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3 years ago

7

A stream of water strikes a stationary turbine blade horizontally, as the drawing illustrates. The incident water stream has a v

elocity 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 53.0 kg/s. Find the magnitude of the average force exerted on the water by the blade.

1 answer:

bulgar [2K]3 years ago

3 0

Answer:

F = 1908 N

Explanation:

GIVEN,

Incident water stream has a velocity = U = +18.0 m/s,

Exiting water stream has a velocity = V = -18.0 m/s.

Mass of water per sec : M / t = 53 kg /s

Change in velocity d v =18 m / s – ( -18 m / s)

= 36 m / s

Magnitude of the average force exerted on the water by the blade

F = ( M / t ) x dv

=( 53 kg /s ) x ( 36 m/s )

F = 1908 N

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