Question

Water flows through a water hose at a rate of Q1 = 860 cm3/s, the diameter of the hose is d1 = 1.85 cm. A nozzle is attached to the water hose. The water leaves the nozzle at a velocity of v2 = 10.8 m/s.
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 flow 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.

Answer 1

Answer:

a) A1 =  $\frac{\pi (d1)^{2} }{4}$

b) A1 = 2.688 cm$^{2}$

c) Q1 = A1 x v1

d) v1 = 3.1994 m/s

e) A2 = $\frac{A1 X v1}{v2}$

f)  A2 = 0.7963cm$^{2}$

Explanation:

a) Area = $\pi r^{2}$

r = $\frac{d}{2}$

thus,

area = $\pi (\frac{d}{2})^{2}$

A1 =  $\frac{\pi (d1)^{2} }{4}[/tex]b) d1 = 1.85 cmsubstituting in the above equation,A1 = [tex]\frac{\pi (d1)^{2} }{4}$

A1 =  $\frac{\pi (1.85)^{2} }{4}$

A1 = 2.688 cm$^{2}$

c) Flow rate = Area x velocity ( refer brainly.com/question/13997998)

Q1 = A1 x v1

d) From the above equation,

v1 = $\frac{Q1}{A1}$ = $\frac{860}{2.688}$ = 319.94 cm/s = 3.1994 m/s

e) Since the flow rate Q1 is constant throughout the hose, Av is a constant.

i.e. A1 x v1 = A2 x v2

thus,

A2 = $\frac{A1 X v1}{v2}$

f) v2 = 10.8 m/s.

substituting the values in the above equation,

A2 = $\frac{2.688 X 3.1994}{10.8}$  = 0.7963cm$^{2}$