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AS Physics P1 MCQs Preparation Thread.

LeQuavina

LeQuavina

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6Astarstudent said:
Energy is always conserved so
Initial energy = final energy
so Ek = new Ek + change in Ep
(mu^2)/2 = (mv^2)/2 + mgh
cancel out all the m ( masses)
(u^2)/2 = (v^2)/2 + gh
28^2 /2 = v^2 / 2+ 9.81x22
392 = v^2 /2 + 215.82
v^2 /2 = 176.18
v^2 = 352.36
v = 18.8
answer is B
Click to expand...
Thanks a lot. :)
 
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6Astarstudent

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Abu mota said:
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s10_qp_11.pdf Q.9,Q.27,Q.29,Q.33
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w10_qp_11.pdf Q.18,Q.34 why arent the wires in the cable considered to be parallel i.e shouldnt we divide the R by 2 not multiply it by 2.
Click to expand...

#34
wire is in a loop ie. supply -> wire -> relay -> wire -> supply -> wire...... etc. so its in series

#18
Energy is always conserved so
Initial energy = final energy
so Ek = new Ek + change in Ep
(mu^2)/2 = (mv^2)/2 + mgh
cancel out all the m ( masses)
(u^2)/2 = (v^2)/2 + gh
28^2 /2 = v^2 / 2+ 9.81x22
392 = v^2 /2 + 215.82
v^2 /2 = 176.18
v^2 = 352.36
v = 18.8
answer is B

for the s10
#9
answer is B because the ball is falling down with constant acceleration so graph will be an reversed parabola
#27
distance between centre and Q, centre and P is both the radius which is the same. so there is no net workdone. answer is A
#29
field lines represent movement of positive charge, since electron is negative so it will mvoe in opposite direction of field line towards observer so answer is D

#33
there are lots of ways for this question
I find V the easiest to use in an explanation
PQR all have same resistance
Q,R have same V while P has 2x of their V so 2V
Power = (V^2)/R
so Power ratio of P Q R =((2v)^2)/R: (V^2)/R : (V^R)/R or 4:1:1
total power is 12, divided to 6 parts so Power of R is 2W
 
Obsidian Fl1ght

Obsidian Fl1ght

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Hey anyone in progressive waves, anti nodal lines... does constructive or destructive interference take place on them?
 
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6Astarstudent

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Suno Yaar

Suno Yaar

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i
6Astarstudent said:
kinematics equation
s = ut + (at^2)/2
since initial speed is 0 so
s = (at^2)/2
substitute the values in
x = (a(t1)^2)/2
x + h = (a(t2)^2)/2
solve the simultaneous equations so
h = (a(t2)^2)/2 - (a(t1)^2)/2 = ((t2)^2 - (t1)^2) x a / 2

a = 2h/((t2)^2 - (t1)^2) so answer is C
Click to expand...
did the same!!!!!!!!!!!!! but the answer is D :/
 
Suno Yaar

Suno Yaar

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oh
6Astarstudent said:
kinematics equation
s = ut + (at^2)/2
since initial speed is 0 so
s = (at^2)/2
substitute the values in
x = (a(t1)^2)/2
x + h = (a(t2)^2)/2
solve the simultaneous equations so
h = (a(t2)^2)/2 - (a(t1)^2)/2 = ((t2)^2 - (t1)^2) x a / 2

a = 2h/((t2)^2 - (t1)^2) so answer is C
Click to expand...
oh lol. your working is right. i just saw your answer/ your working is right and your answer si right. the answer you got is in part D.
you accidentally wrote C lol.
 
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