A small closed bottle is filled with water to adepth of 30 cm .A small hole located 10 cm above the bottom of the bottle is then opened .What is the speed of the water emerge from the hole in m\sex unit

(1/2) m v^2 = m g h

v = sqrt (2 g h) where h = 20 cm = .20 m

The speed of the water will be in m/s units, not m/sex

h=0.30 m, hₒ=0.10 m.

Bernoulli’s equation between the top surface and the exitting stream:
Pₒ+0+ρghₒ=Pₒ+ρv²/2+ ρgh,
v² = 2g(hₒ-h),
v=sqrt{2g(hₒ-h)}=sqrt{2•9.8•(0.30-0.10)} =...

To find the speed at which the water emerges from the hole, we can apply Bernoulli's equation, which relates the pressure, density, and velocity of a fluid.

Bernoulli's equation is given as:

P + (1/2)ρv^2 + ρgh = constant

where:
P is the pressure of the fluid,
ρ is the density of the fluid,
v is the velocity of the fluid,
g is the acceleration due to gravity,
h is the height from the reference point.

In this case, we can take the reference point at the surface of the water in the bottle, h = 0.

Since the bottle is closed, the pressure inside the bottle is atmospheric pressure (P0).

At the surface of the water in the bottle, the velocity is 0 (v = 0).

At the hole, the pressure is still atmospheric pressure (P0). The height from the reference point is h = 10 cm = 0.1 m.

Therefore, Bernoulli's equation becomes:

P0 + (1/2)ρv^2 + ρgh = P0 + 0 + ρgh

Simplifying the equation, we find:

(1/2)ρv^2 = ρgh

v^2 = 2gh

v = √(2gh)

Substituting the values:
g = 9.8 m/s^2
h = 0.1 m

v = √(2 * 9.8 * 0.1) m/s
= √(1.96) m/s
≈ 1.4 m/s

Therefore, the speed at which the water emerges from the hole is approximately 1.4 m/s.