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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Expert Answer

Brittany Patton
Answered 2021-06-11 Author has 27948 answers

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) \(A_{2}=3.80 \times \frac{1.79}{9.2}\)
\(=0.739cm^{2}\)

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