To tackle this problem, we can use principles from fluid dynamics, specifically Torricelli's law, which relates the speed of fluid flowing out of an orifice to the height of the fluid above it. Let's break down the solution step by step.
Finding the Speed of Water Exiting the Hole
According to Torricelli's theorem, the speed \( v \) of fluid flowing out of an opening under the influence of gravity can be calculated using the formula:
v = √(2gh)
In this equation:
- g is the acceleration due to gravity (approximately 9.81 m/s²).
- h is the height of the water column above the hole (in meters).
Given that the hole is 16.0 m below the water level, we can substitute the values into the equation:
v = √(2 × 9.81 m/s² × 16.0 m)
Calculating this gives:
v = √(313.92 m²/s²) ≈ 17.7 m/s
Calculating the Diameter of the Hole
Next, we need to determine the diameter of the hole. We know the flow rate \( Q \) and can relate it to the speed \( v \) and the cross-sectional area \( A \) of the hole using the equation:
Q = A × v
We can express the area \( A \) of a circular hole in terms of its diameter \( d \) as:
A = (π/4) × d²
Substituting this into the flow rate equation gives:
Q = (π/4) × d² × v
Now, we can rearrange this to solve for the diameter \( d \):
d² = (4Q) / (πv)
Substituting the known values:
- Flow rate \( Q = 2.50 × 10^{-3} m³/min = 2.50 × 10^{-3} m³/(60 s) ≈ 4.17 × 10^{-5} m³/s \)
- Speed \( v ≈ 17.7 m/s \)
Now we can calculate:
d² = (4 × 4.17 × 10^{-5} m³/s) / (π × 17.7 m/s)
Calculating this gives:
d² ≈ 1.49 × 10^{-6} m²
Taking the square root to find \( d \):
d ≈ √(1.49 × 10^{-6}) ≈ 0.00122 m
Converting this to millimeters (1 m = 1000 mm):
d ≈ 1.22 mm
Summary of Results
In summary, the speed at which the water exits the hole is approximately 17.7 m/s, and the diameter of the hole is about 1.22 mm. These calculations illustrate how fluid dynamics principles can be applied to real-world scenarios involving fluid flow.
