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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 theta 2) 2avd cos theta 3) a(v^2)d cos theta 4) avd cos theta


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 theta

2) 2avd cos theta

3) a(v^2)d cos theta

4) avd cos theta


Grade:upto college level

3 Answers

abhishek dubey
13 Points
5 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
13 Points
4 years ago
F=2Pav^2cosβ Here P=density of liquida=area of liquidv^2= square of velocityAnd β=theta given angle!!!!
ankit singh
askIITians Faculty 614 Points
one year ago

 

F = dtdP
   = dt2dmVcos60o
   = dt2(ρAdx)Vcos60o
   = 2ρAV2cos60o
   = 103×6×104×122
   =86.4N
thanks and regards ankit singh

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