Question

Water flowing through a garden hose of diameter 2.71 cm fills a 20.0-L bucket in 1.45 min. (a) What is the speed of the water leaving the end of the hose? Your response differs from the correct answer by more than 10%. Double check your calculations. m/s (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? m/s

Answer 1

i)
`v_1 = 0.40 m/s`

ii)
`v_2 = 3.60 m/s`

Step-by-step explanation:

Part A)

As we know that diameter of the hose pipe is 2.71 cm

Now the area of crossection of the pipe will be

`A = \pi ((D)/(2))^2`

`A = \pi ((0.0271)/(2))^2 = 5.77 * 10^(-4) m^2`

Now the flow rate is defined as the rate of volume

It is given as

`Q = (Volume)/(time) = Area * speed`

`(20 L* (10^(-3) m^3)/(1L))/(1.45 * 60 seconds) = 5.77 * 10^(-4) * v`

`v = 0.40 m/s`

Part b)

As per equation of continuity we know

`A_1 v_1 = A_2 v_2`

now we have

`\pi ((d_1)/(2))^2 v_1 = \pi ((d_2)/(2))^2v_2`

`(2.71)^2 (0.40) = ((2.71)/(3))^2 v_2`

`v_2 = 3.60 m/s`