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A stream 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.

Respuesta :

Answer:

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.

onyebuchinnaji

The magnitude of the average force exerted on the water by the blade is 1,404 N.

The given parameters;

  • velocity of the incident water, v = 18 m/s
  • velocity of the exiting water, u = - 18 m/s
  • mass rate of the water, m' = 39 kg/s

The magnitude of the average force exerted on the water by the blade is calculated as follows;

F = ma

[tex]F = \frac{m(v-u)}{t} \\\\F = \frac{m}{t} (v-u)\\\\F = m'(v-u)\\\\F = 39(18 - - 18)\\\\F = 39(18 + 18)\\\\F = 39(36)\\\\F = 1,404 \ N[/tex]

Thus, the magnitude of the average force exerted on the water by the blade is 1,404 N.

Learn more here:https://brainly.com/question/14561704

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