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.

Question

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} = 680 c \frac{m^{3}}{s}

, the diameter of the hose is

d_{1} = 2.2 c m

. 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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Expert Answer

Daphne Broadhurst

Answered 2021-08-13 Author has 109 answers

Step 1
a)

A_{1} = π \times d_{1}^{\frac{2}{4}}

b)

A_{1} = π \times {2.2}^{\frac{2}{4}}

= 3.80 c m^{2}

c)

Q_{1} = A_{1} \times v_{1} \Rightarrow v_{1} = \frac{Q_{1}}{A_{1}}

d)

v_{1} = \frac{680}{3.80}

= 179 c \frac{m}{s}

= 1.79 \frac{m}{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}

= 0.739 c m^{2}

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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.

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