The liquid of density roo flows along a horizontal pipe of uniform area of cross section .What is the force
Answers
Answered by
0
Answer:
ΔP=P
f
−P
i
(change in momentum as Δm of liquid passes through the bend).
Along X direction:
F
x
=
Δt
Δp
=
Δt
m(V
f
−V
i
)
=−
dt
mv
Along Y direction:
F
y
=
Δt
Δp
=
Δt
m(V
f
−V
i
)
=−
dt
mv
Total Rate of change in momentum of liquid = Force on the pipe
=
F
x
2
+F
y
2
=
dt
dm
×
v
2
+v
2
=
2
.
dt
dm v
=
2
vρa
dt
d l
=
2
ρav
2
∴ Force required to hold it in position is
2
aρv
2
in the direction shown otherwise it will bend in opposite direction.
Explanation:
- mark as brain least
Similar questions