Question

A motor pump lifts water from an average
of 10 m at the rate of 5 kg per second. If the
water is coming out of the pump with velocity
2 m/s, the power of the motor is (g = 10 m/s2)
(1) 1020 W
(2) 600 W
(3) 510 W
(4) 400 W​

Answer 1

Answer:

Explanation:

Hope it helps you

Answer 2

Answer:(3)510W.

Explanation: Power delivered by the motor is:-

Power(W)=(per unit time(secs)).

Simply it is:-Power=(Work done/Time).

And work done is change in energy.

So, Work done=(Change in Energy: Both Potential and Kinetic Energy)

in this problem(Because only Kinetic and Potential Energies are Present).

So Now Calculating Energy change per unit time:-

Energy change=ΔK.E+ΔP.E

ΔK.E=(1/2)×m×($v^{2}$-$u^{2}$)(As given in question it is the velocity with which the water comes out is v=2m/s and initially it was at rest u=0m/s).

And mass rate coming out is 5kg/s implies 5kg per sec

ΔK.E=(1/2)×(5kg/s)×($2^{2}$-$0^{2}$).=10J-per second

Now calculating ΔP.E :-

ΔP.E=m×g×Δh=5×10×10=.500J-per second(5kg per second for 10 meters)

Total Energy Change=(ΔK.E+ΔP.E)=510J per second.

Now calculating the power which is:-

Energy consumed per second:-510J/s=510W(Answer).

For more such questions:-https://brainly.in/question/13179958

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