Book edition 10th Edition

Author(s) David Halliday

Pages 1328 pages

ISBN 9781118230718

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Short Answer

About one-third of the body of a person floating in the Dead Sea will be above the waterline. Assuming that the human body density $0.98 g / c m^{3}$ is find the density of the water in the Dead Sea. (Why is it so much greater than $1.0 g / c m^{3} ?$

The density of Dead Sea water is $1.5 g / {cm}^{3} .$

The water of the dead is way saltier than the normal water. Therefore, the density of the water is high.

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Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Listing the given quantities.

The fraction of the body of a person floating in the Dead Sea is $\frac{1}{3} .$

The density of the human body is, $ρ_{p} = 0.98 g / {cm}^{3} .$

Understanding the Archimedes principle.

By using the fraction of the body, which is inside the water, we can find out the ratio of the volume of the seawater displaced by the submerged body $(V_{s})$ to the total volume of the person’s body $(V_{ρ}) .$ Then using Archimedes’ principle, we can find $ρ_{s}$ .

Formula:

$Fraction = 1 - \frac{Vs}{Vp}$

$F_{b} = M_{s} g$

$= ρ_{s} V_{s} g$

$F_{g} = M_{p} g$

$= ρ_{p} V_{p} g$

(a) Calculation of density of Dead Sea water.

The fraction can be written as

$Frac = \frac{V_{p} - Vs}{Vp}$

$Frac = 1 - \frac{Vs}{Vp}$

$\frac{1}{3} = 1 - \frac{V s}{V p}$

$\frac{V s}{V p} = 1 - \frac{1}{3}$

$\frac{V s}{V p} = \frac{2}{3}$

role="math" localid="1657552894793" $V s = \frac{2}{3} V p (1)$

Weight of the sea water displaced by the $\frac{1}{3}$ body.

$W_{s} = M_{s} g$

$W_{s} = ρ_{s} V_{s} g (2)$

Buoyant force acting on the 1/3 body of a person is,

$F_{b} = M_{p} g$

role="math" localid="1657553033464" $F_{b} = ρ_{p} V_{p} g (3)$

According to Archimedes principle, we can equate equation $(2) and (3)$

$ρ_{s} V_{s} g = ρ_{p} V_{P} g$

$ρ_{s} = ρ_{p} \frac{V p}{V s}$

Using equation $(1),$

$ρ_{s} = ρ_{p} \frac{V p}{2_{V p_{3}}}$

$= ρ_{p} \frac{3}{2}$

$= 0.98 g / {cm}^{3} x \frac{3}{2}$

$= 1.47 g / {cm}^{3}$

Therefore, the density of Dead Sea water is $1.5 g / {cm}^{3} .$

The density of the dead sea water is much higher than the normal water. The dead sea is located in the region where temperature is quite high and rainfall is quite low. Therefore, the rate of evaporation in the region is quite high. As a result, the water of the dead is way saltier than the normal water. Therefore, the density of the water is high.

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Fundamentals Of Physics

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Short Answer

About one-third of the body of a person floating in the Dead Sea will be above the waterline. Assuming that the human body density $0.98 g / c m^{3}$ is find the density of the water in the Dead Sea. (Why is it so much greater than $1.0 g / c m^{3} ?$

Step by Step Solution

Listing the given quantities.

Understanding the Archimedes principle.

(a) Calculation of density of Dead Sea water.

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Fundamentals Of Physics

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Short Answer

About one-third of the body of a person floating in the Dead Sea will be above the waterline. Assuming that the human body density0.98 g /cm3 is find the density of the water in the Dead Sea. (Why is it so much greater than 1.0 g/cm3?

Step by Step Solution

Listing the given quantities.

Understanding the Archimedes principle.

(a) Calculation of density of Dead Sea water.

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About one-third of the body of a person floating in the Dead Sea will be above the waterline. Assuming that the human body density $0.98 g / c m^{3}$ is find the density of the water in the Dead Sea. (Why is it so much greater than $1.0 g / c m^{3} ?$