A nozzle of inner radius \(1.00 \mathrm{mm}\) is connected to a hose of inner radius \(8.00 \mathrm{mm} .\) The nozzle shoots out water moving at $25.0 \mathrm{m} / \mathrm{s} .$ (a) At what speed is the water in the hose moving? (b) What is the volume flow rate? (c) What is the mass flow rate?

First, we'll use the conservation of mass principle to determine the speed of the water in the hose. This principle can be expressed as: \(A_1v_1 = A_2v_2\), where \(A_1\) and \(A_2\) are the cross-sectional areas of the hose and nozzle respectively, and \(v_1\) and \(v_2\) are the water velocities in the hose and nozzle respectively. We know the radii and the velocity of the water coming out of the nozzle, so we can solve for \(v_1\). The cross-sectional area of a circle is given by \(A = \pi r^2\). So, the equation becomes: \(\pi r_1^2v_1 = \pi r_2^2v_2\). Since we have all values except \(v_1\), we can now solve for it: \(v_1 = \frac{r_2^2v_2}{r_1^2}\) Plugging in the values: \(v_1 = \frac{(1\,\text{mm})^2(25.0\,\text{m/s})}{(8\,\text{mm})^2}\) \(v_1 = \frac{25\,\text{m/s}}{64} = 0.390625\,\text{m/s}\) So, the speed of the water in the hose is \(0.390625\,\text{m/s}\).

To determine the volume flow rate (Q), we can use the formula: \(Q = Av\), where A is the cross-sectional area and v is the velocity of the fluid. We'll use the area and velocity of the hose to calculate the volume flow rate: \(Q = A_1v_1 = \pi r_1^2v_1\) Plugging in the values: \(Q = \pi (8\,\text{mm})^2 (0.390625\,\text{m/s})\) \(Q = \pi (64\,\text{mm}^2) (0.390625\,\text{m/s})\) \(Q = \pi (64 \times 10^{-6}\,\text{m}^2) (0.390625\,\text{m/s})\) \(Q = 7.854 \times 10^{-5}\,\text{m}^3/\text{s}\) Therefore, the volume flow rate is \(7.854 \times 10^{-5}\,\text{m}^3/\text{s}\).

To find the mass flow rate, we'll use the formula: \(\dot{m} = \rho Q\), where \(\rho\) is the density of water and we have already calculated Q in part (b). The density of water at room temperature is roughly \(1000\,\text{kg/m}^3\). So, we can simply multiply the volume flow rate by the density: \(\dot{m} = (1000\,\text{kg/m}^3)(7.854 \times 10^{-5}\,\text{m}^3/\text{s})\) \(\dot{m} = 0.07854\,\text{kg/s}\) Thus, the mass flow rate is \(0.07854\,\text{kg/s}\).