8.0N
6.8N
8160N
9600N
Correct answer is C
Efficiency = m.A x \(\frac{1}{V.R}\) x 100%
85 = \(\frac{load}{1200}\) x \(\frac{10}{80}\) x \(\frac{100}{1}\)
load = 8160N
1.9J
2.8J
3.45J
0.43J
Correct answer is A
P.E = \(\frac{1}{2}\)Fe
= \(\frac{1}{2}\)\(\frac{F}{K}\)
= \(\frac{1}{2}\) x \(\frac{75 \times 75}{1500}\)
= 1.9J
0.8Ns
1.2Ns
12.0Ns
80.0Ns
Correct answer is B
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
An object moves with uniform speed round a circle. Its acceleration has
constant magnitude and constant direction
constant magnitude and varying direction
varying magnitude but constant direction
varying magnitude and varying direction
Correct answer is B
No explanation has been provided for this answer.
60m
80m
100m
120m
Correct answer is A
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