Question

Water flowing through a garden hose of diameter 2.78 cm fills a 22.0 L bucket in 1.50 min.

(a) What is the speed of the water leaving the end of the hose?

(b) A nozzle is now attached to the end of the hose. If the nozzle diameter is one-third the diameter of the hose, what is the speed of the water leaving the nozzle?

Answer 1

volume flow rate V = volume of water in bucket/time taken = 22*10^-3/

volume of water in bucket = 22 L = 22*10^-3 m^3

time taken t = 1.5 min = 1.5*60 = 90 s

volume flow rate V = 22*10^-3/90 = 0.24*10^-3 m^3/s

(a)

volume flow rate = A1*v1 = pi*r1^2*v1

r1 = radius = 2.78/2 = 1.39 cm = 0.0139 m

0.24*10^-3 = pi*0.139^2*v1

v1 = 0.4 m/s

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(b)

A1*v1 = A2*v2

pi*r1^2*v1 = pi*r2^2*v2

v2 = v1*(r1/r2)^2

r2 = r1/3

v2 = 0.4*(1/9)

v2 = 0.044 m/s