Question
Sat October 13, 2012 By: Amithjose
 

Water jet of 2 cm2 area of cross-section strikes the wall normally with a velocity of 20 m/s and fall freely. The magnitude of force exerted on the wall is: (density of water = 1000 kg m-3)

Expert Reply
Sat October 13, 2012
the rate of change of momentum can be calculated as:
A * v *p
here A is the area of cross section, v is the velocity of water and p is the density of water.
thus will be equal to the force exerted on the wall.
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