The horizontal speed of a jet of water is 100cm/sec and 50 cm 3 of water hits the plate each second. assume that the water moves parallel to the plate after striking it. the force exerted on the stationary plate if it is held perpendicular to the jet of water is?
Answers
Answered by
41
v = 100 cm/s = 1 m/s
Volume of water hitting the plate per sec = 50 cm³ /sec = 50 * 10⁻⁶ m³ /sec
mass of water from the jet = 50 * 10⁻⁶ * 1000 kg/sec = 0.050 kg/sec
The linear momentum of the water from the jet becomes zero after collision.
So force exerted during 1 sec = change in moment of 50 cm³ of water
F * 1 sec = 0.050 * kg /sec * 1 m/sec
F = 0.050 Newtons
Volume of water hitting the plate per sec = 50 cm³ /sec = 50 * 10⁻⁶ m³ /sec
mass of water from the jet = 50 * 10⁻⁶ * 1000 kg/sec = 0.050 kg/sec
The linear momentum of the water from the jet becomes zero after collision.
So force exerted during 1 sec = change in moment of 50 cm³ of water
F * 1 sec = 0.050 * kg /sec * 1 m/sec
F = 0.050 Newtons
kvnmurty:
click on thanks link above please. select brainliest answer
Similar questions