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frez [133]
3 years ago
13

The base of a rectangular tank measures 10 ft by 20 ft. The tank is 16 ft tall, and its top is 10 ft below ground level. The tan

k is full of water weighing 62.4 lb/ft3. How much work does it take to empty the tank by pumping the water to ground level? Give your answer to the nearest ft ∙ lb.
Mathematics
2 answers:
sashaice [31]3 years ago
8 0

Answer:

Step-by-step explanation:

Given a rectangular tank with dimension (10ft by 20ft by 16ft)

Then the volume of the tank is

Volume =length × breadth ×height

Volume=10×20×16=3200ft³

V=3200ft³

Then,

∆F= weight density × volume

∆F= 62.4×3200

∆F= 199,680 lb

Then,

Let the 0-point on the x-axis be at the bottom of the tank, so the level of the water ranges from x = 0 to x = 16ft. (It would just as well to let 0 be ground level and let x range from x = −26ft to x − 0.) Then a slice of water at level x is raised (26− x)ft

Then ∆x=(26-x)ft

Work is given as

W= -∫F∆xdx. From 0 to 16

W= -∫199,680(26-x)dx From 0 to 16

W=-199,680∫26-x dx From 0 to 16

W=-199,680 (26x-x²/2). From 0 to 16

W=-199,680(26×16-0.5×16² -0-0)

W=-199,680(288)

W=-57,507,840J

The work done to empty the tank is 57,507,840J

fredd [130]3 years ago
5 0

Answer:

2 X 10∧6 ft.lb

Step-by-step explanation:

Volume of rectangular tank =  Base area X Height = (10 X 20 X 16)ft³ = 3200ft³

Mass of tank filled with water = 3200ft³ X 62.4 lb/ft³ = 199680lb or 90706.37kg

Hence, work done due to gravity,Δg = m * g * h, where, m = mass of tank, g = gravity = 9.8m/s² and h = height = 10ft or 3.05 meter(m)

∴ Δg = 90706.37 * 9.8 * 3.05 =  2.711 X10∧6 J or 2 X 10∧6 ft.lb

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