Hi there! Have questions about your homework? At Brainly, there are 60 million students who want to help each other learn. Questions are usually answered in less than 10 minutes. Try it for yourself by posting a question! :D

Hi there! Have questions about your homework? At Brainly, there are 60 million students who want to help each other learn. Questions are usually answered in less than 10 minutes. Try it for yourself by posting a question! :D

A tubewell pumps out 2400 kg of water per minute If water is moving with a velocity of 3 ms what is power of the pump

how much work is done if pump runsfor 10 hrs (ii)A car weighing 1120 kg is going up an incline of 1 in 56 at rate of 20 m in 2 s find power of engine if frictional froce is 64 n (iii)a train weighing 100 metric tonne is running on a level track with a uniform speed of 72 kmh if frictional resistance amounts to 0.5 kg per metric tonne find power of engine (iv)a pump can throw 10 quintals of coal per hrfrom a cola mine 120 m depth calculate power of engine in watt assuming that efficiency is 80%

Certified answers contain reliable, trustworthy information vouched for by a hand-picked team of experts. Brainly has millions of high quality answers, all of them carefully moderated by our most trusted community members, but certified answers are the finest of the finest.

2400 kg of water per minute is pumped. initially water is at rest. the change in energy is the work done.

speed of water = 3 meters/second = v So 2400 kg of water comes out in 1 min. Hence 40 kg per second. let t = time. so mass/time = m/t = 40 kg/sec.

The kinetic energy = work done = 1/2 m v² Power = work done / time = 1/2 (m / t) v² = 1/2 * 40 kg/sec * 3² m²/sec² = 180 kg-m/sec²-m = 180 W

work done for 10 hrs = 180 * 10 * 3600 Joules ============================ 2) m = 1120 kg let the angle of slope = Ф => tan Ф = 1/56 As the tan Ф is very small, angle Ф also very small. So Sin Ф is almost equal to tan Ф ie., 1/56. The car travels 20 m along the slope in 2 sec. => velocity = 10 m/s along the inclined plane.

Force exerted by the engine along the slope = m g Sin Ф + Friction force = 1120 kg * 10 m/sec² * 1/56 + 60 N = 260 N Power by the engine = force * velocity = 2600 N-m/s or Watts. =========================== 3) As the truck is moving with uniform velocity and no acceleration, the force exerted by the engine of truck is equal to the frictional force. frictional force = 0.5 kg force / metric tonne * 100 metric tonnes = 50 kg force = 50 * g Newtons = 500 Newtons Power = force * velocity in the same direction = 500 N * 72 * 5/18 m/sec = 10,000 W = 10 kW ============================ 4) height = h = 120m, mass = m = 10 quintals = 1000 kg, time = t = 1 hr = 3600 sec.

change in the potential energy of coal = m g h power at 100% efficiency = energy / time = m g h / t = (m/t) g h = 1000 kg / 3600sec * 10 m/sec² * 120 m = 333.333 Watts power at 80% efficiency = power / efficiency = 333.333 / (80/100) = 416.66 Watts

if you calculate the power with g = 9.8 m/sec² instead of 10 m/sec² then you will get the power as 408.33 Watts.