Daniel B. answered 12/10/23
A retired computer professional to teach math, physics
Let
h = 0.40 m be the depth of the trough,
ρ be the density of water,
g = 9.81 m/s² be gravitation acceleration,
p = ρgh be the pressure at the bottom of the trough,
v (to be calculated) be the velocity of water coming out.
This problem can be solved by conservation of energy of some
small volume V of water located at the opening.
This volume V of water has mass
m = Vρ
Before exiting the trough, that volume V of water has potential pressure energy
Vp = Vρgh
This energy gets converted to kinetic energy upon exit, whose magnitude is
mv²/2 = Vρv²/2
By conservation of energy
Vρv²/2 = Vρgh
v = √(2gh)
Substituting actual numbers,
v = √(2×9.81×0.40) = 2.8 m/s
Notice that the solution depends only on the depth of the trough, not any of the other dimension.
Also it is independent of ρ, meaning, independent of the liquid.