Answer:
The flow rate is [tex]R =2.357 *10^{-4} \ m^3/s[/tex]
Explanation:
From the question we are told that
The velocity is [tex]v = 3.0 \ m/s[/tex]
The diameter of the pipe is [tex]d = 1.0 \ cm = 0.01 \ m[/tex]
The radius of the pipe is mathematically represented as
[tex]r = \frac{d}{2}[/tex]
substituting values
[tex]r = \frac{0.01}{2}[/tex]
[tex]r = 0.005 \ m[/tex]
The flow rate is mathematically represented as
[tex]R = v * A[/tex]
Where is the cross-sectional area of the pipe which is mathematically evaluated as
[tex]A = \pi r^2[/tex]
substituting values
[tex]A = 3.142 * (0.005)^2[/tex]
[tex]A = 7.855 * 10^{-5} \ m^2[/tex]
So
[tex]R = 3.0 * 7.855 *10^{-5}[/tex]
[tex]R = 2.357*10^{-4} \ m^3 /s[/tex]