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Physics MCQs thread.

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Oh sorry, my bad.


It's D because when two particles approach each other, we add the speeds and when they move in the same direction, we subtract their speeds.
 
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zishi said:
Because the half wavelength has an angle of 90 degree. Quarter of it is 45 degrees. The distance between those two points is less than a quarter of wavelength, so it must be less than 45 too.

hey zishi m afraid that half a wavelength has an angle of 180 degree while quarter, 90 degree. So, we can't eliminate the "45 degree option" so easily.

Btw hassam yar, u cud use the fact that the given wave is a sine curve with amplitude y_0. Now imagine moving from say d = 0 degree (although it's a distance) to d = 45 degree. What change in the value of y does it correspond to.................? Obviously ----> y_0 * sin(45 degrees) = y_0/sqrt(2) [Doesn't seem to be the correct option]...... Now consider the other given options similarly and then finally comes the "30 degrees option." ..... Now u know that as y_0 * sin(30 degrees) = y_0/2, so on the given curve as Q is in the 3rd quadrant i.e. y_0 * sin[(180 + 30) degrees] = - y_0/2

So, it's A nt B


Hope that helps


This response corresponds to the last post on Pg#15 of this thread. I culdn't post it on that page...;p
 
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u just simply subtract them but since U2 is in the other direction it has a negative sign ..
u1 - (-u2) = v2- v1
u1 + u2 = v2 - v1
 
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Mobeen said:
http://www.xtremepapers.me/CIE/International%20A%20And%20AS%20Level/9702%20-%20Physics/9702_s06_qp_1.pdf
q 34

E = I ( R + r )
12 = I (1 +3 )
I = 3A

rate of energy = power
P = I^2R
P = 3^2 x 3
P = 27W
 
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How can I solve this?
 

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Xthegreat said:
Mobeen said:
http://www.xtremepapers.me/CIE/International%20A%20And%20AS%20Level/9702%20-%20Physics/9702_s06_qp_1.pdf
q 34

E = I ( R + r )
12 = I (1 +3 )
I = 3A

rate of energy = power
P = I^2R
P = 3^2 x 3
P = 27W
but it asked for "At what rate does the voltage source supply energy to the heater" .. wont we use the resistance of the supply ?
 
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arlery said:
How can I solve this?
Distance between one maximum and the other means half the wavelength, as the distance between a crest and a trough. So λ=30mm. Then by v=fλ, speed of waves = 3.0 x 10^8, ans is C.
 
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by Zishi » Fri Jun 03, 2011 12:00 pm

hassam wrote:Zishi wrote:hassam wrote:b.PNG


Because while its being earthed, and at the same time there's an electric filed between the plates, the charge won't move down the earth. It's a fact, learn it.

zisi this is wat ER says...bt i cnt undrstnd it
Too many candidates of all abilities gave the answer B. It is a very common misconception that an earthed
plate must have no charge on it. The presence of the negatively charged plate would make the potential of
the plate negative were there not some positive charge on it. Also, field lines to a negative charge must start
from a positive charge.


So you've to get your concept right about it. What I said is a fact(as ER says), so learn it. The explanation for it is far beyond scope of A-Level


Hassam yar d u remember the time when we studied O level electrostatic induction in class....? Jux apply those concepts here it's jux a piece of cake (nothing beyond A level's scope whatsoever, no FACTS to learn, jux the use of LOGIC is needed). Accordingly, the extra electrons in the negatively charged upper plate will repel the electrons in the bottom plate, making it electron deficit (positively charged), down to the Earth as the bottom plate is earthed. Consequently, u'll get a charge pattern as shown in part A.

Hop that helps

This response corresponds to the 1st post on Pg#19 ov this thread
 
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