Answer:
The propulsion force at the moment of departure, is 49 N
Explanation:
Given;
diameter of tubes = 5 cm = 0.05 m
volumetric flow rate, V = 50 L/s = 0.005 m³/s
density of water, ρ = 1000 Kg /m³
hydraulic / propulsion force, F = ρVg
where;
ρ is the density of the fluid (water)
V is the volumetric flow rate of water
g is acceleration due to gravity
propulsion force, F = ρVg
propulsion force, F = 1000 x 0.005 x 9.8
propulsion force, F = 49 N
Therefore, the propulsion force at the moment of departure, is 49 N
