Let's analyze the problem step-by-step. --- # Problem Breakdown: The diagram shows a water tank with a cross-sectional areaA = .05\,text{m^2\) and height \(h = 2\, \text{m}\). The water flows out through a small at the bottom and the problem involves calculating the velocity of water flow, pressure, and related aspects. --- # Step 1: Find the velocity of water flowing out of the opening (using Torricelli's theorem) **Given:** - Height of water column, \(h = 2\, \text{m}\) - Acceleration due to gravity, \(g = 9.81 \text{m/s}^2\) **Using Torricelli's theorem:** \[ = \sqrt{2gh} \] **Calculation:** \[ v = \sqrt{2 \times 9.81\, \text{m/s}^2 \times 2\, \text{m}} = \sqrt{39.24} \approx 6.26\, \text{m/s} \] --- # Step 2: Calculate the volumetric flow rate \(Q\) \[ Q = A \times v \] **Given:** - Cross-sectional area \(A = .05\, \text{m}^2\) - Velocity \(v \approx 6.26\, \text{m/s}\) **Calculation:** \[ Q = .05\, \text{m}^2 \times 6.26\, \text{m/s} \approx .313\, \text{m}^3/\text{s} \] --- # Step 3: Find the pressure at the opening (using Bernoulli's equation) **Bernoulli's equation between the water surface and the opening:** \[ P_1 + \frac{1}{2}\rho v_1^2 + \rho g h = P_2 + \frac{1}{2}\rho v_2^2 \] Assuming: - The velocity at the water surface \(v_1 \approx \) (since the surface area is large) - The pressure at the water surface \(P_1 = P_{atm}\) - The pressure at the opening \(P_2 = P_{atm} + \Delta P\) Simplifies to: \[ \rho g h = \frac{1}{2}\rho v^2 \] which is consistent with Torricelli's theorem. --- # Step 4: Calculate the pressure at the opening **Pressure at the opening:** \[ P_{opening} = P_{atm} + \frac{1}{2} \rho v^2 \] Given: - \(\rho = 100\, \text{kg/m}^3\) - \(v = 6.26\, \text{m/s}\) **Calculation:** \[ \frac{1}{2} \times 100\, \text{kg/m}^3 \times (6.26)^2 \approx 100 \times 19.65 \approx 19,650\, \text{Pa} \] So, \[ P_{opening} \approx P_{atm} + 19.65\, \text{kPa} \] --- # Final Summary: | Quantity | Value | |---|---| | Velocity of water outflow | **6.26 m/s** | | Volumetric flow rate | **.313 m³/s** | | Pressure at opening | \(P_{atm} + 19.65\, \text{kPa}\) | --- Let me know if you'd like a detailed explanation of any specific part!