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Solve these numericals:- (PLEASE HELP, URGENTLY NEEDED!) 1) A certain force exerted for 1.2 seconds raises the speed of an object from 1.8 metres per second to 4.2 metres per second. Later the same force is applied for 2 seconds. How much does the speed change in two seconds? 2) A body is released from a height of 20 metres. Calculate the final velocity of the body (take g = 10 metres per second square). |
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Answer» ce = F so acceleration a = F/m a = (v - u) / t = (4.2 - 1.8) / 1.2 = 2 m/sec² SINCE same force is applied again, the same acceleration happens. v = u + a t => v - u = a t = 2 m/sec² * 2SEC = 4 m/sec change in SPEED = 4 m/s2) s = 20 meters. = displacement v² = u² + 2 a s -- equation of motion under gravity, a = + g v² = 0 + 2 g s = 2 * 10 * 20 v = 20 m/sec = FINAL velocity |
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