The work required to fill the tank with water through a hole in the base when the water source is at the base is 783.7425 Joules
The formula for calculating the volume of the hemispherical tank is expressed as [tex]\frac{2}{3} \pi r^3[/tex]
The formula for calculating the density is expressed as:
[tex]Density = \mathb\frac{weight}{volume}[/tex]
Given the following
Weight-density = 62.4 pounds per cubic foot
From the formula;
[tex]Weight = Density \times volume[/tex]
[tex]Weight=\frac{2}{3} \pi r^3 \times 62.4\\Weight= \frac{2}{3} \pi (2)^3 \times 62.4\\Weight= \frac{16}{3} \pi \times 62.4\\Weight= \frac{16}{3} (3.14) \times 62.4\\Weight =1,044.99N[/tex]
The centroid of a hemisphere will be 3/8r above the circular base, so we will raising this weight of water by (3/8)(2 ft) is 3/4 ft
Work done = 1044.99 × 3/4
Workdone = 783.7425 Joules
Hence the work required to fill the tank with water through a hole in the base when the water source is at the base is 783.7425 Joules
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