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

Answer:

i) $v_1 = 0.40 m/s$

ii) $v_2 = 3.60 m/s$

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

$A = \pi (\frac{D}{2})^2$

$A = \pi (\frac{0.0271}{2})^2 = 5.77 \times 10^{-4} m^2$

Now the flow rate is defined as the rate of volume

It is given as

$Q = \frac{Volume}{time} = Area \times speed$

$\frac{20 L\times \frac{10^{-3} m^3}{1L}}{1.45 \times 60 seconds} = 5.77 \times 10^{-4} \times v$

$v = 0.40 m/s$

Part b)

As per equation of continuity we know

$A_1 v_1 = A_2 v_2$

now we have

$\pi (\frac{d_1}{2})^2 v_1 = \pi (\frac{d_2}{2})^2v_2$

$(2.71)^2 (0.40) = (\frac{2.71}{3})^2 v_2$

$v_2 = 3.60 m/s$