A cylindrical container of water has a height of 12 in and opens to atmosphere it squirts horizontally from a hole near the bottom. (Assume for parts a and b that the cross sectional area of this cylinder is much larger than the cross sectional area of the hole). a) What are the values of the pressure at
1. The surface of water
2. Just outside the hole
3. At the bottom of the container, ignoring the velocity at which the water level moving. I
b) Determine the speed at which the water leaves the hole when the water level is 6 cm above the hole. of
c) Now assume that the ratios of the cross sectional areas of the cylinder to that of the hole is 10 times, calculate the velocity at which the surface of water dropping down, when the water level is 6 cm above the hole.

a) 1 & 2The pressure at the surface of water and just outside the hole are both atmospheric pressure. { mean value of 101,325 pascals (roughly 14.6959 pounds per square inch).}

3. P(bottom) = 2987.04 Pa

b) v = 1.0844 m/s

c) v₁  = 0.11 m/s

Explanation:

a)

1) The pressure at the surface of water is same as the atmospheric pressure { mean value of 101,325 pascals (roughly 14.6959 pounds per square inch).}

2) The pressure at just outside the hole is also same as the atmospheric pressure. { mean value of 101,325 pascals (roughly 14.6959 pounds per square inch).}

3) At the bottom of the container, the gauge pressure is given by:

P(bottom) = pgh

where h = 12in = 0.3084m

P(bottom) = 1000 kg/m³ × 9.8 m/s² × 0.3084m

P(bottom) = 2987.04 Pa

b)

The velocity at which the water leaves the hole can be obtained from Bernouilli's equation. Point 1 is at the surface of water, point 2 is just outside the hole;

p₁ + 1/2pv₁² + pgh₁ = p² + 1/2pv₂² + pgh₂

Pressures on both sides are the same and the velocity of the water at the surface is approximately zero (since the hole's cross sectional area is much smaller than the container's). The difference in depths h1 - h2 is just the height of the water level

so

pgh =  1/2pv₂²

v = √(2gh)

we substitute

v = √( 2 × 9.8 m/s² × 0.06 m

v = 1.0844 m/s

c)

Now the velocity of the water level can't be neglected, we can use the continuity equation:

v₂ = (A₁/A²)V₁

so Back to Bernouilli's:

1/2pv₁² + pgh₁ = 1/2 (10v₁)² + pgh₂

phg = 99/2pv₁²

v₁ = √(2/99gh)

v₁  = √(2/99 × 9.8 m/s² × 0.06 m)

v₁  = 0.11 m/s

im guessing the west of station b