2006 - JAMB Physics Past Questions and Answers - page 5

In general, power = IV; 40kw = IV.

∴ 40,000 = 1 x 800

1 = (40.000)/800 = 50A. tune the current through resistor = 50A.

∴ power loss = 1²R = 50² X 2 = 2500 X 2 = 5.0 X 10³W

Power of the lamp = 60W = (V²)/R and the voltage rate = 120v.

R = (V²)/60 = (120 X 120)/60 = 240 Ω, the fuse wire that should ensure current does not exceed this value should be rated 200 Ω

80 days = 4 half lives (i.e 4 x 2odqs = 80 days)

But the fraction always left by any radioactive material after 4half lives is (1/2)⁴ = (1)/16

In the stress - strain graph shown above, X represents the breaking point of the wire

Pressure of the Gas, P = A + h

= 76 + 10

86 cm Hg

For vertical equilibrium, we have

2Ty = 75N

But Ty/T = adj/hyp = cos 60^o

∴ Ty = T Cos 60^o

=> 2Ty = 2Tcos 60^o = 75N

2 x T x 0.5 = 75 => T = 75N

∴ Total tension in the rope = 2T.

= 2 x 75 = 150N

IF the volume of gas is plotted against the reciprocal of pressure, the slope will be given as

Slope = ^V/_1/P = P x V

thus the unit of the slope will be;

= Nm^-2, = V = M³ = NM.

The unit NM (Newton - Meter) is the practical unit of work, which is the product of force (N) and Displacement (M).
Note: It should be noted that the practical unit of work is Newton-Meter, whereas its S.I unit is Joules, just as energy.

Horizontal range = 3 x maxi. height

=> (U²Sin2θ) / g = (3 x U²sin²θ) / 2g

∴ 4sinθcosθ = 3sin²θ

=> 4cosθ = 3sinθ

∴ ^sinθ/_cosθ = ⁴/₃; => tanθ = ⁴/₃ = 1.3333

θ = tan^-1 (1.3333)

= 53^o

In terms of dimension: power , P, = ML²T³

Force F₁ = MLT^-2; Vel, V, LT^-1

From the above, => P = F x V

i.e, ML²T^-3 = MLT^-2 x LT^-1

ML²T^-3

=> 80,000 = F x 32

= 2500N

= 2.50 x 10³N