202010159
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The Kapandu boys dormitory in the UPNG is going up in flames. Rachel, a former student of Physics, who
is now a fire fighter, is part of the team who have rushed onto the scene to try to contain the blaze. As she
holds the fire hose she experiences a force exerted on her by the hose. Provided the diameter of the fire
hose is 70 cm, and the water running through the hose flows at 420 L/min, how much force does she need
to hold the nozzle, if its diameter is 0.75 cm?​

Respuesta :

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

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