Final answer:
The speed of the water coming out of the spigot at the bottom of the vessel can be determined using Bernoulli's equation. By applying the equation and substituting the given values, the speed of the water is calculated to be 9.8 m/s.
Step-by-step explanation:
The speed of water coming out of a spigot at the bottom of a large vessel can be determined using the principle of Bernoulli's equation. Bernoulli's equation relates the pressure, velocity, and height of a fluid at any two points in a system. In this case, we can assume that the pressure at the bottom of the vessel is equal to atmospheric pressure, so we can consider the pressure term to be zero. The equation can be expressed as:
P1 + 1/2ρv12 + ρgh1 = P2 + 1/2ρv22 + ρgh2
Where:
- P1 and P2 are the pressures at points 1 and 2 respectively
- ρ is the density of the fluid (in this case, water)
- v1 and v2 are the velocities of the fluid at points 1 and 2 respectively
- g is the acceleration due to gravity
- h1 and h2 are the heights of the fluid at points 1 and 2 respectively
In this scenario, point 1 is at the bottom of the vessel and point 2 is at the spigot where the water is coming out. Since the water is open to the atmosphere, we can assume that the pressure at point 2 is also atmospheric pressure, which means P2 is also zero. Since the water is coming out of the spigot, its height (h2) is 0. Therefore, the equation simplifies to:
1/2ρv12 + ρgh1 = 0 + 1/2ρv22 + ρgh2
The height difference (y2-y1) mentioned in the question is the difference in height between point 1 and point 2. Substituting the given values:
1/2ρv12 + ρgh1 = 0 + 1/2ρv22 + ρgh2
0 + ρg(y2-y1) = 0 + 1/2ρv22 + 0
ρg(y2-y1) = 1/2ρv22
Cancelling out the common terms:
g(y2-y1) = 1/2v22
Rearranging the equation:
v22 = 2g(y2-y1)
Substituting the given values:
v22 = 2 * 9.8 * 4.9
v22 = 96.04
Taking the square root of both sides:
v2 = 9.8 m/s
Therefore, the speed of the water as it comes out of the spigot is 9.8 m/s.