# Step 1: Find the Volume Flow Rate at the First Point The volume flow rate \( Q \) is given by: \[ Q = \text{area} \times \text{velocity} \] At the first point: - Width \( w_1 = 18.3\, \text{m} \) - Depth \( d_1 = 3.55\, \text{m} \) - Velocity \( v_1 = 2.40\, \text{cm/s} = .024\, \text{m/s} \) \[ \text{Area}_1 = w_1 \times d_1 = 18.3 \times 3.55 = 64.965\, \text{m}^2 \] \[ Q = 64.965 \times .024 = 1.55916\, \text{m}^3/\text{s} \] **Explanation:** We calculated the cross-sectional area of water at the first point and multiplied by the velocity to get the flow rate. --- # Step 2: Use Continuity to Solve for Depth at the Second Point The flow rate must be the same at the second point: - Width \( w_2 = 17.\, \text{m} \) - Depth \( d_2 = ? \) - Velocity \( v_2 = 10.2\, \text{cm/s} = .102\, \text{m/s} \) Let \( Q = \text{Area}_2 \times v_2 \): \[ \text{Area}_2 = w_2 \times d_2 \] \[ Q = w_2 \times d_2 \times v_2 \] \[ 1.55916 = 17. \times d_2 \times .102 \] \[ 1.55916 = 1.734 \times d_2 \] \[ d_2 = \frac{1.55916}{1.734} \approx .899\, \text{m} \] **Explanation:** We set the flow rates equal using the continuity equation, then solved for the unknown depth at the second point. --- # Final Answer **The water is approximately \( \boxed{.90\, \text{m}} \) deep at the second point.**