To solve the problem it is necessary to apply the concepts related to Discharge. The flow rate is the amount of volume of a fluid that circulates in a given time
By definition it is represented as
[tex]Q = vA[/tex]
Where,
v = Velocity
A = Area
However, as the name implies it can also be represented as
[tex]Q = \frac{V}{t}[/tex]
Where,
V = Volume
t = time
In our case we have the values of,
[tex]v = 0.40m/s[/tex]
[tex]A_{hose} = \pi\frac({(5/8)}{2})^2 = 0.3068in^2 \approx 0.00019806412m^2 = 1.98*10^{-4}m[/tex]
[tex]V_{pool} = A*h = \pi (\frac{6.1}{2})^2*1.4 = 40.91m^3[/tex]
Therefore replacing all these data in the previous equation we have
[tex]Q = vA[/tex]
[tex]\frac{V}{t} = vA[/tex]
[tex]\frac{40.91}{t} = 0.4*1.98*10^{-4}[/tex]
[tex]t = \frac{40.91}{0.4*1.98*10^{-4}}[/tex]
[tex]t = 516540.4s[/tex]
[tex]t = 516540.4s( \frac{1min}{60s})(\frac{1hour}{60min})(\frac{1 day}{24hours})[/tex]
[tex]t = 5.97days \approx 6 days[/tex]
It will take about 6 days to fill the pool.
