Answer:
1108.464 N of force
Explanation:
diameter of water hose = 70 cm = 0.7 m
radius = 0.7/2 = 0.35 m
volumetric flow rate Q = 420 L/min
1 L = 0.001 m^3
1 min = 60 s
therefore,
Q = 420 L/min = (420 x 0.001)/60 = 0.007 m^3/s
Area A of fire hose = π[tex]r^{2}[/tex] = 3.142 x [tex]0.35^{2}[/tex] = 0.38 m^2
From continuity equation, Q = AV
where V1 is the velocity of the water through the pipe, and A1 is the area of the pipe.
Q = A1V1
0.007 = 0.38V1
V1 = 0.007/0.38 = 0.018 m/s.
Nozzle diameter = 0.75 cm = 0.0075 m
radius = 0.00375
Area = π[tex]r^{2}[/tex] = 3.142 x [tex]0.00375^{2}[/tex] = 4.42 x [tex]10^{-5}[/tex] m^2
velocity of water through the nozzle will be
V2 = Q/A2 = 0.007 ÷ (4.42 x [tex]10^{-5}[/tex]) = 158.37 m/s
From
F = ρQ(v2 - v1)
Where,
F = force exerted
p = density of water = 1000 kg/m^3
F = 1000 x 0.007 x (158.37 - 0.018) = 1108.464 N of force