1) A car of mass 1000 kg accelerates uniformly from rest to a velocity of 54 km/hour in 5 sec. The average power of the engine during this period in watts is (neglected friction)
{answer should come 22500 W}

2) A man does a given amount o work in 10 sec. Another man does the same amount of work in 20 sec. The ratio of the output power of first man to the second man is
{answer should come 2/1}

3) A ball of mass 0.5 kg i dropped from a height. Find the work done by its weight in one second after the ball is dropped. {answer should come 24 Joules}

4) A stone projected vertically up with a velocity u reaches a maximum height h. When it is at a height of  \frac{3h}{4} from the ground, the ratio o kinetic energy and potential energy at that point is:(consider Potential energy = 0 at the point of projectory)
{answer should come 1:3 }

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Answers

2015-01-13T10:45:19+05:30

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Change in energy = 1/2 * 1000 * (54 * 5/18)^2 = 225,000/2 Joules
Power  = energy / time duration = 225,000/2*5 sec = 22,500 Watts

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Power = Work / time interval
  As work is same, then power is inversely proportional to the time interval.

  P1 / P2 = T2 / T1 = 20 sec/ 10 sec  = 2
==============

work done by gravity in 1 second.
   v = u + g t = 9.8 * 1 = 9.8 m/s
   work done = change in K.E = 1/2 * 0.5 * 9.8^2 = 24.01 Joules
==================

E = PE2 + KE2 = m g (3h/4) + KE2       at s = 3h/4
E = PE3 + KE3 = m g h + 0            at s = h = top
    KE 2 = m g h - m g (3h/4) = mgh/4 
     KE2 / PE2 = (mgh/4) / (3mgh/4)  = 1 : 3


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