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At a point where an irrigation canal having a rectangular cross section is 18.5 $\mathrm{m}$ wide and 3.75 $\mathrm{m}$ deep, the water flows at 2.50 $\mathrm{cm} / \mathrm{s} .$ At a point downstream, but on the same level, the canal is 16.5 $\mathrm{m}$ wide, but the water flows at 11.0 $\mathrm{cm} / \mathrm{s} .$ Howdeep is the canal at this point?

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Physics 101 Mechanics

Chapter 13

Fluid Mechanics

Temperature and Heat

Kim J.

September 27, 2020

Rutgers, The State University of New Jersey

University of Michigan - Ann Arbor

Hope College

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.

02:51

An irrigation canal has a …

01:54

03:05

Irrigation canal You live …

02:08

Okay, So the crux of this problem is recognizing that the flow of water is conserved throughout the canal, and so of this's captured by the Continuity equation, which says that the cross sectional area times the velocity of the water at one point is equal to the cross sectional area times the velocity at another point. So basically, at the area decreases, the water has to flow faster. And this is to prevent, um But like if this equation wasn't satisfied than water would be bunching up at one point or spreading out another point, or is we want the flow to be smooth and continuous. So in the first part of the canal were given that the dimensions or 18.5 and 3.75 So this wolf this combined factors served services are a one, and this serves is our V one and um 16.5 times the unknown that it would be our eight to and the 11 centimeters per second would be R V two. So since we want to solve this equation for a H, it's simply going to be 18.5 times 3.75 times 2.5 over 16.5 times 11. Now it might seem inconsistent that we're using centimeters for the velocity and meters for the dimensions of the canal. But because we divide the one by V two, the's length units cancel out, and we don't actually have to worry about doing a conversion.

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