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(a) What is the density of a woman who floats in freshwater with 4.00% of her volume above the surface? This could be measured by placing her in a tank with marks on the side to measure how much water she displaces when floating and when held under water (briefly). (b) What percent of her volume is above the surface when she floats in seawater?

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College Physics

1st Edition
Paul Peter Urone + 1 other
Publisher: OpenStax College
ISBN: 9781938168000

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BuyFindarrow_forward

College Physics

1st Edition
Paul Peter Urone + 1 other
Publisher: OpenStax College
ISBN: 9781938168000
Chapter 11, Problem 46PE
Textbook Problem
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(a) What is the density of a woman who floats in freshwater with 4.00% of her volume above the surface? This could be measured by placing her in a tank with marks on the side to measure how much water she displaces when floating and when held under water (briefly). (b) What percent of her volume is above the surface when she floats in seawater?

(a)

To determine

The density of the body of women

Explanation of Solution

Given:

Density of water dw=1gmcm3Percentage of volume of the body of women above the surface of water =4%

Formula used:

In floating condition,

Weight of body = Weight of displaced water

Calculation:

Let, percentage of total volume of the body of a women (VS) =100%

So, the percentage of volume of the body below the surface of water (VW) =1004

  =96 %

Volume of the body below the surface of water = Volume of water displaced = Vw

According to Archimedes' principle,

Weight of body = Weight of displaced water

  VSd

(b)

To determine

Percentage of volume of the body of women above the surface of sea water

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Chapter 11 Solutions

College Physics
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