Answer :
The speed of the water leaving the end of the hose [tex]{0.405 }\frac{ \mathrm{~m}}{ \mathrm{s}} $[/tex]
As per the given data in question:
Initial diameter of the hose [tex]D_1=2.80 \mathrm{~cm}$[/tex]
Volume of the bucket [tex]\mathrm{V}=21.0 \mathrm{~L}=0.021 \mathrm{~m}^3$[/tex]
Time = 1.30 minutes.
The cross-sectional area of the hose is
[tex]\mathrm{A}_1=\frac{\pi \mathrm{D}_1{ }^2}{4} \\{A}_1=\frac{\pi\left(2.80 \times 10^{-2} \mathrm{~m}\right)^2 }{4}\\{A}_1=6.16 \times 10^{-4} \mathrm{~m}^2$[/tex]
The volume flow rate
[tex]$\dot{V}=\frac{V}{t}[/tex]
[tex]=\frac{0.021 \mathrm{~m}^3}{1.30 \times 60 \mathrm{~s}}\\=2.50 \times 10^{-4} \frac{\mathrm{~m}^3}{\mathrm{s}$}[/tex]
Speed over water is only required to be used for collision avoidance and not necessarily for navigation. By using speed over ground, a navigator is more aware of the situation than otherwise. For example, if the ship is drifting towards a danger, navigator will know it better if he has speed over ground in radar.
Then, the speed of the water leaving the end of the hose is
[tex]$\mathrm{v}_1=\frac{\dot{V}}{\mathrm{A}_1} =\frac{\left(2.50 \times 10^{-4} \mathrm{~m}^3 / \mathrm{s}\right)}{\left(6.16 \times 10^{-4} \mathrm{~m}^2\right)} ={0.405 }\frac{ \mathrm{~m}}{ \mathrm{s}} $[/tex]
Therefore the value of speed of the water leaving the end of the hose is 0.405 m/s
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