What power of an engine is required to pump 2450 N of water per second from a well 50 m deep to the surface? (a) 2.45*10^3 watts (b) 2.45*10^4 watts (c) 2.45*10^5 watts (d) none of these

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What power of an engine is required to pump 2450 N of water per second from a well 50 m deep to the surface? (a) 2.45*10^3 watts (b) 2.45*10^4 watts (c) 2.45*10^5 watts (d) none of these

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@Michele_Laino please can you solve it?
okay so work is Fd so the answer is 2450 N times 50 m
wait... per second

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ignore my last answer
I think you're right because a watt is a N-m/s, so the units work out
we have to consider the manometric head H, which is about 50 m, so the requested power P is: \[\Large P = F\left( {H + h} \right) = 2450\left( {50 + 50} \right) = ...?\] being h the deepness of the well
sorry H is equal to the ratio between the pressure difference and the specific weight of the water and H+h is the manometric head or manometric prevalence
would you please tell me a little about manometric head,50 m deepness is okay to understand but what about additional 50m that you added
since the formula for computing the power, is: \[\large P = \frac{{FH}}{t}\] where H is the total prevalence, whose formula is: \[H = h + \frac{{\Delta p}}{\gamma }\] \Delta p is the difference between the pressures
|dw:1434218860993:dw|
|dw:1434218975568:dw|
so: \[\frac{{\Delta p}}{\gamma } = 5 \times 10.33 = 51.65\;meters\]
please keep in mind that 10.33 meters is the maximum aspiration height for a pump
and: 1 athmosphere is equivalent to 10.33 meters of water
so you think that 2.45*10^5 would be the correct option?
yes!
are you confirmed liano?
. i'd go with d) but i'd like the OP to post the answer when that becomes apparent
yes! I confirm my answer above!
i asked this to my friend he say to me that you can apply formula \[P=mgh/t\] since W=mgh \[P=W/t\]
so after this answer is none of these,the (d) option
i am so much confused about the answer,but i think what you are telling is also correct!

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