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AS Physics P1 MCQs Preparation Thread.

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Why are we assuming that the initial and final velocity are 0?? It dusn't say it is dropped from rest?

question says ball is dropped....which means its currently in rest,
had it been moving, it could not be dropped i guess
what you say???
 
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For this velocity-time graph :
Untitled1.png

Whats the displacement-time graph? My answer was B but displacement cant be negative right? although later it was moving in opposite direction ...

Untitled.png

Edit: Option A looked like this :

Untitled2.png
 
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it was a curve begining from s=0 and ending at s=0 also.............
Yah dude how can that be ryt when Displacement becomes zero two times , Look at the velocity time graph , it goes to max displacement then 0 and then max displacement in opposite direction and then zero...
 
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btw for variant 1 ppl...there was a question in da last questions which included da LDR and a fixd resistor in which voltmeter was parallel 2 da fxd resistor and not 2 da LDR...then they said 4 a high voltage what should da conditions be of light 4 resistor and ldr......wat was da answer :S
 
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btw for variant 1 ppl...there was a question in da last questions which included da LDR and a fixd resistor in which voltmeter was parallel 2 da fxd resistor and not 2 da LDR...then they said 4 a high voltage what should da conditions be of light 4 resistor and ldr......wat was da answer :S
hmmm wasnt it an LDR and a variable resistor??:confused:
i wrote A as the answer : high light intensity, high resistance for second column
not sure though:/
 
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hey i've s0me questions on physics too, they are
may/jun 02 Q13 in the attachment below and
oct/nov 03 Q14
i just don't get which distance to take between the two forces to calculate the torque:unsure:
 

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hey i've s0me questions on physics too, they are
may/jun 02 Q13 in the attachment below and
oct/nov 03 Q14
i just don't get which distance to take between the two forces to calculate the torque:unsure:


for question 13 ... for finding torque we always have to take the perpendicular distance between the two forces. So taking it needs to find it out.
Do so by using the theory of right angled triangle.

sin50 = per./hyp.

sin50 = per./0.30
solve to get the perpendicular distance here :)
 
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for question 13 ... for finding torque we always have to take the perpendicular distance between the two forces. So taking it needs to find it out.
Do so by using the theory of right angled triangle.

sin50 = per./hyp.

sin50 = per./0.30
solve to get the perpendicular distance here :)
thXx,,,it really helped;)
 
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I have some problems..!

October-November 2011. Paper 11. Question no. 29. Question no. 30. I am not getting it at all! Can anyone please explain it?

May-June 2011 Paper11. Question no. 13. Also help me in this one. Followed by question no. 14
 
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HELP
I NEED QUESTION 5 THIS PAPER
 

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anyone cud help me with
QUESTION 5
 

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