Provided the diameter of the fire hose is 7.0 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?

the amount of force  needed to hold the nozzle is 1100 N

Explanation:

Given that;

The diameter of the fire horse is 7.0cm

The radius of the fire horse is d/2 = 7.0/2 = 3.5 cm = 0.035 m

the water running through the hose flows at 420 L/min = 0.42 m³/ 60 sec

The diameter of the nozzle = 0.75 cm

The radius of the nozzle = d/2 = 0.75/2 = 0.375 cm = 0.00375 m

Amount of force needed to hold the nozzle = ??

Using equation of continuity product

where

= cross sectional area of the hose

= velocity of water flow at hose

= cross sectional area of the nozzle

= velocity of water flow at nozzle

Making the subject of the formula; we have

Also making the subject of the formula:

Also; from newton's law; we all know that F = ma

where ;

m = mass

a = acceleration

So ;

Also ; mass = density × volume

Replacing  by from above ; so

where the product of cross section area of velocity is equal to the volume of water and its time flow;

SO;

Replacing that into what we have above; we have:

Where;

A = πr²

F = 1100 N

Thus; the amount of force  needed to hold the nozzle is 1100 N

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