1991 - JAMB Physics Past Questions and Answers - page 1

horizontal component = W cos(90^o - (\theta))

= W sin

Time from tower to top, t = (\frac{u - v}{g})

= (\frac{20 - 0}{10})

= 2s

distance from tower to top

x = ut - (\frac{1}{2})gt²

= (20 x 2) - ((\frac{1}{2}) x 10 x 2 x 2)

= 20m

Time from the top to the ground, t = 6 - 2

= 4s

distance from top to ground,

x = ut - (\frac{1}{2})gt²

= (0 x 4) + ((\frac{1}{2}) x 10 x 4 x 4)

= 80cm

therefore, height of tower = 80 - 20

= 60m

Initial momentum = 0.1 x 10 = 1Ns

final velocity = -2ms^-1

Final momentum = -2 x 0.1 = -0.2Ns

Change in momentum = -0.2 - 1 = -1.2Ns

P.E = (\frac{1}{2})Fe

= (\frac{1}{2})(\frac{F}{K})

= (\frac{1}{2}) x (\frac{75 \times 75}{1500})

= 1.9J

Efficiency = m.A x (\frac{1}{V.R}) x 100%

85 = (\frac{load}{1200}) x (\frac{10}{80}) x (\frac{100}{1})

load = 8160N

Efficiency = m.A x (\frac{1}{V.R}) X 100%

75 = (\frac{20 \times 10}{effort}) x (\frac{1}{\frac{g}{sin\theta}}) x 100%

Effort = 133.3N

let the length be Lcm

(\frac{5 - 3}{25 - L}) = (\frac{10 - 5}{30 - 25})

L = 23cm

R.D = (\frac{\text{ mass of liquid displaced}}{\text {mass of water displaced}})

2 = (\frac{m - 10}{m - 15})

m = 20g