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14 August, 09:34

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.

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Answers (2)
  1. 14 August, 11:57
    0
    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.
  2. 14 August, 12:11
    0
    1404 N.

    Explanation:

    Force = rate of change of momentum

    d/dt (mv₁ - mv₂)

    If v₂ = - v₁

    rate of change of momentum

    = d/dt (mv₁+mv₁)

    =2x v₁ dm / dt (Here velocity of water v₁ throughout is constant)

    Force = 2 xv₁ dm / dt

    Given,

    v₁ = 18 m/s

    dm / dt = rate of flow of mass of water

    = 39 kg / s

    Putting the values in the equation above

    Force = 2xv₁ dm / dt

    = 2x18 x 39

    = 1404 N.
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