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
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 determine the speed at which the water emerges from the hole in meters per second, we can use Torricelli's law, which relates the speed of the efflux of a fluid through a small hole from a container to the height of the fluid above the hole.
The formula for the speed of efflux is:
v = √(2gh)
Where:
- v is the speed of efflux
- g is the acceleration due to gravity (approximately 9.8 m/s²)
- h is the height of the fluid above the hole
In this case, the height of the water above the hole is 30 cm, but we need to convert it to meters to maintain consistent units throughout the calculation. There are 100 cm in 1 meter, so:
h = 30 cm ÷ 100 cm/m = 0.3 m
Substituting the values into the formula:
v = √(2 × 9.8 m/s² × 0.3 m)
v = √(5.88 m²/s²)
v ≈ 2.43 m/s
Therefore, the speed at which the water emerges from the hole in meters per second is approximately 2.43 m/s.