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(III) Estimate the density of the water 5.4 $\mathrm{km}$ deep in the sea. By what fraction does it differ from the density at the surface?

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$$1.05 \times 10^{3} \mathrm{kg} / \mathrm{m}^{3}$$

Physics 101 Mechanics

Chapter 13

Fluids

Fluid Mechanics

Rutgers, The State University of New Jersey

Simon Fraser University

University of Sheffield

University of Winnipeg

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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So we want to find the density of seawater given, given we have a depth of 5.4 kilometers. So we can say that the change in volume divided by the original volume would be equal to the negative change in pressure divided by BEA, the book module ISS s. So we can say that the change in pressure would be equal to the density times to the gravity, the acceleration due to gravity, rather times a change in height. But in this case, the depth. And we can say that the mass is gonna be equal to the density times the volume or we considered the original density times the original volume. This is simply saying that the mass is going to the massive any ah, unit, any unit, massive water is going to be the same whether you're at the surface of the sea or again 5.4 kilometers beneath the surface. Uh, all this is saying is that the mass is going to be constant. So using this relationship, we can say that the density at that at that particular depth is going to be equal to the density at the surface times the volume of the surface divided by the volume at the surface, plus the change in volume as you descended down 5.4 kilometers. Ah, this is gonna be equal to the original the density of the surface times the volume of the surface divided by the volume of the surface, plus the negative the O rather plus negative volume of the surface times the change in pressure divided by the book module ISS. So at this point, this is going to be equal to the density at the surface divided by one minus the density of the surface times the acceleration due to gravity times h the depth divided by BEA, the book module ISS and at this point, we can actually solve so p. The density, at 5.4 kilometers will be equal to 1000 25 kilograms per cubic meter. This would be the density at the surface and then it would be divided by one minus 10 25 kilograms per cubic meter times 9.8 meters per second squared times 5400 clomp of 55.4 kilometers or 5.4 times 10 to the third meters and then this is all going to be divided by the book module ISS of 2.0 times 10 to the ninth Newtons per meter squared. And so the density is gonna be equal to 1054 kilograms per cubic meter. And again, this would be at 54 5.4 kilometers or 5400 meters. Ah, we want to find the ratio. So the density divided by the density at the surface would be equal to 10 54 divided by 10. 25 and this is equaling 1.28 Ah, Therefore the ah density, it's going to be 2.8% more dense. The density of water will increase by 2.8% at a depth equaling 5400 meters. So this would be here second part of the answer. That is the end of the solution. Thank you for watching

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