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The horizontal pipe shown in Figure 13.47 has a cross-sectional area of 40.0 $\mathrm{cm}^{2}$ at the wider portions and 10.0 $\mathrm{cm}^{2}$ at the constriction. Water is flowing in the pipe, and the discharge from the pipe is $6.00 \times$$10^{-3} \mathrm{m}^{3} / \mathrm{s}(6.00 \mathrm{L} / \mathrm{s}) .$ Find(a) the flow speeds at the wide and the narrow portions; (b) the pressure difportions; (c) the difference difporcury columns in the U-shaped tube.

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c)1$\rightarrow$ represents the wide cross sectional area2$\rightarrow$ represents the narrow cross sectional areathe pressure at the point at the bottom of the length h is equal for both branches.$$p_{1}=p_{2}+\rho_{\text {mercury}} g h$$$$h=\frac{p_{1}-p_{2}}{\rho_{\text {mercury}} g}=\frac{1.688 \times 10^{4} P a}{\left(13.6 \times 10^{3} k g / m^{3}\right)\left(9.8 m / s^{2}\right)}=0.1267 m$$

Physics 101 Mechanics

Chapter 13

Fluid Mechanics

Temperature and Heat

University of Washington

Simon Fraser University

McMaster University

Lectures

03:45

In physics, a fluid is a substance that continually deforms (flows) under an applied shear stress. Fluids are a subset of the phases of matter and include liquids, gases, plasmas and, to some extent, plastic solids.

09:49

A fluid is a substance that continually deforms (flows) under an applied shear stress. Fluids are a subset of the phases of matter and include liquids, gases and plasmas. Fluids display properties such as flow, pressure, and tension, which can be described with a fluid model. For example, liquids form a surface which exerts a force on other objects in contact with it, and is the basis for the forces of capillarity and cohesion. Fluids are a continuum (or "continuous" in some sense) which means that they cannot be strictly separated into separate pieces. However, there are theoretical limits to the divisibility of fluids. Fluids are in contrast to solids, which are able to sustain a shear stress with no tendency to continue deforming.

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all right, so I'm part of this problem. We want to find the velocities of the water in the two different parts of the pipe can were given that the flow reyes six times sent the negative three meters Q per second. So part is just a matter of playing continuity. Ueno that a one times everyone has to be equal to the flow rate. And this is true for any cross sectional area in velocity. So the first portion of the pipe were given that the cross sectional area is 40 square centimeters or 40 times 10 to the negative for square meters, and this times everyone is equal to six times in the negative three. So to solve for the one, we simply divide both sides by 40 times in the negative for likewise in the other portion of the tube. The cross sectional area is now 10 square centimeters or tend to the negative four square meters, and this times the new velocity will be equal to the same flow rate. So to solve for me, too, we just divide both sides by 10 times 10 to the negative for for Part B were asked about the pressure difference between the two portions of the pipe. And so for this we want to apply for new lease equation. And because the two portions of the piper at the same height weaken drop thie ro gh term. So if we solve this equation for the pressure difference, what we get is that he won minus P too is equal to one have row um V two squared minus fee one squared where rose the density of water at 10 to the three kilograms per meter. Cute and V two is our solution for part A and B one also comes from her, eh? Lastly in part. See, we want to know what the height difference is of the mercury. And basically the pressure generated by the height difference in the mercury has to be equal to the pressure difference coming from the flows TV in equilibrium. So the P one minus p two in this equation comes from our answer to part B. We were given that the density of mercury is 13.6 times 10 to the three kilograms per meter cubed and so we simply solve this for for H, and that involves the division by gravitational acceleration which is 9.8 meters per second squared

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