Calculating Liters of Water Raised From Well

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Homework Help Overview

The discussion revolves around calculating the volume of water a horse can raise from a well, given specific parameters such as the density of water, the depth of the well, and the duration of work. The subject area includes concepts from physics related to work, energy, and unit conversions.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants explore the energy produced by the horse and the energy required to raise water. There are discussions about unit conversions and the application of formulas related to work and power. Some participants question the initial assumptions and calculations regarding the depth of the well and the conversion of horsepower to watts.

Discussion Status

The discussion is ongoing, with participants providing insights and calculations. Some have offered guidance on the relationships between power, work, and energy, while others are questioning the accuracy of the calculations and the assumptions made. Multiple interpretations of the problem are being explored.

Contextual Notes

There are constraints regarding unit conversions and the need to convert measurements to standard units. Participants are also discussing the implications of assuming all energy from the horse is used for raising water.

jezelee
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here is the question...water w/ density 1.0 X 10 (3) kg/m cubed, a well 20 feet deep, and a horse worked for 8 hours. How many liters of water did the horse raise from the well? I am ok in unit conversions, but I need help in figuring out steps in solving this problem.
 
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What do you think? How much energy does the horse produce in those 8 hours? (Assume its power is 1 horse power. :wink: And that all its energy goes into raising water.) And how much energy is required to raise a liter of water 20 feet?

(Be sure to convert to standard units--meters, kilograms, watts, joules.)
 
First, the depth of the well was meters, not feet. Second, I'm still not sure if I'm heading in the right direction. If water is 1 kg per liter, the work= F dx, and F= ma...then W= (1kg)(9.8 m/s sq.)(20m) = 196 J for each liter rased from the well. correct? If power is W/ dt, and I let the variable n equal the total number of liters raised, I set up an equation of (196)(n)/ 28800 sec = 1Watt. Solving for n gives me ~146.9 liters. Is this correct?
 
If my memory is correct, for liquids, P = Q * density * height, where Q is flow in volume/unit time.
 
You're still missing the horsepower. How many watts is 1 horsepower?
 

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