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$\bullet$ In the Challenger Deep of the Marianas Trench, the depth of seawater is 10.9 $\mathrm{km}$ and the pressure is $1.16 \times 10^{8}$ Pa (about 1150 atmospheres). (a) If a cubic meter of water is taken to this depth from the surface (where the normal atmospheric pressure is about $1.0 \times 10^{3} \mathrm{Pa}$ , what is the change in its volume? Assume that the bulk modulus for seawater is the same as for freshwater $\left(2.2 \times 10^{9} \mathrm{Pa}\right) .$ (b) At the surface, seawater has a density of $1.03 \times 10^{3} \mathrm{kg} / \mathrm{m}^{3} .$ What is the density of sea-water at the depth of the Challenger Deep?

$1.0827 \times 10^{3} \frac{k g}{m^{3}}$

Physics 101 Mechanics

Chapter 11

Elasticity and Periodic Motion

Equilibrium and Elasticity

Periodic Motion

University of Sheffield

University of Winnipeg

McMaster University

Lectures

04:12

In physics, potential ener…

02:18

In physics, an oscillation…

02:01

In the Challenger Deep of …

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01:59

The deepest point in the o…

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07:05

03:32

The deepest point known in…

02:12

The deepest place in the o…

05:26

Calculate the pressure due…

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00:59

The greatest ocean depths …

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02:50

(a) Calculate the absolute…

02:06

At a depth of $10.9 \mathr…

05:18

Submarines change their de…

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The Mariana trench is loca…

02:34

At the surface of the ocea…

00:57

06:04

02:02

(a) For seawater of densit…

02:09

I guess let's take a look at the question 1, so the question 1 was asking us to find out the changing volume. So we know for the both majus, which is equal to the ratio of the changing pressure to bottom string, and this will give us b equal to minus delta p over delta v over v 0 point delta. P is changing pressure, which is 1 minus p. 2.1. Is the pressure when the sea water reaches 10.2 kilometers here and has already 10.9 kilometers and p 2 is the normal pressure for the sea water, which is 1 times 10 to the power 3? Pascal, if we calculate delta p, which is a changing pressure, we'll figure, we will get something that is 1.1610 to power, a pascal minus 1.1 times 10 to the power 3 pascal. This will give us 1.15999 times 10 to the power of 8. Pascal okay, so remember we're looking able to figure out the changing pressure and changing volume, which is will give us minus delta p times v 0 over both marsus and the v 0 is a cub meter of water, which is 1 meter cube and the b. The book measure with book measures was given as 2.2 times 10 to the power of 9 pascal okay. So now we can figure out the change in volume, which is minus 1.159 times 10 to the power of 8 pascal equals 1.1599 times 1 meter cube over 2.2 times 10 to the power of 9 pascal. This will give us minus 0.0527 meter cube and that's the value, the changing volume. Okay, so lets take the question b. The question b was asking us: what's the density of the water and the depth of the challenge? Deep, while that's a real prime, we don't know, but we know real prime, which is the density at challenge. Deep is equal to rho, plus delta ropo. Rho here is the initial density of the sea water, which is given as 1.03 times 10 to the power 3 kilogram per meter cube delta, always the changing temperature, i'm sorry a changing density, okay, and we know that when the changing volume decrease because from the previous Question we get a negative value changing bond so when the changing volume is decrease, that means the density will increase okay. So then, this will give us delta delta rho, which is simply equal to 0.0527 times 10 to the power 3 kilogram per meter cube okay, since the change in wording was negative, it means the changing density is positive. Okay, so as to some algebra here, which is a prime, is simply equal to 1.03 times 10 to the power 3 kilogram per meter cube plus 0.0527 times 10 to the power 3 and 10 to the power 3 kilogram per meter cube. This will give us 1.0827 times 10 to the power of 3 kilogram per meter cube. So this is the density of the sea water at a challenger, deep level. Okay- and thank you-

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