A water tower supplies water through the plumbing in a house. A 2.54 -cm- diameter faucet in the house can fill a cylindrical container with a diameter of \(44 \mathrm{cm}\) and a height of \(52 \mathrm{cm}\) in 12 s. How high above the faucet is the top of the water in the tower? (Assume that the diameter of the tower is so large compared to that of the faucet that the water at the top of the tower does not move. \()\)

To find the volume of the cylindrical container, use the formula: \(V_\text{container} = πr_\text{container}^2h_\text{container}\) Where \(V_\text{container}\) is the volume, \(r_\text{container}\) is the radius, and \(h_\text{container}\) is the height of the container. First, calculate \(r_\text{container}\) by dividing the diameter by 2: \(r_\text{container} = \frac{44\,\text{cm}}{2} = 22\,\text{cm}\) Then, calculate the volume of the container: \(V_\text{container} = π(22\,\text{cm})^2(52\,\text{cm}) = 79289.0\,\text{cm}^3\)

Flow rate is given by the volume of water per unit time. Since it takes \(12\,\text{s}\) to fill the container, the flow rate (Q) is: \(Q = \frac{V_\text{container}}{t}\) Where \(t\) is the time taken to fill the container. \(Q = \frac{79289.0\,\text{cm}^3}{12\,\text{s}} = 6607.4\,\text{cm}^3/\text{s}\)

Use the formula for the area of a circle to find the cross-sectional area of the faucet: \(A_\text{faucet} = πr_\text{faucet}^2\) Where \(A_\text{faucet}\) is the cross-sectional area and \(r_\text{faucet}\) is the radius of the faucet. First, calculate \(r_\text{faucet}\) by converting to centimeters and dividing the diameter by 2: \(r_\text{faucet} = \frac{2.54\,\text{cm}}{2} = 1.27\,\text{cm}\) Then, calculate the cross-sectional area of the faucet: \(A_\text{faucet} = π(1.27\,\text{cm})^2 = 5.0701\,\text{cm}^2\)

Use the formula for the velocity of the water flow: \(v_\text{faucet} = \frac{Q}{A_\text{faucet}}\) \(v_\text{faucet} = \frac{6607.4\,\text{cm}^3/\text{s}}{5.0701\,\text{cm}^2} = 1302.3\,\text{cm/s}\)

Bernoulli's equation for this problem can be written as: \(P_\text{faucet} + \frac{1}{2}ρv_\text{faucet}^2 = ρgh_\text{tower}\) Where \(P_\text{faucet}\) is the pressure at the faucet (equal to atmospheric pressure, so it can be ignored), \(ρ\) is the density of water (assume to be \(1000\,\text{kg/m}^3\) or \(1\,\text{g/cm}^3\)), \(v_\text{faucet}\) is the velocity of the water, \(g\) is the acceleration due to gravity (\(981\,\text{cm/s}^2\)), and \(h_\text{tower}\) is the height of the water above the faucet. Rearrange the equation to solve for \(h_\text{tower}\): \(h_\text{tower} = \frac{v_\text{faucet}^2}{2g}\) Convert the velocity of the faucet to meters per second and plug in the values: \(h_\text{tower} = \frac{(1302.3\,\text{cm/s})^2}{2(981\,\text{cm/s}^2)} = 86.4\,\text{cm}\) The height of the water in the tower above the faucet is approximately 86.4 cm.