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AS Physics P1 MCQs Preparation Thread.

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HELP ME ON THESE QUESTIONS PLS
15 22 32 of n10 variant 12
16 19 34 of J11 variant 12

waiting for quick reply pls with details on this and my previous asked questions on page 38 :) pls lesco where are u :( :( :( :(
Q15)

At terminal velocity, weight = air resistance
so mg = kv

And v = mg/k

Now substitute v with mg/k in 1/2 mv^2 formula to get the answer.

Q22) You want high sensitivity for small masses, i.e. a big change in extension for smaller masses. This means a lower value of 'k' is needed for the smaller masses. The rigid box is there to prevent external forces like wind from pushing the masses on the spring. Also, when you have larger masses on the spring, then you need a higher value of 'k', i.e. a small extension for a large load. So the answer would be A.

Q32)

A and B are wrong because the e.m.f. can't decrease, its the energy being provided by the battery.

Also, VT = E - Ir (terminal voltage i.e. voltage across the battery is equal to e.m.f. - internal resistance * current).

From this equation, we can confirm that answer C is correct.

S11/12

Q16) Assume the length of the rod to be 1m. You know that the weight lies exactly in the middle, so each side is 0.5m apart. If you take the wall to be the pivot, then you can form an equation,

F sin 30 * 1 = 0.5 * 10
F = 10.0 N

Q19)

Work = force * distance in direction of force
10 = 0.005 * force
force = 2000 N

Q34) The answer to this one is a fact, you have to learn it. The e.m.f. depends on both r and R.

A is wrong because using the equation VT = E - Ir, the terminal voltage will change.
B is wrong because the current would decrease, not increase.
C is wrong because the e.m.f. of the battery can't change.
D is a correct statement. You can verify this yourself using P = I^2 * R
 
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please can someone give a detailed explanation for this, i just don't get it....a diagram would be appreciated too!.... thanx soo much and God blessView attachment 12424
here...dey say dat the detected signal decreases till X and then increases so there has to b a dark fringe at X and den a bright fringe after dat...
so der is destructive interference taking place at X so the path difference has to be (n+1/2)lambda
so here S2X-S1X=1/2 lambda=lambda/2 ==>C
 
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Please it's the 2nd time am asking :'(
At X
Initial Momentum = mcar x v + msand x 0 ----------> speed of sand is zero since it was dropped from rest
Final Momentum = 2m___ --->
Initial = Final Momentum so the new speed is v/2 ms^-1
-At Y
Initial Momentum = 2mv/2
Final Momentum = ( mssand x v/2 ) + (mcar x ___) ------> the speed of sand dropped is v/2 since it was moving with the car and dropped from it
Initial = Final Momentum ----> speed of car should be v/2
I hope now it clears your confusion :)

***** Now let me show it to you as conservation of kinetic energy
At X:-
Initial K.e = 1/2 mcar v^2 + 1/2 msand x 0
Final K.e = 1/2 (2m) ___^2 so speed must have decreased!
At Y:-
Initial K.e = 1/2 (2m) v^2
Final K.e = 1/2 msand (v)^2 + 1/2 mcar (v)^2

If momentum hasnt satisfied you, I hope kinetic energy does the job
 
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Hi peoplee, finally! Very happy with the progress!! Soo, i was thinking if anyone has all the graphs that we need to know compiled on one or two papers so we all can download and memorize them. Does anyone of you have it?????
 
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here...dey say dat the detected signal decreases till X and then increases so there has to b a dark fringe at X and den a bright fringe after dat...
so der is destructive interference taking place at X so the path difference has to be (n+1/2)lambda
so here S2X-S1X=1/2 lambda=lambda/2 ==>C
owkay tx so wats the formula if there was a constructive interference???
 
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can someone please provide a detailed explanation on how to use a micrometer screw gauge? Thanx!Screen shot 2012-06-11 at 11.49.08 AM.png
 
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View attachment 12537

How is I1>I2 ? calculation anyone ? the answer is A.

Because the resistors are in parallel,

The p.d across the 3ohm resistor is V1

and The p.d across the 2ohm resistor is V2

But you already deduced that V1 > V2

Since V = IR, I = V/R

As V1 is higher than V2, the current I1 is higher than I2 according to equation above...

I think this is it :)
 
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question 18.png

the equation 9-(sin(17.45)*20kN)] * 40 = 120000 J should be the useful energy and not the heat dissipated..am i right? But the answer is A? why?
 
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http://www.xtremepapers.com/papers/...and AS Level/Physics (9702)/9702_w04_qp_1.pdf
http://www.xtremepapers.com/papers/...A and AS Level/Physics (9702)/9702_w04_er.pdf
Please can someone explain Q18 i get B but answer is A. and Q20 and for Q 33 i get B 2I but correct answer is C 4I. How?

http://www.xtremepapers.com/papers/...nd AS Level/Physics (9702)/9702_w11_qp_11.pdf
http://www.xtremepapers.com/papers/...nd AS Level/Physics (9702)/9702_w11_ms_11.pdf
Q10 Why cant answer be A?
Q18,26,28???
Q29 How is a stationary wave formed in Q since both ends are open?
Q30??
Q33 Why cant answer be C?
Q36???
 
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