Flow rate:

$\overline{){\mathbf{Q}}{\mathbf{=}}{\mathbf{A}}{\mathbf{v}}}$

Where Q is the flow rate, A is the cross-sectional area, and v is the velocity.

**(a)**

A_{1} = πr_{1}^{2} = (πd_{1}^{2})/4

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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