2005 - JAMB Physics Past Questions and Answers - page 4
The electric force experienced in an electric field of intensity E is given by:
F = qE: where q = the test charge.
E = F/Q = 0.4/10-10 = 0.4 X 10-10
= 4.0 X 10<sup>9</sup> NC<sup>-1</sup>
Users' Answers & CommentsThe charge must be opposite in sign, since the positive charges will deflect to the negative plate, and the negative charge to the positive plate
Users' Answers & Comments[Resistivity of the wire = 4 x 10-7 Ω m]
since resistance R = p (l)/a
= (4 x 10-7 x 20)/8.0 x 10-6
R = 1.0 Ω
Users' Answers & Comments


Normal reaction R = 3 kg x 10 = 30N
coefficient of friction , μ = 0.5
∴ acting frictional force = μR = 30 x 0.5
Pulling/sliding force F = 6kg x 10 = 60N.
∴ Net downward force = ma = 60 - 15.0
=> (3 + 6)a = 45
∴ 9a = 45
a = 5m/s2
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If the string is in equilibrium, the the tension in the string Q = 4N + 5N = 9N
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Since the force is inclined at an angle 60o to the horizontal, the effective component of te force along the horizontal is given by F x cos 60o
= 100 x 0.5 = 50N.
∴ work done = F x cos 60 x displacement
= 50 x 8.0m
= 400.0J
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The fig above shows a conventional transistor, showing the three basic features, the Base, the Emitter and the collector. Thus comparing this with the figure in the question it implies that P = Emitter, Q = Base and R = collector
Users' Answers & CommentsIn the reaction above, X is
235U + 1n → 144Ba + 90Kr + 2x9205636
In the reaction above, balancing the left with the right, show that X is neutron
Users' Answers & Comments