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Some user posted it here...
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Physics: Post your doubts here!
- Thread starter XPFMember
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it is given in the question 'the sphere of radius R.....' and the equation stated above is of a sphere and we are using the same equation in the calculation 4 rain drop. so it can not be cone or hemisphere or anything else
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i found it really helpful but agar text pura hojtaaa ...
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^ You may ask the user who posted it... Link : http://www.xtremepapers.com/community/threads/physics-practical-notes.7715/#post-102210
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thanks =)
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Anytime... 
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for as mcq paper june 2007 question no 10 .how does area under graph provide us the force as f=mxt .? and why have u subtracted t1 from t2?when t2 is out of the area under graph.i am confused :/
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really need help with electromagnetism and transformers..please if you can send me notes I'll be very grateful
i am really sorry. i replied to the question before but i stated the wrong formula. the correct ans is
f = change in momentum / change in time
f = (p1 - p2) / (t2 - t1)
in parallel circuits, voltage is the same in every branch so v2 = v3
total voltage will be v1 + v2 = v
or v1 + v3 = v
v - v1 = v3
http://www.xtremepapers.com/CIE/International A And AS Level/9702 - Physics/9702_s07_qp_4.pdf
Q7.PART C how to approach this question.......
Q7.PART C how to approach this question.......
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when light intensity increases the ldr resistance decreases, but then the ldr will be on at day time. we need to use it in night time when light intnesity is low in the potential divider circuit?
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How do you solve Q12, Q13 and Q14 of the November 2009 variation 11 paper?
is it option B? it is moving at constant speed. so forward force = backward force
forward force = (1 x 10^3)sin30
work = f*d
(1x10^3)sin30 * 5 = 2.5 x 10^3
The only confusion was whether to use sin or cos but i get it now.So yeah thanks very much.
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If it's parallel to the slope then use sin. If it's in the opposite direction of the contact force, then use cos.
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For q14, the initial speed is v. The projectile is launched at 45 degrees to the horizontal with air resistance being negligible. So at the the highest point the projectile has no vertical component for speed, only a horizontal one.
Horizontal component= v cos 45
K.E. at highest point= 1/2 x m x (v cos 45)^2= 1/2 x m x 1/2v=1/4mv
You should be able to find the correct ratio, and final answer should be (A)0.5E
For q13, there is only tension in the top belt. That's the only line across which the force acts.
Torque=force x perpendicular distance to pivot from line of action of force.
We are given torque at Q to be 3.0Nm.
So tension in upper part of belt=3.0Nm/0.05m=60N
This tension acts on P so answer is D, but to confirm.
Torque at P= 60N x 0.075m= 4.5Nm
D is answer.
ur welcum