Answer: 1.135 L/s; 1.35 kg/s, 22.57 m/s
Explanation:
Given
Volume of bucket [tex]V=15\ gal\approx 56.78\ L[/tex]
time to fill it [tex]t=50\ s[/tex]
Volume flow rate
[tex]\dot{V}=\dfrac{56.78}{50}=1.135\ L/s\approx 1.135\times 10^{-3}\ m^3/s[/tex]
The inner diameter of the hose [tex]D=1.5\ cm[/tex]
diameter of the nozzle exit [tex]d=0.8\ cm[/tex]
we can volume flow rate as
[tex]\Rightarrow \dot{V}=Av\quad \quad \text{v=average velocity through nozzle exit}\\\\\Rightarrow 1.135\times 10^{-3}=\frac{\pi }{4}d^2\times v\\\\\Rightarrow 1.135\times 10^{-3}=\frac{\pi }{4}(0.8\times 10^{-2})^2\times v\\\\\Rightarrow v=\dfrac{4\times 1.135\times 10^{-3}}{\pi \times 64\times 10^{-6}}=22.57\ m/s[/tex]
Mass flow rate
[tex]\Rightarrow \dot{m}=\rho \times \dot{V}\\\Rightarrow \dot{m}=1\ kg/L\times 1.135\ L/s=1.35\ kg/s[/tex]
