To determine an athlete's body fat, she is weighed first in air and then again while she's completely underwater. it is found that she weighs 690 n when weighed in air and 48.0 n when weighed underwater. what is her average density?

Respuesta :

The difference in weight is due to the displacement of water (the buoyancy of water is acting on the athlete thus giving her smaller weight).

The amount of weight displaced or the amount of buoyant force is:

Fb = 690 N - 48 N

Fb = 642 N

From newtons law, F = m*g. Using this formula, we can calculate for the mass of water displaced:

m of water displaced = 642N / 9.8m/s^2

m of water displaced = 65.5 kg

Assuming a water density of 1 kg/L, and using the formula volume = mass/density:

V of water displaced = 65.5kg / 1kg/L = 65.5 L

The volume of water displaced is equal to the volume of athlete. Therefore:

V of athlete = 65.5 L

The mass of athlete can also be calculated using, F = m*g

m of athlete = 690 N/ 9.8m/s^2

m of athlete = 70.41 kg

 

Knowing the volume and mass of athlete, her average density is therefore:

average density = 70.41 kg / 65.5 L

average density = 1.07 kg/L = 1.07 g/mL

Her average density is about 1070 kg/m³

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Further explanation

Let's recall Buoyant Force formula as follows:

[tex]\boxed{ F = \rho g V}[/tex]

where:

F = buoyant force ( N )

ρ = density of  fluid ( kg/m³ )

g = gravitational acceleration ( m/s² )

V = displaced body volume of liquid ( m³ )

Let us now tackle the problem!

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Given:

weight in air = w = 690 N

weight in underwater = w' = 48.0 N

density of water = ρ_w = 1000 kg/m³

Asked:

average density = ρ = ?

Solution:

Firstly, we will calculate the volume of the athlete's body:

[tex]w' = w - F[/tex]

[tex]w' = w - \rho_w g V[/tex]

[tex]\rho_w g V = w - w'[/tex]

[tex]\boxed {V = ( w - w' ) \div ( \rho_w g )}[/tex] → Equation 1

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Next, we could calculate the average density of the athlete's body:

[tex]\rho = m \div V[/tex]

[tex]\rho = m \div [ ( w - w' ) \div ( \rho_w g ) ][/tex] ← Equation 1

[tex]\rho = ( \rho_w m g) \div ( w - w' )[/tex]

[tex]\rho = ( \rho_w w ) \div ( w - w' )[/tex]

[tex]\rho = ( 1000 \times 690 ) \div ( 690 - 48.0 )[/tex]

[tex]\boxed {\rho \approx 1.07 \times 10^3 \texttt{ kg/m}^3}[/tex]

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Learn more

  • Buoyant Force : https://brainly.com/question/13922022
  • Kinetic Energy : https://brainly.com/question/692781
  • Volume of Gas : https://brainly.com/question/12893622
  • Impulse : https://brainly.com/question/12855855
  • Gravity : https://brainly.com/question/1724648

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Answer details

Grade: High School

Subject: Physics

Chapter: Pressure

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