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Q5need help
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w05_qp_1.pdf
5, 12, 13, 16, 23, 24, 25, 28, 29
Volume= (5*2*1) = 10
uncertainty V= (0.01/5) + (0.01/2) +( (0.01/1) * 10) = 0.17
So volume = 10 +- 0.17
And mass = 25 +- 0.1
using formula uncertainty of density/density= (uncertinity of volume/volume) +(uncertainty of mass/mass)
Uncertainty in density = (0.17/10) + (0.1/25) * 2.5 = 0.05
Q12
the 100N will act at the middle of the beam
that is at 0.5 m
and from pivot it is 0.1 m
so 100*0.1 +o.6*10 is the anticlockwise force =16
0.4*20 is the clock wise force =8
for the beam to be uniform anticlockwise=clockwise
but here
anticlockwise is more and clockwise less
so the 20 will be added in such a way that it acts clockwise
16=(8)+(20*x)
x=0.4
so 0.4 from the pivot D
Q13
torque=45 Non zero
and force=30+30=60 non zero
Q16
weight=mg=10^3
force = mg sin tetha
=10^3 sin 30 = 500 N
Work = force * distance moved in direction of force
=500 * 5 = 2500 J
Q23 amplitude is the displacement from MEAN position A and C are wrong
D is wrong because the wave length is marked wrong
that is the time period
Q24 information in the question
I proportional to a^2 and I proportional to f^2
I=a^2 I=f^2
For P
Io=x0^2
Io=(1/t0)^2
For Q
I1=(2x0)^2=4xo^2 =Io*4
I1=(1/2t0)^2=1/4to^2 =I0/4
finally Io=I1
Q25
v=f*lamda
In WATER
1500/150=lamda=10
In AIR
300/150=2
Q28
x=lamda*D/a
3*10^-3 = 700*10-9*D/a
a/D=7/30000
x=350*10^-9*D/2a
x=350*10^-9/((7/3000)*2)
Q29 d*sin(tetha)=n*lamda
tetha will be 60/2 as it is suppose to be the angle made by the normal to the granting
1.15 * 10^-6 sin 30 = 1 * λ
λ = 575 nm
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