school exam (lower 6)
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school exam (lower 6)
1(a) Given the expression X=ρAѵ where ρ ,A, ѵ represent density,
crosssectional area and velocity respectively. What quantity does X
represent?(2m)
Solution:
[ρAѵ]=ML3xL2xLT1
=M/T
X=The rate of change of mass
(b) A quantity Z is defined according to the following expression
Z= πpr4/8ПL
Where k is a dimensionless constant and p,r,П,and L represent the
pressure difference between the ends of a tube, radius of tube,
viscosity and the length of tube. What physical quantity does Z
represent?
Solution: (2m)
R.H.S=[ML1T2][L4]/[ML1T1][L]
=L3T1
Z=rate of change of volume
2(a) Define the term acceleration.
Acceleration is the rate of change of velocity.
(b)A force F of 600 N is applied to accelerate two bodies that are in contact with each other as shown in the diagram below.
F
→
If the friction between the bodies and the floor is negligible, what force does the 100kg mass exert on the 200kg mass?(3m)
Solution: F−F y on X =mxa
600−F y on X=100(2)
F y on X=400N
3(a )State the conditions for the equilibrium of a particle.(1m)
The net force acting on a particle must be equal to zero.
4 A particle of mass m moves in a horizontal circle of radius r at
constant tangential speed ѵ, as shown in the figure below.
B
A
When the particle moves from point A to point B where AB is the diameter, determine
(a) The work done by centripetal force. (2m)
Centripetal force is perpendicular to the angular displacement. Hence, the work done =o
(b) The change of linear momentum of the particle(3m)
ΔP=P2−P1=mv2−mv1
=m(v1)− mv1
=2 mv1
5(a)A wheel has moment of inertia 0.04kg m2 about an axis which passes
through the centre of the wheel and is perpendicular to the plane of
the wheel. It rotates at angular velocity 600 rpm. At t=0, a retarding
force is applied on the wheel, which accelerates at constant rate and
stops at t=10s.Determine
(a) The angular acceleration of the wheel (1m)
W0=600X2π/60=20πrads1
α=w w0/t=020π/10=2πrads2
(b) The number of revolutions turned through by wheel in 10s.(2m)
W2=W02+2αѳ
0=(20π)2+2(2π)ѳ
Ѳ=400π2/4π=100π
100π/2π=50 rev
(c ) the torque acting on the wheel while it decelerates.(2m)
τ=Iα=0.04(2π)=0.25Nm
crosssectional area and velocity respectively. What quantity does X
represent?(2m)
Solution:
[ρAѵ]=ML3xL2xLT1
=M/T
X=The rate of change of mass
(b) A quantity Z is defined according to the following expression
Z= πpr4/8ПL
Where k is a dimensionless constant and p,r,П,and L represent the
pressure difference between the ends of a tube, radius of tube,
viscosity and the length of tube. What physical quantity does Z
represent?
Solution: (2m)
R.H.S=[ML1T2][L4]/[ML1T1][L]
=L3T1
Z=rate of change of volume
2(a) Define the term acceleration.
Acceleration is the rate of change of velocity.
(b)A force F of 600 N is applied to accelerate two bodies that are in contact with each other as shown in the diagram below.
F
→
If the friction between the bodies and the floor is negligible, what force does the 100kg mass exert on the 200kg mass?(3m)
Solution: F−F y on X =mxa
600−F y on X=100(2)
F y on X=400N
3(a )State the conditions for the equilibrium of a particle.(1m)
The net force acting on a particle must be equal to zero.
4 A particle of mass m moves in a horizontal circle of radius r at
constant tangential speed ѵ, as shown in the figure below.
B
A
When the particle moves from point A to point B where AB is the diameter, determine
(a) The work done by centripetal force. (2m)
Centripetal force is perpendicular to the angular displacement. Hence, the work done =o
(b) The change of linear momentum of the particle(3m)
ΔP=P2−P1=mv2−mv1
=m(v1)− mv1
=2 mv1
5(a)A wheel has moment of inertia 0.04kg m2 about an axis which passes
through the centre of the wheel and is perpendicular to the plane of
the wheel. It rotates at angular velocity 600 rpm. At t=0, a retarding
force is applied on the wheel, which accelerates at constant rate and
stops at t=10s.Determine
(a) The angular acceleration of the wheel (1m)
W0=600X2π/60=20πrads1
α=w w0/t=020π/10=2πrads2
(b) The number of revolutions turned through by wheel in 10s.(2m)
W2=W02+2αѳ
0=(20π)2+2(2π)ѳ
Ѳ=400π2/4π=100π
100π/2π=50 rev
(c ) the torque acting on the wheel while it decelerates.(2m)
τ=Iα=0.04(2π)=0.25Nm
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