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AS Physics P1 MCQs Preparation Thread.

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This question is tricky in that at first sight the amplitude, which is equal to the maximum displacement, may appear to be reached after 2.5 seconds, but in reality,
on a displacement time graph, at the maximum displacement you can see that the gradient of the graph is zero, i.e. the velocity is zero so the answer for the right hand side is in reality 5 and 15 - these numbers correspond to 2 different times when the displacement is maximum, i.e. when the amplitude is zero. On the right hand column, we need to select the acceleration at point Q. The acceleration at Q is equal to the gradient of the velocity time graph which is flat, parallel to the x-axis and is thus equal to zero. This means that the answer is D.

Good luck for your exams!
Sorry did not understand it.Can someone plz explain it better?
 
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q27.png

the answers B. But how? any description will be appreciated.thank you.

and can anyone tell the effects of changing any one one value will have on other values in d.sin(teta)=n.lamda
 
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Questions: 10, 11 , 21, 24, 26, 27, 31, 33 ,36


PLEASE HELP URGENTLY!!!! AND EXPLAIN AS WELL, I CAN GET THE ANSWERS FROM A MARK SCHEME I NEED EXPLANATIONS!! THANK YOU GOD BLESS
 

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Q3=>compare units. A is the answer as _/g(lamda) = _/ms-2 * m = _/m2s-2 = ms-1 . (unit for speed)

Q7=> we know that V=u+at ; with v being final velocity u being initial velocity and at when an object is under a constant acceleration for t seconds. ANS:C

Q8=> V2=u2 + 2 as where u=0 a=9.81 s=40 take out v at 40 and 30.subtract them both.you will get v=3.75.the from v=u+at find t for u=0. ANS: 0.38

Q12: velocity after collision = 2*4+1*4/6 = 2ms-1 we know K.E=0.5 m v2 = 0.5*6*(2)^2 =12J

Q13=> moment= F * perpendicular distance bw the two the two forces. 8*sin(60)*0.60

36=> voltage lost as a result of internal resistance = 2 ohms * 0.5 Amp = 1 volt hence terminat pd =3-1=2 V and output power= 2 Volts * 0.5 Amp = 1 W
 
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