During time Δt the hose emits a tube of water and the tube length can be calculated using
l = vΔt,
where l – the length, v – the velocity of the jet of water and Δt – the time interval.
Also, the tube’s area of cross-section A is given by equation
A = πr^2.
The mass m of water in the tube is
m = ρV,
where ρ – the density of water (1000 kg/m^3) and V is the volume of the tube of water
which equals A*l or πr^2*vΔt.
Hence, the mass m of water in the tube is ρ*πr^2*vΔt
The mass loss per second is
dm/dt = ρ*πr^2*vΔt/Δt = ρ*πr^2*v.
Since the force F = d(mv)/dt
F = m*dv/dt + v*dm/dt.
The velocity v is constant then dv/dt = 0 and
F = v*dm/dt.
F = v*dm/dt = v*ρ*πr^2*v = ρπ*(vr)^2.
Now we can sabstitute the values and calculate the force
F = 1000*π*(20*0.04)^2 ≈ 2010.62 N