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Answer :
The work required to pump all the water out of the tank is approximately 25.51 million joules.
To find the work required to pump all the water out of the tank, we need to calculate the total gravitational potential energy of the water in the tank and then convert it into work.
The gravitational potential energy [tex]\( U \)[/tex] of an object at height [tex]\( h \)[/tex] in a gravitational field with acceleration due to gravity [tex]\( g \)[/tex] is given by:
[tex]\[ U = mgh \][/tex]
Where:
[tex]- \( m \)[/tex] is the mass of the object,
[tex]- \( g \)[/tex] is the acceleration due to gravity, and
[tex]- \( h \)[/tex] is the height.
For a fluid, we use the formula [tex]\( m = \rho V \),[/tex] where [tex]\( \rho \)[/tex] is the density of the fluid and [tex]\( V \)[/tex] is the volume of the fluid.
In this case, the tank is shaped like a cone. The volume of a cone is given by the formula:
[tex]\[ V = \frac{1}{3} \pi r^2 h \][/tex]
Where:
[tex]- \( r \)[/tex] is the radius of the base of the cone,
[tex]- \( h \)[/tex] is the height of the cone.
Given that [tex]\( r = 4 \)[/tex] meters and [tex]\( h = 7 \)[/tex] meters, we can calculate the volume of the cone.
[tex]\[ V = \frac{1}{3} \pi (4^2)(7) = \frac{1}{3} \pi (16)(7) = \frac{112}{3} \pi \][/tex]
Now, the mass of the water is:
[tex]\[ m = \rho V = 9800 \times \frac{112}{3} \pi \][/tex]
The gravitational potential energy of the water is then:
[tex]\[ U = mgh = 9800 \times \frac{112}{3} \pi \times 7 \][/tex]
Now, the work done in pumping out the water is equal to the change in potential energy, which is the initial potential energy minus the final potential energy (when the tank is empty).
[tex]\[ \text{Work} = U_{\text{initial}} - U_{\text{final}} \][/tex]
Since the final potential energy is 0 (when the tank is empty), the work done is just the initial potential energy:
[tex]\[ \text{Work} = 9800 \times \frac{112}{3} \pi \times 7 \][/tex]
Now, let's calculate this value.
First, let's calculate the mass of the water:
[tex]\[ m = 9800 \times \frac{112}{3} \pi \][/tex]
[tex]\[ m \approx 367201.582 \, \text{kg} \][/tex]
Now, let's calculate the initial potential energy:
[tex]\[ U_{\text{initial}} = mgh \][/tex]
[tex]\[ U_{\text{initial}} = 367201.582 \times 9.8 \times 7 \][/tex]
[tex]\[ U_{\text{initial}} \approx 25508881.43 \, \text{J} \][/tex]
So, the work required to pump all the water out of the tank is approximately [tex]\( 25,508,881.43 \, \text{J} \).[/tex]
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Rewritten by : Barada
The work required to pump all the water out of the tank is [tex]$\frac{35}{2}\pi \times 10^4$[/tex] joules.
To solve this problem, we will use the concept of work done against gravity to lift a volume of water out of the tank. The work done to lift a small volume of water a height [tex]$h$[/tex] is given by the product of the weight of the water and the height it is lifted. Since the density of water is [tex]$10^4$N/m³[/tex], the weight of a small volume [tex]$dV$[/tex] of water is [tex]$10^4 dV$[/tex] newtons.
The tank is shaped like a cone with height [tex]$H = 7$[/tex] meters and base radius [tex]$R = 3$[/tex] meters. The radius of the water surface decreases linearly with height as we pump the water out. At a height [tex]$h$[/tex] from the bottom of the tank, the radius [tex]$r$[/tex] of the water surface is given by the similar triangles formed by the water level and the sides of the cone:
[tex]\[ r = \frac{R}{H}h \][/tex]
Now, we consider a small disk of water at height [tex]$h$[/tex] with thickness [tex]$dh$[/tex]. The volume of this disk is the volume of a cylinder with radius [tex]$r$[/tex]and height [tex]$dh$[/tex]:
[tex]\[ dV = \pi r^2 dh \][/tex]
Substituting [tex]$r$[/tex] from the previous equation, we get:
[tex]\[ dV = \pi \left(\frac{R}{H}h\right)^2 dh \][/tex]
[tex]\[ dV = \pi \left(\frac{3}{7}h\right)^2 dh \][/tex]
[tex]\[ dV = \frac{9\pi}{49}h^2 dh \][/tex]
The work done to lift this disk to the top of the tank is the weight of the water times the height [tex]$H - h$[/tex]:
[tex]\[ dW = 10^4 \times dV \times (H - h) \][/tex]
[tex]\[ dW = 10^4 \times \frac{9\pi}{49}h^2 dh \times (7 - h) \][/tex]
To find the total work $W$ required to pump out all the water, we integrate [tex]$dW$[/tex] from [tex]$h = 0$ to $h = H$[/tex]:
[tex]\[ W = \int_{0}^{7} 10^4 \times \frac{9\pi}{49}h^2 (7 - h) dh \][/tex]
[tex]\[ W = 10^4 \times \frac{9\pi}{49} \int_{0}^{7} h^2 (7 - h) dh \][/tex]
[tex]\[ W = 10^4 \times \frac{9\pi}{49} \int_{0}^{7} (7h^2 - h^3) dh \][/tex]
[tex]\[ W = 10^4 \times \frac{9\pi}{49} \left[ \frac{7h^3}{3} - \frac{h^4}{4} \right]_{0}^{7} \][/tex]
[tex]\[ W = 10^4 \times \frac{9\pi}{49} \left[ \frac{7 \times 7^3}{3} - \frac{7^4}{4} \right] \][/tex]
[tex]\[ W = 10^4 \times \frac{9\pi}{49} \left[ \frac{343 \times 7}{3} - \frac{2401}{4} \right] \][/tex]
[tex]\[ W = 10^4 \times \frac{9\pi}{49} \left[ \frac{2401}{3} - \frac{2401}{4} \right] \][/tex]
[tex]\[ W = 10^4 \times \frac{9\pi}{49} \left[ \frac{4 \times 2401 - 3 \times 2401}{12} \right] \][/tex]
[tex]\[ W = 10^4 \times \frac{9\pi}{49} \left[ \frac{2401}{12} \right] \][/tex]
[tex]\[ W = 10^4 \times \frac{9\pi \times 2401}{49 \times 12} \][/tex]
[tex]\[ W = 10^4 \times \frac{21609\pi}{588} \][/tex]
[tex]\[ W = 10^4 \times \frac{35}{2}\pi \][/tex]
Therefore, the work required to pump all the water out of the tank is [tex]$\frac{35}{2}\pi \times 10^4$[/tex] joules.
