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According to Archimedes' principle, the mass of a floating object equals the mass of the fluid displaced by the object. A 150-lbm swimmer is floating in a nearby pool; 95% of his or her body's volume is in the water while 5% of his or her body's volume is above water. Determine the density of the swimmer's body. The density of water is 0.036lbm/in^3.

Respuesta :

Answer:

Density of the swimmer = [tex]0.0342\ lbm/in^3[/tex].

Explanation:

Assuming,

  • [tex]\rho[/tex] = density of the swimmer.
  • [tex]\rho_w[/tex] = density of the water.
  • [tex]m[/tex] = mass of the swimmer.
  • [tex]m_w[/tex] = mass of the water displaced by the swimmer.
  • [tex]V_w[/tex] = volume of the displaced water.
  • [tex]V[/tex] = volume of the swimmer.

Given:

  • [tex]m=150\ lbm.[/tex]
  • [tex]\rho_w = 0.036\ lbm/in^3.[/tex]

The density of an object is defined as the mass of the object per unit volume.

Therefore,

[tex]\rho =\dfrac{m}{V}\ \Rightarrow m = \rho V\ \ .........\ (1).[/tex]

Since only 95% of the body of the swimmer is inside the water, therefore,

[tex]V_w = 95\%\ \text{of}\ V=\dfrac{95}{100}\times V = 0.95V.[/tex]

According to Archimedes' principle,

[tex]m=m_w\\[/tex]

Using (1),

[tex]\rho V=\rho_w V_w\\\rho V = 0.036\ lbm/in^3\times 0.95 V\\\rho=0.036\times 0.95\ lbm/in^3=0.0342\ lbm/in^3.[/tex]

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