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Physics MCQs thread.

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Can someone please explain the rules of applying different formulas for calculating % uncertainty? e.g. OCT 2003 Q4.
 
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arlery said:
http://www.xtremepapers.me/CIE/International%20A%20And%20AS%20Level/9702%20-%20Physics/9702_s10_qp_12.pdf

q. 16 why is the answer B 5.5 kW?

power = Force * velocity.
Crane's weight = 1000
mass weight = 10000

velocity of both = .5
power by crane = 1000 * 0.5 joules =550 W
power by mass = 10000 * .5 = 5000 W

total power = 5500 W = 5.5 kW
 
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mickysharif said:
ok thanks guys, what about this next one.. I hate electricity questions..
Let's suppose the current is I. Using parallel/series, P gets I, and Q and are both get I/2. Power across P will therefore be given as I^2*R, and that across Q will be (I/2)^2*R. Same for R. This will simplify to I^2/4*R.
So we can see the ratio for the powers across P:Q:R are 4:1:1.

Using ratio theorem, power across R can be found.

1/6 * 12 = 2.
 
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ivorydale said:
can someone tell how the answer to Q.9 is B in this paper:
http://www.xtremepapers.me/CIE/Internat ... 7_qp_1.pdf

max height (displacement) by object is when v= 0
so thats when t=3
when t=3 , displacement = area under graph = 45

then ball starts falling down in opposite direction... Displacement will be negative
so for next two seconds, at t = 5, area under graph = 2O
displacement for next 2 seconds is -20

total displacement = displacement for 5 seconds = 45 + (-20) = 25 m

B answer
 
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ivorydale said:
can someone tell how the answer to Q.9 is B in this paper:
http://www.xtremepapers.me/CIE/Internat ... 7_qp_1.pdf
Draw a rough sketch of the path the ball might have travelled. V=0 at the highest point, so label t=3 at the peak of your sketch. t=5 comes b/w t=3 and ground level. So you will realize that to get displacement from the ground, u need to subtract area b/w t=3 to t=5 from t=0 to t=3.
 
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bloooooo said:
http://www.xtremepapers.me/CIE/International%20A%20And%20AS%20Level/9702%20-%20Physics/9702_w10_qp_11.pdf
PLZ HELP ME WITH QUESTION 34 of this paper!!!! URGENTLY!!!!!

each wire has 0.005 ohm per meter resistance.
Resistance for 800 m for each wire = 800 * .005 = 4 ohm
Two wires (they are in series)
total R = 8 ohm

V for wires = IR = 8* .6 = 4.8

total V = 16+4.8= 20.8
so minimum V is C ..
 
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pspwxpfan said:
ivorydale said:
can someone tell how the answer to Q.9 is B in this paper:
http://www.xtremepapers.me/CIE/Internat ... 7_qp_1.pdf
Draw a rough sketch of the path the ball might have travelled. V=0 at the highest point, so label t=3 at the peak of your sketch. t=5 comes b/w t=3 and ground level. So you will realize that to get displacement from the ground, u need to subtract area b/w t=3 to t=5 from t=0 to t=3.
thnx I got my ans :)
 
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bloooooo said:
http://www.xtremepapers.me/CIE/International%20A%20And%20AS%20Level/9702%20-%20Physics/9702_w10_qp_11.pdf
PLZ HELP ME WITH QUESTION 34 of this paper!!!! URGENTLY!!!!!
The most common issue with that is ppl forget that there are two wires, so instead of 800m, 1600m should be taken as the length. This will give 0.005*1600 =8 as total resistance of wire.
V=IR will give V=(0.6)(8) =4.8v, which are used in overcoming resistance of the wire.
Add this to 16, and you get the answer, 20.8
 
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The answer to that is C where as i chosee b when i was doing that paper . Why is it C? I guess because they are connected in Parallel
 
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gexceln said:
Thought you guys might find this interesting.

they have equal resistances...
The second one is cut into smaller wires of equal length
but when u put all those together, it just forms the first one single wire, as both have same volume..
 
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