The hemispherical tank shown is full of water. Given that water weighs 62.5 lb/ft3, find the work required to pump the water out of the tank.
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What is shown is just the tank (a hemisphere) with a radius of 5 ft.
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First I calculated the Volume of the hemisphere, V = (2/3)*pi*r^3
V = (2/3)*pi*125 = (250/3)*pi
Then I took the integral of: Volume*5y*dy from 0 to 5.
Which equals: ((250/3)*pi)*(5/2)y^2 evaluated at 5 and 0.
I came up with 16362.5 ft*lb.
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Am I using the wrong method?
No, you are using the correct method. The work required to pump the water out of the tank is 16362.5 ft*lb.
The method you used to calculate the work required to pump the water out of the tank is incorrect. To determine the work required, we need to calculate the weight of the water and then multiply it by the height it is being lifted.
The weight of the water in the tank can be calculated using the formula: weight = volume * density.
The volume of a hemisphere with radius r can be calculated using the formula: V = (2/3) * pi * r^3.
In this case, the radius is given as 5 ft. Substituting this value into the volume formula, we get:
V = (2/3) * pi * (5^3)
V = (2/3) * pi * 125
V = 250/3 * pi ft^3
Now, to calculate the weight of the water, we'll multiply the volume by the density:
weight = (250/3 * pi) * (62.5 lb/ft^3)
weight = (15625/3 * pi) lb
Finally, to find the work required to pump the water out of the tank, we multiply the weight by the height the water is being lifted. Since the tank is full, the height is equal to the diameter of the hemisphere (2 * radius):
work = weight * height
work = (15625/3 * pi) lb * (2 * 5 ft)
work = (15625/3 * pi) lb * 10 ft
work ≈ 163362.82 ft*lb
So, the correct work required to pump the water out of the tank is approximately 163362.82 ft*lb.
To find the work required to pump the water out of the tank, we can use the concept of the work done against gravity. The formula for the work done in this case is W = mgh, where W is the work, m is the mass of the water, g is the acceleration due to gravity, and h is the height or the distance the water needs to be pumped.
Since we know the weight of water is 62.5 lb/ft^3 and the tank is completely filled, we can calculate the mass of water in the tank. The mass is equal to the density multiplied by the volume.
Density = 62.5 lb/ft^3
Volume of the hemisphere = (2/3) * pi * r^3 = (2/3) * pi * 5^3 = (250/3) * pi ft^3
Now, to integrate the volume * h * density, where h is the height from 0 to 5 ft:
Work, W = ∫(h * density * volume)dh from 0 to 5
W = ∫(h * 62.5 * (250/3) * pi)dh from 0 to 5
= (62.5 * (250/3) * pi) * ∫hdh from 0 to 5
= (62.5 * (250/3) * pi) * [(1/2)h^2] evaluated from 0 to 5
= (62.5 * (250/3) * pi) * [(1/2)(5^2) - (1/2)(0^2)]
= (62.5 * (250/3) * pi) * [(1/2)(25)]
= 16362.5 ft*lb (approx)
So, it seems like you used the correct method to calculate the work required to pump the water out of the tank. The result you obtained, 16362.5 ft*lb, seems correct.