision their speeds are exchanged. So after the collisions, the moving one well be strationar. D is the answer.
 
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change in momentum is p1-p2
p1=mv
p2=m-v (cuz after collsion it'll go left, so we take a negative sign!)
p1-p2
mv - m-v
=2mv to the left it'll bounce back to the left side!!
 
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Hateexams93 said:
:( why i'm always getting stuck in this kind of questions , can some1 help with this
change in momentum= m ( v - u )
change in momentum= m ( -v - v )
change in momentum= -2mv , the sign indicates the direction so it is 2mu to the left.
 
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robotic94 said:
http://www.xtremepapers.me/CIE/International%20A%20And%20AS%20Level/9702%20-%20Physics/9702_w06_qp_1.pdf

q4,
q9 (why C and why not A)
q17

thankyou :)

Q4: Frequency = 1/t where t = time period in seconds.

you've to multiply time base count by the number of boxes in which a cycle completes. That will be time period of the wave in milliseconds. 1/(time in ms) = 1000/(your product)

This will give you the frequency. Check all the options for it. B will be the answer.

Q9: Well, vertical component of velocity is not constant and has an acceleration of -g, that's why it's C.

Q17: efficiency = output/input times 100 = 7/100 times 100 = 7%
 
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Hateexams93 said:
yubakkk said:
ur welcome man
do u hate xam??
lol i'm a girl .. :%) and right now i hate everything..its so unfair ..every1 is done with their exams except of me ((
o sory for telling man
be positive u wil get sucess
 
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