**Problem Statement:**Water is flowing through a pipe that has an area of \(0.5 \, m^2\) with a velocity of 1 m/s. The pipe opens up to an area of \(1 \, m^2\). What is the new velocity of the water?**Solution Explanation:**We can use the principle of conservation of mass for incompressible fluids, often referred to as the equation of continuity. This principle states that the mass flow rate must remain constant in a closed system. For fluids, this is given by:\[ A_1 \times v_1 = A_2 \times v_2 \]Where:- \(A_1 = 0.5 \, m^2\) is the initial cross-sectional area of the pipe.- \(v_1 = 1 \, m/s\) is the initial velocity of the fluid.- \(A_2 = 1 \, m^2\) is the final cross-sectional area of the pipe.- \(v_2\) is the final velocity of the fluid, which we need to find.Rearranging the equation to solve for \(v_2\), we get:\[ v_2 = \frac{A_1 \times v_1}{A_2} \]Substituting in the known values:\[ v_2 = \frac{0.5 \times 1}{1} \]\[ v_2 = 0.5 \, m/s \]Therefore, the new velocity of the water is \(0.5 \, m/s\).

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Answered: Water is flowing through a pipe that…

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

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Water is flowing through a pipe that has an area of 0.5m² with a velocity of 1 m/s. The pipe opens up to an area of 1m². What is the new velocity of the water?