The question states: A firemans hose has a nozzle of area 5cm^2 and the hose has a cross sectional area of 360cm^2. The hose ejects water at 25m/s and at a rate of 12L/s What is the force that the fireman must exert to keep the hose steady?(adsbygoogle = window.adsbygoogle || []).push({});

My attempt:

What I have done so far is to calculate the speed of water as it travels through the parts of the hose.

In the nozzle it travels at 24m/s (12000/500) and in the hose it is travelling at (12000/36000) = 0.33ms^-1, and as it is ejected, it is travelling at 25m/s.

So it goes from travelling at 0.33m/s to 24m/s to 25m/s. Hence, the force is

m(v-u)/t = 12kg/s * (25-0.33) = 296N

So the fireman exerts a force of 296N in the direction of the flow of water.

Is this correct? It seems to simplistic to ignore the 24m/s in the nozzle, yet it seems to work.

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# Homework Help: Force of water out of a hose

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