Answer:
v2 = 65 m/s
the speed of the water leaving the nozzle is 65 m/s
Step-by-step explanation:
Given;
Water flows at 0.65 m/s through a 3.0 cm diameter hose that terminates in a 0.3 cm diameter nozzle
Initial speed v1 = 0.65 m/s
diameter d1 = 3.0 cm
diameter (nozzle) d2 = 0.3 cm
The volumetric flow rates in both the hose and the nozzle are the same.
V1 = V2 ........1
Volumetric flow rate V = cross sectional area × speed of flow
V = Av
Area = (πd^2)/4
V = v(πd^2)/4 ....2
Substituting equation 2 to 1;
v1(πd1^2)/4 = v2(πd2^2)/4
v1d1^2 = v2d2^2
v2 = (v1d1^2)/d2^2
Substituting the given values;
v2 = (0.65 × 3^2)/0.3^2
v2 = 65 m/s
the speed of the water leaving the nozzle is 65 m/s