To solve the problem, we'll apply the **Bernoulli's equation**, which is a principle of fluid dynamics that relates the pressure and velocity at two points in a streamline. ### Given Data: - Pressure difference \( P_1 - P_2 = 1.4 \) cm of mercury - Speed at the point of greater cross-section \( v_1 = 60 \) cm/s - We need to find the speed \( v_2 \) at the other point. ### Bernoulli's Equation for Horizontal Flow: For horizontal flow, Bernoulli's equation is: \[ P_1 + \frac{1}{2} \rho v_1^2 = P_2 + \frac{1}{2} \rho v_2^2 \] Where: - \( P_1 \) and \( P_2 \) are the pressures at points 1 and 2. - \( v_1 \) and \( v_2 \) are the velocities at points 1 and 2. - \( \rho \) is the density of water. ### Step 1: Convert Pressure Difference to Pascals The pressure difference is given in cm of mercury, which we need to convert to pascals. The conversion factor is: \[ 1 \text{ cm of mercury} = 1333.22 \text{ Pa} \] So, \[ P_1 - P_2 = 1.4 \times 1333.22 = 1866.51 \text{ Pa} \] ### Step 2: Simplify Bernoulli's Equation Substitute the pressure difference into the Bernoulli equation: \[ \frac{1}{2} \rho v_2^2 - \frac{1}{2} \rho v_1^2 = P_1 - P_2 \] Factor out \( \frac{1}{2} \rho \): \[ \frac{1}{2} \rho (v_2^2 - v_1^2) = 1866.51 \text{ Pa} \] ### Step 3: Solve for \( v_2^2 \) \[ v_2^2 = v_1^2 + \frac{2 \times 1866.51}{\rho} \] For water, \( \rho = 1000 \) kg/m\(^3\), so: \[ v_2^2 = (60 \text{ cm/s})^2 + \frac{2 \times 1866.51}{1000} \] Convert 60 cm/s to m/s: \[ v_1 = 0.6 \text{ m/s} \] So: \[ v_2^2 = (0.6)^2 + \frac{2 \times 1866.51}{1000} \] \[ v_2^2 = 0.36 + 3.733 = 4.093 \text{ m}^2/\text{s}^2 \] ### Step 4: Calculate \( v_2 \) \[ v_2 = \sqrt{4.093} = 2.02 \text{ m/s} \] Convert back to cm/s: \[ v_2 = 202 \text{ cm/s} \] ### Final Answer: The speed of the flow of water at the other power is 2.02ms
