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The rectangular tank shown here, with its top at ground level, is used to catch runoff water. Assume that the water weighs 
62.5lb//ft^(3).
a. How much work does it take to empty the tank by pumping the water back to ground level once the tank is full?
b. If the water is pumped to ground level with a (10/11)-horsepower (hp) motor (work output 
500ft-lb//sec ), how long will it take to empty the tank (to the nearest minute)?
c. Show that the pump in part (b) will lower the water level 
10ft (halfway) during the first 18 minutes of pumping.
d. What are the answers to parts (a) and (b) in a location where water weighs

62.31lb//ft^(3)?62.61lb//ft^(3)?
a. Set up an integral to find the work done.

W= 
◻
How much work does it take to empty the tank?

◻
b. How long will it take to empty the tank?

◻ minutes (Round to the nearest minute as needed.)
c. How much work does it take to lower the water level halfway?

◻

The rectangular tank shown here, with its top at ground level, is used to catch runoff water. Assume that the water weighs 62.5lb/ft3 62.5 \mathrm{lb} / \mathrm{ft}^{3} .\newlinea. How much work does it take to empty the tank by pumping the water back to ground level once the tank is full?\newlineb. If the water is pumped to ground level with a (1010/1111)-horsepower (hp) motor (work output 500ftlb/sec 500 \mathrm{ft}-\mathrm{lb} / \mathrm{sec} ), how long will it take to empty the tank (to the nearest minute)?\newlinec. Show that the pump in part (b) will lower the water level 10ft 10 \mathrm{ft} (halfway) during the first 1818 minutes of pumping.\newlined. What are the answers to parts (a) and (b) in a location where water weighs\newline62.31lb/ft3?62.61lb/ft3? 62.31 \mathrm{lb} / \mathrm{ft}^{3} ? 62.61 \mathrm{lb} / \mathrm{ft}^{3} ? \newlinea. Set up an integral to find the work done.\newlineW= \mathrm{W}= \square \newlineHow much work does it take to empty the tank?\newline \square \newlineb. How long will it take to empty the tank?\newline \square minutes (Round to the nearest minute as needed.)\newlinec. How much work does it take to lower the water level halfway?\newline \square

Full solution

Q. The rectangular tank shown here, with its top at ground level, is used to catch runoff water. Assume that the water weighs 62.5lb/ft3 62.5 \mathrm{lb} / \mathrm{ft}^{3} .\newlinea. How much work does it take to empty the tank by pumping the water back to ground level once the tank is full?\newlineb. If the water is pumped to ground level with a (1010/1111)-horsepower (hp) motor (work output 500ftlb/sec 500 \mathrm{ft}-\mathrm{lb} / \mathrm{sec} ), how long will it take to empty the tank (to the nearest minute)?\newlinec. Show that the pump in part (b) will lower the water level 10ft 10 \mathrm{ft} (halfway) during the first 1818 minutes of pumping.\newlined. What are the answers to parts (a) and (b) in a location where water weighs\newline62.31lb/ft3?62.61lb/ft3? 62.31 \mathrm{lb} / \mathrm{ft}^{3} ? 62.61 \mathrm{lb} / \mathrm{ft}^{3} ? \newlinea. Set up an integral to find the work done.\newlineW= \mathrm{W}= \square \newlineHow much work does it take to empty the tank?\newline \square \newlineb. How long will it take to empty the tank?\newline \square minutes (Round to the nearest minute as needed.)\newlinec. How much work does it take to lower the water level halfway?\newline \square
  1. Define tank dimensions and weight: Step 11: Define the dimensions of the tank and the weight of the water.\newlineAssume the tank is a rectangular prism with dimensions length (LL), width (WW), and height (HH). The weight of water is given as 62.5lb/ft362.5 \, \text{lb/ft}^3.
  2. Calculate water volume: Step 22: Calculate the volume of water in the tank when full. Volume, V=L×W×HV = L \times W \times H cubic feet.
  3. Calculate total weight: Step 33: Calculate the total weight of the water in the tank.\newlineTotal weight = Volume * Weight of water = LWH62.5lb.L * W * H * 62.5 \, \text{lb}.
  4. Set up integral for work done: Step 44: Set up the integral to find the work done to pump the water to ground level.\newlineWork done, W=0H(62.5LW(Hy))dyW = \int_{0}^{H} (62.5 \cdot L \cdot W \cdot (H - y)) \, dy, where yy is the depth from the top of the tank.
  5. Solve integral for total work: Step 55: Solve the integral to find the total work done. \newlineW=62.5×L×W×0H(Hy)dy=62.5×L×W×[H×yy22]0H=62.5×L×W×[H2H22]=62.5×L×W×H22W = 62.5 \times L \times W \times \int_{0}^{H} (H - y) dy = 62.5 \times L \times W \times [H\times y - \frac{y^2}{2}]_{0}^{H} = 62.5 \times L \times W \times [H^2 - \frac{H^2}{2}] = 62.5 \times L \times W \times \frac{H^2}{2} ft-lb.

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