The given information is:
- The tank has the shape of a cylinder
- The dimensions of the cylinder are 5 ft long and diameter 2.2 ft.
- The tank is drained at a rate of 2.1 ft^3 per minute.
The volume of the tank is given by the formula:
[tex]V=\pi *(\frac{d}{2})^2*h[/tex]
Where d is the diameter and h is the height.
By replacing the known values we obtain the initial volume:
[tex]\begin{gathered} V=3.14*(\frac{2.2ft}{2})^2*5ft \\ V=3.14*(1.1ft)^2*5ft \\ V=3.14*1.21ft^2*5ft \\ V=18.997ft^3 \end{gathered}[/tex]
As the drain rate is 2.1 ft^3 per minute, the time that is needed to empty the tank is:
[tex]\frac{18.997ft^3}{2.1ft^3\text{ / min}}=9.04\text{ min}[/tex]
The answer is 9 minutes.
