 # 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. Ramsey 2021-08-12 Answered
Water flows through a water hose at a rate of ${Q}_{1}=680c\frac{{m}^{3}}{s}$, 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 ${v}_{2}=9.2\frac{m}{s}$.
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) ${A}_{1}=\pi ×{d}_{1}^{\frac{2}{4}}$
b) ${A}_{1}=\pi ×{2.2}^{\frac{2}{4}}$
$=3.80c{m}^{2}$
c) ${Q}_{1}={A}_{1}×{v}_{1}⇒{v}_{1}=\frac{{Q}_{1}}{{A}_{1}}$
d) ${v}_{1}=\frac{680}{3.80}$
$=179c\frac{m}{s}$
$=1.79\frac{m}{s}$
e) Using continuty equation, ${A}_{2}×{v}_{2}={A}_{1}×{v}_{1}$
${A}_{2}={A}_{1}×\frac{{v}_{1}}{{v}_{2}}$
f) ${A}_{2}\right\}=3.80×\frac{1.79}{9.2}$
$=0.739c{m}^{2}$