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What is the density, in kilograms per cubic meter, 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).
What percent of her volume is above the surface when she floats in seawater?

Respuesta :

davidlcastiblanco

Answer:

a) [tex]P_W=960kg/m^3[/tex]

b) %v=6.34

Explanation:

To this problem important information you can get in textbooks or internet is the density of freshwater and seawater

Freshwater

[tex]P_f =1x10^3 kg/m^3[/tex]

Seawater

[tex]P_f =1025 kg/m^3[/tex]

Now using the equation of fraction submerged that compared the density of both elements the woman and the freshwater

a).

[tex]F_s=\frac{P_W}{P_f}[/tex]

Solve to density of the woman

[tex]P_W=P_f*F_s[/tex]

[tex]F_s=1.0-0.04=0.96[/tex]

[tex]P_W=1x10^3 kg/m^3*0.96=960kg/m^3[/tex]

Now the percent in sea water is:

[tex]F_s=\frac{P_W}{P_s}=\frac{960 kg/m^3}{1025 kg/m^3}= 0.9366[/tex]

[tex]F_s=1.0-0.9366=0.0634*100[/tex]

%v= 6.34

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