Answer:
3.0549 MJ
Explanation:
You want to know the work done to empty a half-full spherical tank of radius 3 m containing liquid of density 900 kg/m³ through a 2 m spout at the top of the tank.
Mass
The mass of the liquid is the product of its volume and its density. The volume of the hemisphere is ...
V = 2/3πr³ = 2/3π(3³) = 18π . . . . . . cubic meters
Then the mass of the liquid is ...
m = ρV = (900 kg/m³)(18π m³) = 16200π kg
Height
The centroid of the liquid is 3/8r below the center of the tank. The mass of liquid must be moved from that location to a point 2 m above the top of the tank. The total height the liquid must be moved is ...
3/8(3 m) + 3m + 2m = 6.125 m
Work
The work done to lift the liquid is ...
W = Fd = (m·g)d = (16200π kg)(9.8 m/s²)(6.125 m) = 972405π J
W ≈ 3,054,900.4 J ≈ 3.0549 MJ
The work required is about 3.0549 megajoules.
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Additional comment
We could find the incremental work to move a layer of liquid dy thick to the necessary height, then integrate over the depth of liquid. It is much easier to look up the centroid location for a hemisphere, and use that in the calculation.
We have used "m" to represent both mass and meters. We trust you can determine which is which from the context.
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