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barxatty [35]
3 years ago
13

(1 point) A rectangular tank that is 3 feet long, 9 feet wide and 12 feet deep is filled with a heavy liquid that weighs 110 pou

nds per cubic foot. In each part below, assume that the tank is initially full. Your answers must include the correct units. (You may enter lbf or lb*ft for ft-lb.) (a) How much work is done pumping all of the liquid out over the top of the tank
Physics
1 answer:
Aloiza [94]3 years ago
3 0

Answer:

Explanation:

Work in pumping water from the tank is given as

W = ∫ y dF. From a to b

Where dF is the differential weight of the thin layer of liquid in the tank, y is the height of the differential layer

a is the lower limit of the height

b is the upper limit of the height.

We know that, .

F = ρVg

Where F is the weight

ρ is the density of water

V is the volume of water in tank

g is the acceleration due to gravity

Then,

dF = ρg ( Ady)

We know that the density and the acceleration due to gravity is constant, also the base area of the tank is constant, only the height that changes.

Then,

ρg = 62.4 lbs/ft³

Area = L×B = 3 × 9 = 27ft²

dF = ρg ( Ady)

dF = 1684.8dy

The height reduces from 12ft to 0ft

Then,

W = ∫ y dF. From a to b

W = ∫ 1684.8y dy From 0 to 12

W = 1684.8y²/2 from 0 to 12

W = 842.4 [y²] from y = 0 to y = 12

W = 842.4 (12²-0²)

W = 121,305.6 lb-ft

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