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A ball is thrown upward with the speed of 19.6 m/s, find the speed of ball after 3 seconds.
V = u + at
u = 19.6, a = g = –9.8 m/s^{2}, t = 5 s
v = 19.6 + (–9.8) × 3
v = – 9.8 m/s
A truck falls from a bridge into the river in 0.5 s. let g = 10 ms–2, what is the height of the bridge from the water surface?
h = ut +(½)at^{2}
Initial velocity , u = 0, t = 0.5s, a = 10 ms^{–2}
h = 0 × 0.5 + (½)× 10 × (½) × (½)
= 1..25 m
Two objects of mass 200 kg and 800 kg separated by a distance of 50 m. Find the gravitational force between the two bodies.
F =G× mM/r^{2}
= 6.67 × 10^{–11} × {(200× 800)/(50×50)} 4.26×10^{9} N
When a ball is thrown vertically upwards, it goes through a distance of 19.6 m. Find the initial velocity of the ball.
u = ?, v = 0, g = –9.8 m/s^{2} as ball goes up h 19.6 m
v^{2} = u^{2} + 2gh
0^{2} = u^{2} + 2(–9.8 ) × 19.6
⇒ u2 = (19.6)^{2}
⇒ u = 19.6 m/s
Find the value of acceleration due to gravity on the surface of the moon, given mass of the moon = 7.4 ×1022 kg; radius of the moon = 1740 km; and G = 6.7 × 10–11 Nm2 /kg2
g = GM/R^{2}= 6.7× 10^{11} ×7.4 × 10^{22 }/^{ }(1.74×10^{6} )^{2} ^{ }( R = 1740)
km = 1740 × 1000 m = 1.74 × 106 m
⇒ g = 1.63 m/s^{2}
If the distance between two bodies is doubled, the force of attraction F between them will be
The force of attraction F is inversely proportional to the square of distance between two objects. Distance is doubled, force gets one–fourth.
An object weighs 10 N in air, when immersed fully in a liquid it weighs only 8 N. The weight of the liquid displaced by the object will be
The weight of liquid displaced = loss in weight = 10N – 8N = 2N
Relative density is purely a ratio of two similar quantities (masses), so it has no units.
What is the final velocity of a body moving against gravity when it attains the maximum height ?
When a moving body attains the maximum height, final velocity becomes zero.
A body whose weight is 120 kg on the earth. Find its weight on the surface of moon
g_{moon} = (1/ 6) g_{earth}
∴ W_{moon} = (1/ 6) W_{earth} = (1/ 6) × 120 = 20
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