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 Q1=680cm3s, the diameter of the hose is d1=2.2cm. A nozzle is attached to the water hose. The water leaves the nozzle at a velocity of v2=9.2ms.
a) Enter an expression for the cross-sectional area of the hose, A1, in terms of its diameter, d1
b) Calculate the numerical value of A1, in square centimeters.
c) Enter an expression for the speed of the water in the hose, v1, in terms of the volume floe rate Q1 and cross-sectional area A1
d) Calculate the speed of the water in the hose, v1 in meters per second.
e) Enter an expression for the cross-sectional area of the nozzle, A2, in terms of v1,v2 and A1
f) Calculate the cross-sectional area of the nozzle, A2 in square centimeters.
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Expert Answer

Daphne Broadhurst
Answered 2021-08-13 Author has 109 answers
Step 1
a) A1=π×d124
b) A1=π×2.224
=3.80cm2
c) Q1=A1×v1v1=Q1A1
d) v1=6803.80
=179cms
=1.79ms
e) Using continuty equation, A2×v2=A1×v1
A2=A1×v1v2
f) A2}=3.80×1.799.2
=0.739cm2
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