❇️HEYA❇️
An electric pump is 60 efficient and is rated 2 HP Calculate the maximum amount of water it can lift through a height of 5 min 40 s.
Take g = 10 m/s2 and HP = 750 W]
✔️✔️help please ✔️✔️
Answers
Answered by
8
Let the amonit of water be m kg.
Given Comditions-
Power = 2 HP.
= 2 × 750
= 1500 W. As per as the question, electric pump is 60% efficient. Thus, Power of the pump = (60/100) × 1500. = 900 W.
Height = 5 m
Time = 10 seconds.
Acceleration due to gravity = 10 m/s^2
Using the formula,
Work done = mgh
= m × 10 × 5
= 50 × m.
Now, Power = Work/Time
900 = 50 × m/10
m = 180 kg.
Thus, the amount of water is 180 kg.
Answered by
6
✴ola!!!!✴
1 HP=750W
So, 2HP = 750 W × 2
2HP = 1500W
Power = (60/100)× 1500 = 900 W
=> Work done by electric pump in lifted height = mgh = m×10s^-2 × 5 = 50 m^2s^-2
=> m×50m^2s^-2 = 900
m= 900/50
=> 180kg answer✔✔
tysm❤
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