Asked by Denise
I really need help with this problem.
A conical vessel is 12 feet across the top and 15 feet deep. If it contains a liquid weighing p lbs/ft^3 (p=62.5 lbs/ft^3)to a depth of 10 feet. Find the work done in pumping the liquid to a height of 3 feet above the vessel.
A conical vessel is 12 feet across the top and 15 feet deep. If it contains a liquid weighing p lbs/ft^3 (p=62.5 lbs/ft^3)to a depth of 10 feet. Find the work done in pumping the liquid to a height of 3 feet above the vessel.
Answers
Answered by
drwls
The center of gravity of the cone is 1/3 of the height of 5m from the bottom, which is 10 feet from the top. To raise all the liquid 3 feet above the top, tha average mass is raised 3 + 10 = 13 feet.
The work required is the weight of the water in the conical tank,
W = (1/3)*pi*R^2*H*g
multiplied by 13 feet.
The answer will be in ft-lb.
The work required is the weight of the water in the conical tank,
W = (1/3)*pi*R^2*H*g
multiplied by 13 feet.
The answer will be in ft-lb.
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