2014 - WAEC Physics Past Questions and Answers - page 1
1
A ball falling through a viscous liquid is acted upon by
A
upthrust only
B
upthrust and the ball's weight
C
the ball's weight and viscous force
D
upthrust, the ball's weight and viscous force
correct option: d
Users' Answers & Comments2
An object of mass m moves with a uniform speed v round a circular path of radius. If its angular speed is \(\omega\), the magnitude of the centripetal force acting on it is
A
\(m \omega^2 r\)
B
\(\frac{mv^2}{r}\)
C
\(m \omega r\)
D
\(\frac{mv^2}{r^2}\)
correct option: b
Users' Answers & Comments3
When an elastic material is stretched by a force, the energy stored in it is
A
kinetic
B
potential
C
thermal
D
electrical
correct option: b
Users' Answers & Comments4
When the linear momentum of a body is constant, the net force acting on it
A
is zero
B
increases
C
decreases
D
remains constant
correct option: a
Users' Answers & Comments5
If no net force acts on an object, the object maintains a state of rest or constant speed in a straight line. The above is a statement of Newton's
A
first law of motion
B
second law of motion
C
third law of motion
D
law of universal gravitation
correct option: a
Users' Answers & Comments6
Which of the following pairs of physical quantities is made up of vectors?
A
speed and displacement
B
mass and force
C
dislacement and acceleration
D
momentum and length
correct option: c
Users' Answers & Comments7
The knowledge of surface tension is applied in the following areas except in
A
manufacturing of rain coats
B
production of palm oil
C
floating of a needle on water
D
washing of plates with soapy water
correct option: b
Users' Answers & Comments8
A body is projected vertically upwards with a speed of 10 ms-1 from a point 2m above the ground. Calculate the total time taken for the body to reach the ground. [g = 10ms-2]
A
1.00s
B
2.00s
C
2.18s
D
3.00s
9
A car starts from rest and covers a distance of 40m in 10s. Calculate the magnitude of its acceleration
A
0.25ms-2
B
0.80ms-2
C
3.20ms-2
D
4.00ms-2
correct option: b
S = ut + \(\frac{1}{2} at^2\)
40 = 0 x 10 + \(\frac{1}{2}\) x a x 102
50a = 40
a = \(\frac{40}{50}\)
= 0.8ms-2
Users' Answers & Comments40 = 0 x 10 + \(\frac{1}{2}\) x a x 102
50a = 40
a = \(\frac{40}{50}\)
= 0.8ms-2
10
The diagram above illustrates a uniform metre rule which is balanced on a pivot by some masses. Calculate the value of M
A
500.0g
B
400.0g
C
44.4g
D
25.0g
correct option: d
clockwise moment = Anti-clock wise moment
= M x 40 = 100 x 10
M = \(\frac{1000}{40}\)
= 25g
Users' Answers & Comments= M x 40 = 100 x 10
M = \(\frac{1000}{40}\)
= 25g