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CBSE Class 9 Answered

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)
Asked by | 16 Dec, 2011, 06:48: PM
Expert Answer
as the cylinder strikes the wall in 1 sec, force is applied.
 
the volume of cyl of water striking in 1 sec=0.0002*20=0.004 m3  (area of cr=0.0002 m2, ht=20m)
mass striking in 1 sec=m=0.004*density
=4 kg
force exerted=rate of change of momentum=mv/t=4*20    as the velocity changes from 20 m/s to 0 in 1 sec.
force exerted=80 N
Answered by | 19 Dec, 2011, 11:04: AM
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