Astream of water strikes a stationary turbine blade horizontally, as the drawing illustrates. the incident water stream has a velocity of +18.0 m/s, while the exiting water stream has a velocity of -18.0 m/s. the mass of water per second that strikes the blade is 39.0 kg/s. find the magnitude of the average force exerted on the water by the blade.

1404 N

Explanation:

speed of incident water, u = 18 m/s

speed of exiting water, v = - 18 m/s

mass per unit time, m / t = 39 kg/s

According to the second law of Newton's

force = rate of change of momentum

F = m (v - u) / t

F = 39 x ( -18 - 18)

F = - 1404 N

Thus, the force exerted on the water by the blade is 1404 N.

force is mass*acceleration.

we have a very small mass: 11.4*0.001 kg

if we want 0.035n of force, we need an acceleration:

0.035 = 11.4*0.001*a

a = 0.035/(11.4*0.001) m/s^2