2013 - WAEC Physics Past Questions and Answers - page 5
41
Which of the following phenomena supports the theory that waves have a particle nature?
A
electron diffraction
B
photoelectric effect
C
diffraction of light
D
x-ray interference
correct option: c
Users' Answers & Comments42
A block of wood of mass 5kg is pulled on a platform by a force of 40N as illustrated in the diagram above. If the frictional force, F experienced by the block is 12N, calculate the magnitude of the acceleration of the acceleration of the block
A
2.4ms-2
B
5.6ms-2
C
8.0ms-2
D
2.4ms-2
correct option: b
F + Ma = P
a = \(\frac{P - F}{m}\)
= \(\frac{40 - 12}{5} = 5.6ms^{-2}\)
Users' Answers & Commentsa = \(\frac{P - F}{m}\)
= \(\frac{40 - 12}{5} = 5.6ms^{-2}\)
43
Four co-planar forces of magnitudes 10N, 17N, 6N and 20N act at point O as shown in the above diagram. Determine the magnitude of the resultant force
A
53.N
B
21.0N
C
7.0N
D
5.0N
correct option: d
Vertical resultant = 10 - 6 = 4N
Horizontal resultant = 20 - 17 = 3N
R = \(\sqrt{3^2 + 4^2}\)
= 5N
Users' Answers & CommentsHorizontal resultant = 20 - 17 = 3N
R = \(\sqrt{3^2 + 4^2}\)
= 5N
44
The outlet of a bicycle pump is used to inflate a football as illustrated in the diagram above. Why does the pressure of the air inside the pump increases as the pimp handle is slowly pushed downward at constant temperature?
A
momentum of the air molecules are increased
B
volume occupied by the molecules are increased
C
frequency of collision of the air molecules with the walls of the pump is increased
D
more air molecules are collidin with one another in the system
correct option: c
Users' Answers & Comments45
The diagram above illustrates a waveform. Which of the point on the waveform are in phase?
A
P and R
B
R and T
C
P and T
D
Q and S
correct option: d
Users' Answers & Comments46
An electric circuit is connected as illustrated above. Determine the equivalent e.m.f and current flowing through the circuit respectively, neglecting the internal resistance of the cells
A
2V, 1.0A
B
2V, 4.0A
C
6V, 0.3a
D
6V, 3.0a
correct option: a
cells in V = voltage of the cell
= 2V
I = \(\frac{v}{R} = \frac{2}{3}\)
= 1A
Users' Answers & Comments= 2V
I = \(\frac{v}{R} = \frac{2}{3}\)
= 1A