• We need your support!

    We are currently struggling to cover the operational costs of Xtremepapers, as a result we might have to shut this website down. Please donate if we have helped you and help make a difference in other students' lives!
    Click here to Donate Now (View Announcement)

AS Physics P1 MCQs Preparation Thread.

J

JD REBORN

Messages
143
Reaction score
27
Points
38
sagar65265 said:
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!
Click to expand...
Sorry did not understand it.Can someone plz explain it better?
 
SalmanslK

SalmanslK

Messages
83
Reaction score
47
Points
28
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
 
SalmanslK

SalmanslK

Messages
83
Reaction score
47
Points
28
K

kinglynx

Messages
61
Reaction score
29
Points
18
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
 

Attachments

  • Summer Physics 2011 (P13) QP.pdf
    611 KB · Views: 3
SalmanslK

SalmanslK

Messages
83
Reaction score
47
Points
28
xxfarhaxx said:
Sum1???
Q- 3, 7, 8, 12, 13, 22, 36
http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s07_qp_1.pdf
http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s07_ms_1.pdf
Click to expand...

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
 
You must log in or register to reply here.
Top