Question

a stream of water strikes a stationary turbine blade horizontally, as the drawing illustrates. the incident water stream has a velocity of +16.0 m/s, while the exiting water stream has a velocity of -16.0 m/s. the mass of water per second that strikes the blade is 22.0 kg/s. find the magnitude of the average force exerted on the water by the blade.

Explanation:

Step1: Recall the impulse - momentum theorem

The impulse - momentum theorem states that \(F_{avg}\Delta t=\Delta p = m(v_f - v_i)\). We can rewrite it for the force per unit time. Since \(F_{avg}=\frac{\Delta p}{\Delta t}\) and \(\frac{m}{\Delta t}\) is given as \(22.0\space kg/s\).

Step2: Substitute the values of velocities and mass per second into the formula

We know that \(v_i = + 16.0\space m/s\), \(v_f=-16.0\space m/s\) and \(\frac{m}{\Delta t}=22.0\space kg/s\). Then \(F_{avg}=\frac{m}{\Delta t}(v_f - v_i)\).
Substitute the values: \(F_{avg}=(22.0\space kg/s)\times(-16.0 - 16.0)\space m/s\).

$$
LATEXBLOCK0
$$

The magnitude of the force is \(|F_{avg}|=(22.0)\times(32.0)\space N\)

Answer:

\(704\space N\)