Answer:
0.153 m/s
Step-by-step explanation:
The flowrate Q = 0.625 x 10-3 m^3-/s
The diameter of the nozzle d = 5.19 x 10^-3 m
the velocity V = ?
The cross-sectional area of the flow A =
![\pi d^(2)/4](https://img.qammunity.org/2021/formulas/physics/college/y0ojfhqlf89c9s1e93staveribbo7n29xm.png)
==> (3.142 x 5.19 x 10^-3)/4 = 4.077 x 10^-3 m^2
From the continuity equation,
Q = AV
V = Q/A = (0.625 x 10-3)/(4.077 x 10^-3) = 0.153 m/s