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mochoa4
23.06.2019 •
Mathematics
Ahemispherical tank of radius 2 feet is positioned so that its base is circular. how much work is required to fill the tank with water through a hole in the base when the water source is at the base? (the weight-density of water is 62.4 pounds per cubic foot.)
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Ответ:
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![\frac{2}{3} \pi r^3](/tpl/images/0009/6016/c6ae6.png)
The formula for calculating the density is expressed as:
Given the following
Weight-density = 62.4 pounds per cubic foot
From the formula;
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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Ответ:
784 ft·lb
Step-by-step explanation:
The amount of work required is equivalent to the work required to raise the total weight of water to the height of the centroid of the hemisphere.
The weight of the water is ...
Volume × density = (2/3)πr³ × 62.4 lb/ft³
= (2/3)π(2 ft)³(62.4 lb/ft³) ≈ 1045.5 lb
A reference page tells us that the centroid of a hemisphere is 3/8r above the circular base, so we're raising this weight of water by (3/8)(2 ft) = 3/4 ft.
The work done is ...
(3/4 ft)(1045.5 lb) ≈ 784 ft·lb
Ответ: