**(a)**

Cross-section of the hose is a circle.

Area of a circle is:

A = πr^{2}

In this case, r = d_{1}/2

1. Water flows through a water hose at a rate of *Q*_{1} = 620 cm^{3}/s, the diameter of the hose is *d*_{1} = 2.48 cm. A nozzle is attached to the water hose. The water leaves the nozzle at a velocity of *v*_{2} = 14.8 m/s.

a. Enter an expression for the cross-sectional area of the hose, *A*_{1}, in terms of its diameter, *d*_{1}.

A_{1}=

b. Calculate the numerical value of *A*_{1}, in square centimeters.

A_{1}=

c. Enter an expression for the speed of the water in the hose, *v*_{1}, in terms of the volume flow rate *Q*_{1} and cross-sectional area *A*_{1}.

v_{1}=

d.Calculate the speed of the water in the hose, *v*_{1} in meters per second.

v_{1}=

e.Enter an expression for the cross-sectional area of the nozzle, *A*_{2}, in terms of *v*_{1}, *v*_{2} and *A*_{1}.

A_{2}=

f. Calculate the cross-sectional area of the nozzle, *A*_{2} in square centimeters.

A_{2}=

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