a jet of water with cross sectional area 'a' is striking against a wall at an angle theta with normal and rebounds elastically.If the velocity of water jet is 'v' and density is'd' the normal force on wall is:1) 2a(v^2)d cos theta2) 2avd cos theta3) a(v^2)d cos theta4)avd cos theta
Amit Saxena , 10 Years ago
Grade upto college level
3 Answers
abhishek dubey
Last Activity: 8 Years ago
inthe following case volume of water striking the wall per sec=av (a=cross section of the cylinder,v=velocity of water ie length of cyl striking per sec.
but in this problem as water is making angle
,volume of water striking per sec=av sin?
mass striking /sec=avpsin?
as bouncing back, so momentum change /sec=force=2mv ;m=mass v= vel
=2av2 pcostheta.
Suhail Fayaz
Last Activity: 7 Years ago
F=2Pav^2cosβ Here P=density of liquida=area of liquidv^2= square of velocityAnd β=theta given angle!!!!
ankit singh
Last Activity: 4 Years ago
F = dtdP
= dt2dmVcos60o
= dt2(ρAdx)Vcos60o
= 2ρAV2cos60o
= 103×6×10−4×122
=86.4N
thanks and regards ankit singh
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