Respuesta :
Answer:
Step-by-step explanation:
Given
height of tank [tex]h=5\ ft[/tex]
Width of tank [tex]w=3\ ft[/tex]
length of tank [tex]L=2\ ft[/tex]
suppose a layer of water at height h of thickness dh from bottom needed to be pump out
So distance moved by this layer to come out of tank is [tex]\Delta h=5-h[/tex]
weight density of water [tex]\rho =50\ pounds/ft[/tex]
Force required to hold this layer up [tex]F_s=2\times 3\times \Delta h\times 50=300\Delta h[/tex]
Work done to remove the water
[tex]W=\int_{0}^{5}300\Delta hdh[/tex]
[tex]W=\int_{0}^{5}300\left ( 5-h\right )dh[/tex]
[tex]W=3750\ Pound-ft[/tex]
The work does it take to pump all of the liquid out of the top of the tank is 3750-pound feet.
Given that
A tank in the shape of a right rectangular prism has a height of 5 feet, width of 3 feet, and length of 2 feet.
It is full of a liquid weighing 50 pounds per cubic foot.
We have to determine
How much work does it take to pump all of the liquid out of the top of the tank?
According to the question
A tank in the shape of a right rectangular prism has a height of 5 feet, width of 3 feet, and length of 2 feet.
The distance moved by this layer to come out of the tank is;
[tex]\rm \triangle h = 5-h[/tex]
The force required to hold this layer up is;
[tex]\rm Force = \triangle Height \times Width \times length \times liquid \ weighing\\ \\ Force = 2 \times 3 \times (5-h) \times 50\\ \\ Force = 300 (5-h)[/tex]
The work does it take to pump all of the liquid out of the top of the tank is calculated by;
[tex]\rm Work = \int\limits^5_0 {300(5-h)} \, dh\\ \\ Work = 300(5\int\limits^5_0 {} \, dh - \int\limits^5_0 {h} \, dh) \\ \\ Work = 300(5[h]^5_0- [\dfrac{h^2}{2}]^5_0)\\ \\ Work = 300(5(5-0)-\dfrac{5^2}{2}-\dfrac{0^2}{2})\\ \\ Work = 300(25-\dfrac{25}{2})\\ \\ Work = 300\times \dfrac{50-25}{2}\\ \\ Work = 300 \times \dfrac{25}{2}\\ \\ Work = 300 \times (12.5)\\ \\ Work = 3750 \ pounds \ feet[/tex]
Hence, The work does it take to pump all of the liquid out of the top of the tank is 3750-pound feet.
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