Question

Water from a fire hose is directed horizontally against a wall at a rate of 53.9 kg/s and a speed of 38.9 m/s. Calculate the magnitude of the force exerted on the wall (in N), assuming the water's horizontal momentum is reduced to zero.

Answer 1

`F_(avg)=2096.71N`

Step-by-step explanation:

As from the given data that:

m/Δt=53.9kg/s

The numerator is the mass and denominator is the change in time

Solve for change in velocity we have:

Δv=v₂-v₁

`=0-38.9m/s\\=-38.9m/s`

Δv= -38.9m/s

From Newtons second law we know that:

F=ma

We can write this equation as:

Favg=(m/Δt)Δv

Substitute the given values

So

`F_(avg)=(53.9kg/s)(-38.9m/s)\\F_(avg)=-2096.71N`

Using the average velocity formula.The answer will be positive, the negative only implies that force is coming away from the wall

`F_(avg)=2096.71N`