force = rate of change of momentum
change of velocity = 15+17.4 = 32.4 m/s
rate of change of momentum = mass per second * change of velocity
= 33.6 kg/s * 32.4 m/s
= 1089 kg m/s^2 or Newtons
A stream of water strikes a stationary turbine blade horizontally, as the drawing illustrates. The incident water stream has a velocity of + 17.4 m/s, while the exiting water stream has a velocity of – 15.0 m/s. The mass of water per second that strikes the blade is 33.6 kg/s. Find the magnitude of the average force exerted on the water by the blade.
3 answers
21
Favg∆t=m∆V
mass=33.6 kg/s
∆V=(-15-17.4)
Forces acting on the blade or positive and forces acting on the water are negative. Still, they are equal and opposite.
Favg=(33.6kg/s)(-32.8m/s)
Favg=-1088.63 kg/s^2
final answer: Favg= =-1100 N
mass=33.6 kg/s
∆V=(-15-17.4)
Forces acting on the blade or positive and forces acting on the water are negative. Still, they are equal and opposite.
Favg=(33.6kg/s)(-32.8m/s)
Favg=-1088.63 kg/s^2
final answer: Favg= =-1100 N
0 answers