The second image shows how I worked it out. I'm suspecting I miscalculated on very last step. Please help me see where I'm missing this one. Thanks A fountain sends a stream of water straight up into the air to a maximum height of 4.21 m. The effective area of the pipe feeding the fountain is 4.07 x 10^-4 m². Neglecting air resistance and any viscous effects, determine how many gallons per minute are being used by the fountain. (1 gal = 3.79 x 10^-3 m³)Number: [Text box for input]Units: [Dropdown menu for selection] **Given Data:**- The maximum height of the water is: \( H = 4.21 \, \text{m} \)- The cross-sectional area of the pipe is: \( A = 4.07 \times 10^{-4} \, \text{m}^2 \)The expression for the velocity to attain the maximum height is given by:\( V = \sqrt{2gh} \)Substitute the value in the above equation:\( \sqrt{2 \times 9.81 \, \text{m/s}^2 \times 4.21 \, \text{m}} = 9.088 \, \text{m/s} \)**Step 2:**The expression for the flow rate of the water is: \( Q = A \times V \)Substitute the value in the above equation:\( Q = 4.07 \times 10^{-4} \, \text{m}^2 \times 9.088 \, \text{m/s} \)\( Q = 36.99 \times 10^{-4} \, \text{m}^3/\text{s} \times \frac{60 \, \text{s}}{1 \, \text{min}} \times \frac{1 \, \text{gal}}{3.79 \times 10^{-3} \, \text{m}^3} \)\( Q = 585.59 \times 10^1 \, \text{gal/min} \)

Name: The second image shows how I worked it out. I'm suspecting I miscalcul
Uploaded: 2024-03-20T06:51:10.000Z
Channel: Bartleby
Description: The second image shows how I worked it out. I'm suspecting I miscalculated on very last step. Please help me see where I'm missing this one. Thanks A fountain sends a stream of water straight up into the air to a maximum height of 4.21 m. The effecti