there are two holes one each along the opposite sides of wide rectangular tank.the cross section of each hole is 0.01 m and the vertical distance between two holes is one metre.the tank is filled with water the net force on the tank in newton when the water flows out of the holes is ( density of water = 1000 kgm^-3)
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Bernoulli's equation will give us the speed of water coming out the hole
(density)×g×h + Patm + 0 = (density)×g×0 + Patm + (1/2)×(density)×(v^2)
h=1m. density =1000kgm^-3
v = √(20)m/s
Force exerted by water coming out is
F=(density)×(area of hole)×(v^2)
area of hole =0.01m^2
F = 200 N
(density)×g×h + Patm + 0 = (density)×g×0 + Patm + (1/2)×(density)×(v^2)
h=1m. density =1000kgm^-3
v = √(20)m/s
Force exerted by water coming out is
F=(density)×(area of hole)×(v^2)
area of hole =0.01m^2
F = 200 N
abk16:
two wires of length l,radius r and length 2l and radius 2r respectively having same youngs modulus are hung with a weight mg.net elongation is
please solve this
and secondly pls explain me the step in which you balance force and velocity how its possible
force exerted by fluid
momentum is conserved in this case ,. and dP/dt = F
so P = (mass of fluid flowed ) ×( velocity of fluid) velocity of fluid will come from bernoulli's equation. I hope you know this
P = m × v. F = dP/dt = (dm/dt) × v
by equation of continuity dm/dt = density × area × velocity
putting that in above equation we will get F = density × area × velocity^2
ab to samajh me aaya
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