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AS Physics P1 MCQs Preparation Thread.

Nibz

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Can someone please help me with the questions 11 and 32 of May/June 2003 Physics P1?:(

Q.11
The wheel exerts a force on the road in left direction <-
The wheel in turn exerts a force in the right direction ->
The normal reaction force is upwards ↑
Net force exerted by the road is the one in option B, i.e the diagonal of rect.

Q32. V = IR
7.5 / 15 = I
I = 0.5A
 
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14 A box of weight 200 N is pushed so that it moves at a steady speed along a ramp, through a
height of 1.5 m. The ramp makes an angle of 30° with the ground. The frictional force on the box
is 150 N while the box is moving.

What is the work done by the person?
PLZ ANYONE HELP ME WID W_10 QP 12 PART 14??? PLZZZZ
 
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14 A box of weight 200 N is pushed so that it moves at a steady speed along a ramp, through a
height of 1.5 m. The ramp makes an angle of 30° with the ground. The frictional force on the box
is 150 N while the box is moving.

What is the work done by the person?
PLZ ANYONE HELP ME WID W_10 QP 12 PART 14??? PLZZZZ

work done by the person will be used to over come the frictional force and to move the object on an inclined plane... so this will bee equal to 150(3) + 100(3) = 750.... work done is force * distance... inclined distance moved is 3m... 1.5/.5 =3...
 
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any thing for J 2004 go to the AS pysics p1 mcq YEARLY ONLY thread and ull find all 2004 answered..for j 2005
Q 13: moment = F x perpendicular distance...clockwise = anti clock wise...so ( 10 x 2) +(20 x 3)= (5 x 2) ...answer is 30 so A
Q 16: SRY CANT HELP
Q27: speed of electromagnetic waves is 3 x 10^8 ms-1...distance between 2 adjacent minima or maxima is 0.5 lambda so...0.5 lambda=0.015m...lambda = 0.03...f= v/lambda= 3x 10^8 / 0.03 = 1 x 10^10 so C
Q32:wait a little more time and ill answer
fr Q27 hw do u gt 0.03??? tnx a million fr da othr ;)
 
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work done by the person will be used to over come the frictional force and to move the object on an inclined plane... so this will bee equal to 150(3) + 100(3) = 750.... work done is force * distance... inclined distance moved is 3m... 1.5/.5 =3...
THANK U(y)
 
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Nov 2005

Q3: D... E=kg m^2 s^-2..power=kg m^2 s^-3...V=P/I...so kg m^2 s^-3 A^-1= V
Q9: D...u must understand it as it came from the equation m1u1 + m2u2 = m1v1 + m2v2..as the total INITIAL momentum before collision is equal to the total final momentum after collison..same meaning but different words
Q11: A...in vaccum..the only force acting is gravity..so only XV as it is vertical
Q15: A...potential energy increases as height increases..and he didnt say it fell to the ground so C is wrong
Q20: work done is equal to the strain energy which is the area under the graph so B is the answer
Q28: slits seperation is inversely proportional to the fringe spacing..so slit space is doubled so spce is halfed so 3/2=1.5..then wavelength is proportional to fringe seperation..so wavelength halfed hence fring seperation halfed so 1.5/2= 0.75 so A
wish i helped
thnx alot fr ths..bt i posted da wrong ppr..its actaly o/n/o4...
Q- 3 9 11 15 20 28

http://www.xtremepapers.com/papers/...and AS Level/Physics (9702)/9702_w04_qp_1.pdf
http://www.xtremepapers.com/papers/...and AS Level/Physics (9702)/9702_w04_ms_1.pdf
in xtremely sryyyyyy!!! :(
 
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can u pls solve them plsmates i am posting dis another time plssssssssssssssssssss help me on these can anyone pls help me out on these question plssss 6 24 30 39 :) pls

i have no idea abt 6TH one...

24... young modulus is a ratio and is constant for any particular material... it is clearly stated that same steel was used again so young modulus will be same again. that is E

30.... if u consider the electron coming out of the field, its horizontal component is u.. this is because the field is in vertical direction.. this will only affects its vertical velocity but the horizontal will remain the same.... this gives you cos0=u/v or V=u/cos0

39....uranium has mass no. of abt 135... so its total mass will be mass no. into mass of one proton... that is 1.67*10^-27... so 235*1.67*10^-27 into 235 gives you 3.9*10^-25... so approximately 10^-25 will be answer...
 
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work done by the person will be used to over come the frictional force and to move the object on an inclined plane... so this will bee equal to 150(3) + 100(3) = 750.... work done is force * distance... inclined distance moved is 3m... 1.5/.5 =3...
BUT U DIDNT TELL ME THT HOW DID U GET THIS 150 AND 100?? I DNT GET IT CAN U EXPLAIN IT IN DETAIL PLZZ:confused:
 
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BUT U DIDNT TELL ME THT HOW DID U GET THIS 150 AND 100?? I DNT GET IT CAN U EXPLAIN IT IN DETAIL PLZZ:confused:
oh sorry mate... 150 was the frictional force given to us... and the force acting inclined to the motion was mgsin30... 200sin30= 100N...
 
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w_10 qp 12
22 To determine the mass of food in a pan, a scale is used that has high sensitivity for small masses
but low sensitivity for large masses.
To do this, two springs are used, each with a different spring constant k. One of the springs has a
low spring constant and the other has a high spring constant.
Which arrangement of springs would be suitable?
 
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i have no idea abt 6TH one...

24... young modulus is a ratio and is constant for any particular material... it is clearly stated that same steel was used again so young modulus will be same again. that is E

30.... if u consider the electron coming out of the field, its horizontal component is u.. this is because the field is in vertical direction.. this will only affects its vertical velocity but the horizontal will remain the same.... this gives you cos0=u/v or V=u/cos0

39....uranium has mass no. of abt 135... so its total mass will be mass no. into mass of one proton... that is 1.67*10^-27... so 235*1.67*10^-27 into 235 gives you 3.9*10^-25... so approximately 10^-25 will be answer...
thanks alot may Allah bless u pls my iother questions too pls :)
 
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omfg, will someone please answer about that nuclear decay question!!!! how the hell are neutrons conserved?
 
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fr Q27 hw do u gt 0.03??? tnx a million fr da othr ;)
0.5lambda = 0.015
so 0.015/2 = 0.03
and for that wrong papers...go to the thread named ( As physics P1 MCQ YEARLY ONLY) i just poted the whole november 2004 P1 EXAM.. And we r working there year by year...we r in June 2005 now..wish i helped
 
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and these too pls :) u r help will be appreciated :) q 5 13 15 18

5... sir without any logic of acceleration... displacement can never decrease... if an object is falling down it must be covering some distance... so option D only remains...

13.. anticlockwise is equal to clockwise direction.... so F1D1=F2D2.. 900*.20 = F2 * 1.20.. F2 is 150N...
15... when half of the water will flow.. mass will be halved and the height of the container will be halved... so (m/2)(h/2)(g)... (mgh/4)
18.... i am not so sure but i think when heated, right hand level will increase by h and left side will decrease by height h..... so total change will be 2h... so pressure will be 2mgh... i think...
 
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