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Sagot :
Image of the tank has been attached.
From online sources, we will use radius r = 9 meters with a spout of height h = 3 meters.
Answer:
W = 75800 KJ
Explanation:
Let's assume that we inscribe a cylinder which is x units from the center of the sphere and also has a height denoted by ∆x.
Now, At the center of the sphere with radius r, the equation of the sphere is given as:
x² + y² + z² = r².
Now, at the point location where the cylinder is x units from the center point, the width(y) of the sphere will be gotten as:
0² + y² + x² = 9²
y² + x² = 81
y² = 81 - x²
Thus;
y = √(81 - x²)
So y = r is the radius of the cylinder.
Thus; volume of the cylinder is given as:
V = πr²h
∆V = π(81 - x²)∆x
Integrating, we have;
V = π(81 - x²)
The weight of the water is given as:
F = mg
We know that density is;
ρ = m/V
So, m = ρV
Thus;
F = ρVg
F = (900)[π(81 - x²)](9.8)
F = 8820π(81 - x²)
Work needed to move this water up is;
W = Fd
The water will move up x to units to the center which is 9 units up to the top of the sphere, and 3 units to get to the top of the spout. Thus, d = 9 + 3 + x = 12 + x
Thus;
W = 8820π(81 - x²)(12 + x).
Due to the fact that x varies from -9 to 0 at the bottom half of the sphere, the required work will be;:
W = ∫8820π(81 - x²)(9 + x)dx (from x = -9 to 0)
Using online integral calculator, we have;
W = 8820π × 2733.75
W = 75749.061 KJ
To 3 significant figures = 75800 KJ
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