This problem involves fluid dynamics and can be solved using Torricelli's Law, derived from Bernoulli's Equation. Here's how we solve it step by step:

Given:

• Diameter of the hole, $d = 5.00 \, \text{cm} = 0.05 \, \text{m}$
• Radius of the hole, $r = \frac{d}{2} = 0.025 \, \text{m}$
• Height difference, $h = 15.0 \, \text{m}$
• Density of water, $\rho = 1000 \, \text{kg/m}^3$
• Acceleration due to gravity, $g = 9.8 \, \text{m/s}^2$
• Atmospheric pressure cancels out since both sides are exposed to it.

Step 1: Find the speed of water ($v$) exiting the hole.

From Torricelli’s Law, the speed of water exiting the hole is: $v = \sqrt{2 g h}$ Substitute the given values: $v = \sqrt{2 \cdot 9.8 \cdot 15.0}$ $v = \sqrt{294} \approx 17.15 \, \text{m/s}$

Step 2: Find the mass flow rate ($\dot{m}$).

The mass flow rate is given by: $\dot{m} = \rho \cdot A \cdot v$ where:

• $A$ is the cross-sectional area of the hole: $A = \pi r^2 = \pi (0.025)^2 = \pi \cdot 0.000625 \approx 0.00196 \, \text{m}^2$ Substitute into the mass flow rate equation: $\dot{m} = 1000 \cdot 0.00196 \cdot 17.15$ $\dot{m} \approx 33.6 \, \text{kg/s}$

Final Answers:

1. Speed of water exiting the hole: $v \approx 17.15 \, \text{m/s}$
2. Mass flow rate of water: $\dot{m} \approx 33.6 \, \text{kg/s}$

Would you like further clarification or additional details on any part of this solution? Here are some related questions for practice:

1. How would the speed of water change if the height $h$ was doubled?
2. What would happen to the mass flow rate if the diameter of the hole was reduced to $3.00 \, \text{cm}$?
3. How does atmospheric pressure influence the result?
4. What assumptions are we making about the tank in applying Torricelli’s Law?
5. How would this setup differ if the fluid were oil ($\rho = 850 \, \text{kg/m}^3$) instead of water?

Tip: Always check if atmospheric pressure cancels out in fluid flow problems; this can simplify calculations significantly!