91影视

To find the volume of the cylindrical container, use the formula: \(V_\text{container} = 蟺r_\text{container}^2h_\text{container}\) 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: \(V_\text{container} = 蟺(22\,\text{cm})^2(52\,\text{cm}) = 79289.0\,\text{cm}^3\)

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: \(A_\text{faucet} = 蟺r_\text{faucet}^2\) 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: \(A_\text{faucet} = 蟺(1.27\,\text{cm})^2 = 5.0701\,\text{cm}^2\)

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: \(P_\text{faucet} + \frac{1}{2}蟻v_\text{faucet}^2 = 蟻gh_\text{tower}\) Where \(P_\text{faucet}\) is the pressure at the faucet (equal to atmospheric pressure, so it can be ignored), \(蟻\) 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.