Answer:
i) [tex]v_1 = 0.40 m/s[/tex]
ii) [tex]v_2 = 3.60 m/s[/tex]
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
[tex]A = \pi (\frac{D}{2})^2[/tex]
[tex]A = \pi (\frac{0.0271}{2})^2 = 5.77 \times 10^{-4} m^2[/tex]
Now the flow rate is defined as the rate of volume
It is given as
[tex]Q = \frac{Volume}{time} = Area \times speed[/tex]
[tex]\frac{20 L\times \frac{10^{-3} m^3}{1L}}{1.45 \times 60 seconds} = 5.77 \times 10^{-4} \times v[/tex]
[tex]v = 0.40 m/s[/tex]
Part b)
As per equation of continuity we know
[tex]A_1 v_1 = A_2 v_2[/tex]
now we have
[tex]\pi (\frac{d_1}{2})^2 v_1 = \pi (\frac{d_2}{2})^2v_2[/tex]
[tex](2.71)^2 (0.40) = (\frac{2.71}{3})^2 v_2[/tex]
[tex]v_2 = 3.60 m/s[/tex]
