physics

[ruby92]

M/j 2003 paper1
34,40,18,30,35

[ruby92]

o/n 2003 paper 1
12,7(why isnt it A how is it B?arent we supposed to find the gradient?)and 37,16

[astarmathsandphysics]

34.0 to 7,5 cos 0 to 5/6 of resistaNCE can be between P and Q
30.D since 2*2/3 -1/3 =1 charges must add up to 1
18.D since KE is propritional to v^2 F=Energy =braking force*distance=1/2mv^2
30.average current =(100+20)/2=60mA
charge=It=60mA*8=480mC
45. E is uniform between paralell plates

[astarmathsandphysics]

7 C horizontal acceleration is 0 so horizontal vel is constant=ucos alpha
apply v=u+at vertically
12. 8*2+2*4=6v v=4

[ruby92]

2.why not A? how do we know the direction of the resultant force?
could you please explain Q's 15 and 35 in more detail

Also o/n 2002 paper 1; 36

no 12(o/n 2003) 

[astarmathsandphysics]

which paper

[ruby92]

m/j 2002 paper 1

[ruby92]

M/j 2004 paper 1
8 (why D why not C)
9
31(why D and why not C)

O/N 2004 paper 1
20, 18,11,31,33,26.

[astarmathsandphysics]

can you post question and paper and year in same post
mj 2004 p1 q8 D - as ball speeds up acc falls as air resistance increases]

9. Work vertically s=ut+1/2at^2 u=0 cos he vertically speed is zero
1.25=0+1/2*10*t^2
t=0.5
work horizontally s=ut 10=u*0.5 u=20
31. R=pl/A and A=kd^2 so double the diameter mean divide resistance by 4. V=IR and V is same for wires in parallel so current in P is 4x cuurent in Q

[astarmathsandphysics]

20. A cos P=pgh
so 2xp(P)+xp(Q)=2xp(Q) so xp(P)=2p(Q) so p(P)1/2p(Q)
18.work dine=increase in potential energy+heat generated
9000*40=20000*12+H
H=120000J A
11.B m(v-u)=m(-v-v)=-2mv
31. C use P=VI
33.R=pl/A double the length means double R and double the diameter means 1/4 R so R is halved overall
V=IR so double V and hal;ve R mean I is 4x bigger
26 !6 I=kA^2

[ruby92]

O/n 2003 paper 2 CIE physics
Q3 (C) part iii 2. how is the torque of this also 45?

Q5 of the same papers part c

m/j 2004 paper 2
1 (b)

[ruby92]

Posted by: astarmathsandphysics
can you post question and paper and year in same post
mj 2004 p1 q8 D - as ball speeds up acc falls as air resistance increases]

9. Work vertically s=ut+1/2at^2 u=0 cos he vertically speed is zero
1.25=0+1/2*10*t^2
t=0.5
work horizontally s=ut 10=u*0.5 u=20
31. R=pl/A and A=kd^2 so double the diameter mean divide resistance by 4. V=IR and V is same for wires in parallel so current in P is 4x cuurent in Q

i dont get 31 

[ruby92]

M/j 2004 paper 2 physics
5 b
3 B ii
1 b

o/n 2004 paper 2
1 b ii

m/j 2005
3 b we dont convert g to kg?

[astarmathsandphysics]

3bii)s=1/2at^2 a=s/(1/2t^2)=2*gradient of graph=
on2004p2 %error =2*0.02/0.50*100=8%
mj2005 you should convert to kg it is good practice but units of grams cancels cos factor of 1000 occurs on both sides

[ruby92]

O/n 2005 paper 2
3 c