 Water flows through a water hose at a rate of Q_{1}=680cm^{3}/s, the diameter of the hose is d_{1}=2.2cm. A nozzle is attached to the water hose. The e1s2kat26 2021-06-10 Answered
Water flows through a water hose at a rate of $$Q_{1}=680cm^{3}/s$$, the diameter of the hose is $$d_{1}=2.2cm$$. A nozzle is attached to the water hose. The water leaves the nozzle at a velocity of $$v_{2}=9.2m/s$$.
a) Enter an expression for the cross-sectional area of the hose, $$A_{1}$$, in terms of its diameter, $$d_{1}$$
b) Calculate the numerical value of $$A_{1},$$ in square centimeters.
c) Enter an expression for the speed of the water in the hose, $$v_{1}$$, in terms of the volume floe rate $$Q_{1}$$ and cross-sectional area $$A_{1}$$
d) Calculate the speed of the water in the hose, $$v_{1}$$ in meters per second.
e) Enter an expression for the cross-sectional area of the nozzle, $$A_{2}$$, in terms of $$v_{1},v_{2}$$ and $$A_{1}$$
f) Calculate the cross-sectional area of the nozzle, $$A_{2}$$ in square centimeters.

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Step 1
a) $$A_{1}=\pi\times d_{1}^{\frac{2}{4}}$$
b) $$A_{1}=\pi\times2.2^{\frac{2}{4}}$$
$$=3.80cm^{2}$$
c) $$Q_{1}=A_{1}\times v_{1}\Rightarrow v_{1}=\frac{Q_{1}}{A_{1}}$$
d) $$v_{1}=\frac{680}{3.80}$$
$$=179cm/s$$
$$=1.79m/s$$
e) Using continuty equation, $$A_{2}\times v_{2}=A_{1}\times v_{1}$$
$$A_{2}=A_{1}\times\frac{v_{1}}{v_{2}}$$
f) $$A_{2}=3.80 \times \frac{1.79}{9.2}$$
$$=0.739cm^{2}$$