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.

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

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

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

f=ma newton's second law 7x8=56n

we're given that v₀ = 0 and a = 5.4 m/s². therefore, since v = at + v₀ for constant acceleration the equation for velocity is

after 28s,

m/s

which i would round to 150 m/s as my final answer