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An open barge has the dimensions shown in Figure 13.44 . If the barge is made out of 4.0 -cm-thick steel plate on each of its four sides and its bottom, what mass of coal can the barge carry in fresh water without sinking? Is there enough room in the barge to hold this amount of coal? (The density of coal is about 1500 $\mathrm{kg} / \mathrm{m}^{3} . )$

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$V_{\text {coal}}=6547.456 \mathrm{m}^{3}$this volume is less than the volume of the barge 10560 $\mathrm{m}^{3}$ so, the conl can fit easily in the barge

Physics 101 Mechanics

Chapter 13

Fluid Mechanics

Temperature and Heat

Ana H.

September 30, 2020

Rutgers, The State University of New Jersey

University of Michigan - Ann Arbor

University of Washington

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.

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all right. In this problem, we're given a barge of various, uh, excuse me of given dimensions and the barge is made out of steel. We were given the density of steel 7.8 times tend to the three kilograms per meter cubed. And we want to know how much coal, um, the barge can carry before it sinks. So the first thing that we're going to need is the mass of the steel that the barge is made out of, and that's going to be given by the volume of steel times, the density of steel and the volume comes from just adding up all the panels of the boat. So we have two panels of area 22 by 12. We have two panels of area 40 by 12 and the bottom panel has area of 40 by 22. Then we need to multiply by the thickness of the steel 0.4 meters and lastly by its density. Now, if we want to find the condition for the maximum amount of coal that it can hold, we know that the barge will flow until its density becomes equal to the density of water and its density is going to be given by the mass of the steel it holds, plus the mass of the cold that it holds divided by X value. And we set that equal Teo ro water to find the maximum amount of cold. So and Cole is equal to row water times v barge minus I'm steel. Where am steel? We had calculated before, and the barge eyes simply the dimensions of it. 22 times, 12 times 40. And the density of water, of course, is tend to the three kilograms per meter cubed. Um, now we want to compare the this volume of cool to the volume of the barge. That's a simple calculation. V. Cole is equal to the mass of the coal divided by its density, which is given in the problem. And if you go through, the calculation correctly, should find that the coal is less than the barge, so it all fits nicely.

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