anonymous
  • anonymous
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
Physics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
jamiebookeater
  • jamiebookeater
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
@Michele_Laino please can you solve it?
anonymous
  • anonymous
okay so work is Fd so the answer is 2450 N times 50 m
anonymous
  • anonymous
wait... per second

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
ignore my last answer
anonymous
  • anonymous
I think you're right because a watt is a N-m/s, so the units work out
Michele_Laino
  • Michele_Laino
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
Michele_Laino
  • Michele_Laino
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
anonymous
  • anonymous
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
Michele_Laino
  • Michele_Laino
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
Michele_Laino
  • Michele_Laino
|dw:1434218860993:dw|
Michele_Laino
  • Michele_Laino
|dw:1434218975568:dw|
Michele_Laino
  • Michele_Laino
so: \[\frac{{\Delta p}}{\gamma } = 5 \times 10.33 = 51.65\;meters\]
Michele_Laino
  • Michele_Laino
please keep in mind that 10.33 meters is the maximum aspiration height for a pump
Michele_Laino
  • Michele_Laino
and: 1 athmosphere is equivalent to 10.33 meters of water
anonymous
  • anonymous
so you think that 2.45*10^5 would be the correct option?
Michele_Laino
  • Michele_Laino
yes!
anonymous
  • anonymous
are you confirmed liano?
IrishBoy123
  • IrishBoy123
. i'd go with d) but i'd like the OP to post the answer when that becomes apparent
Michele_Laino
  • Michele_Laino
yes! I confirm my answer above!
anonymous
  • anonymous
i asked this to my friend he say to me that you can apply formula \[P=mgh/t\] since W=mgh \[P=W/t\]
anonymous
  • anonymous
so after this answer is none of these,the (d) option
anonymous
  • anonymous
i am so much confused about the answer,but i think what you are telling is also correct!

Looking for something else?

Not the answer you are looking for? Search for more explanations.