Asked by Anonymous
A pump slowly lifts 3.0*10^6 kg of water an average distance of 80.0m in 12minutes. 1) how much work is done in the first 12 minutes? 2)How much power is delivered by the pumps?
Answers
Answered by
Henry
F = 3*10^6kg * 9.8N/kg = 2.94*10^7 N.
1. Work = F*d = 2.94*10^7*80=2.352*10^9
Joules.
2. P = W/T = 2.352*10^9 / 720s. = 3.27*10^6 Joules/s. = 3.27*10^6 Watts.
1. Work = F*d = 2.94*10^7*80=2.352*10^9
Joules.
2. P = W/T = 2.352*10^9 / 720s. = 3.27*10^6 Joules/s. = 3.27*10^6 Watts.
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