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 10⁹ NC^-1

The charge must be opposite in sign, since the positive charges will deflect to the negative plate, and the negative charge to the positive plate

since resistance R = p (l)/a
= (4 x 10^-7 x 20)/8.0 x 10^-6
R = 1.0 Ω

F = \(\frac{17}{sin 60} = \frac{17}{0.866}\)

= 20N

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

If the string is in equilibrium, the the tension in the string Q = 4N + 5N = 9N

Since the force is inclined at an angle 60^o to the horizontal, the effective component of te force along the horizontal is given by F x cos 60^o
= 100 x 0.5 = 50N.
∴ work done = F x cos 60 x displacement
= 50 x 8.0m
= 400.0J

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

235U + 1n → 144Ba + 90Kr + 2x9205636
In the reaction above, balancing the left with the right, show that X is neutron