Water is pumped steadily out of a flooded basement at a speed of 6.1 m/s through a uniform hose of radius 0.89 cm. the hose passes out through a window to a street ditch 2.9 m above the waterline. what is the power of the pump?

47.83W

Explanation:

Suppose that Δm of water is pumped in time Δt

The potential energy of the water increases by the steady pumping of  water.

                                        ΔU = (Δm)gh

where,

h is the vertical distance through which it is lifted, and

This increases the kinetic energy by

                        ΔK = (Δm)v²

where v is its final speed

The workdone  is given by;

                          ΔW = ΔU + ΔK

                    ⇒ ΔW = (Δm)gh + (Δm)v²

and its power is given by;

                                 P = ΔW/Δt

                           ⇒ P = Δm/Δt (gh + (Δm)v²)

The rate of mass flow is Δm/Δt = ρ × A × v

                        ∴  Rate of mass flow = ρAv

where,

ρ is the density of water

A is the area of the hose

A = πr²

  A = π (0.0089)² = 2.49 × 10⁻⁴ m²

Rate of mass flow ρAv = (1000kg/m³) x (2.49 × 10⁻⁴ m²) x (6.1 m/s)

                                       = 1.52 kg/s

The power of the pump is;

P = ρAv (gh + (Δm)v²)

P = (1.52kg/s)( (9.8m/s² × 2.9m) + 1)

P = 1.52 × 31.47

P = 47. 83W

the answer is b i think

momentum of soft ball three times less than baskey ball