What power is needed to lift the water to the surface?

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SUMMARY

The discussion centers on calculating the power required to lift water from a 25-meter deep well using an electric pump that pulls water at a rate of 0.25 m/s and discharges it at a speed of 8.5 m/s. The initial calculations provided were incorrect, particularly in the time and power formulas. The correct approach involves using the density of water to determine mass flow rate and applying gravitational potential energy to find the work done, which is essential for accurate power calculations.

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UNknown 2010
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An electric pump pulls water at a rate of 0.25 m /s
from a well that is 25 m deep. The water leaves the
pump at a speed of 8.5 m/s.
1. What power is needed to lift the water to the surface?
2.The necessary increase in power the pump to make the water flowing through it quickly 8.5m/s

my answer
1. t= (25^3)\0.25=62500 seconds
P=w\t=Fd\t= ((25)(25))\(62500)=1\10000 W

2. P=FV= (.25)(8.5)=2.125 W
 
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at a rate of 0.25 m /s
Is this correct or do you mean .25 cubic meters per second?
A volume or mass per second is required to solve the problem.
If it is .25 cubic meters, you can use the known density of water to find the mass. Then you can use a gravitational potential energy formula to find the work done in one second to lift that mass of water. Then it will be easy to find the power.

The time calc you did does not make sense. For one thing, it comes out in units of m^2 times seconds.
 

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