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MrRa [10]
3 years ago
12

To determine an athlete's body fat, she is weighed first in air and then again while she's completely underwater. it is found th

at she weighs 690 n when weighed in air and 48.0 n when weighed underwater. what is her average density?

Physics
2 answers:
Makovka662 [10]3 years ago
6 0

Her average density is about 1070 kg/m³

\texttt{ }

<h3>Further explanation</h3>

Let's recall Buoyant Force formula as follows:

\boxed{ F = \rho g V}

<em>where:</em>

<em>F = buoyant force ( N )</em>

<em>ρ = density of  fluid ( kg/m³ )</em>

<em>g = gravitational acceleration ( m/s² )</em>

<em>V = displaced body volume of liquid ( m³ )</em>

Let us now tackle the problem!

\texttt{ }

<u>Given:</u>

weight in air = w = 690 N

weight in underwater = w' = 48.0 N

density of water = ρ_w = 1000 kg/m³

<u>Asked:</u>

average density = ρ = ?

<u>Solution:</u>

<em>Firstly, we will calculate the volume of the athlete's body:</em>

w' = w - F

w' = w - \rho_w g V

\rho_w g V = w - w'

\boxed {V = ( w - w' ) \div ( \rho_w g )} <em>→ Equation 1</em>

\texttt{ }

<em>Next, we could calculate the average density of the athlete's body:</em>

\rho = m \div V

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

\rho = ( \rho_w m g) \div ( w - w' )

\rho = ( \rho_w w ) \div ( w - w' )

\rho = ( 1000 \times 690 ) \div ( 690 - 48.0 )

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

\texttt{ }

<h3>Learn more</h3>
  • Buoyant Force : brainly.com/question/13922022
  • Kinetic Energy : brainly.com/question/692781
  • Volume of Gas : brainly.com/question/12893622
  • Impulse : brainly.com/question/12855855
  • Gravity : brainly.com/question/1724648

\texttt{ }

<h3>Answer details</h3>

Grade: High School

Subject: Physics

Chapter: Pressure

il63 [147K]3 years ago
5 0

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).<span>

The amount of weight displaced or the amount of buoyant force is: </span>

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

<span>average density = 1.07 kg/L = 1.07 g/mL</span>

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