/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Q37P (II) On dry land, an athlete wei... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 10: Q37P (page 260)

(II) On dry land, an athlete weighs\({\rm{70}}{\rm{.2 kg}}\). The same athlete, when submerged in a swimming pool and hanging from a scale, has an 鈥渁pparent weight鈥� of \({\rm{3}}{\rm{.4 kg}}\). Using Example 10-8 as a guide, (a) find the total volume V of the submerged athlete. (b) Assume that when submerged, the athlete鈥檚 body contains a residual volume \({V_{\rm{R}}} = 1.3 \times {10^{ - 3}}{\rm{ }}{{\rm{m}}^3}\)of air (mainly in the lungs). Taking \(V - {V_{\rm{R}}}\)to be the actual volume of the athlete鈥檚 body, find the body鈥檚 specific gravity, \(SG\). (c) What is the athlete鈥檚 percent body fat assuming it is given by the formula \(\left( {495/SG} \right) - 450\)?

Short Answer

Expert verified

(a) Thetotal volume of the submerged athlete is \(6.68 \times {10^{ - 2}}\;{{\rm{m}}^3}\).

(b) Thespecific gravity of athlete鈥檚 body, when submerged is \(1.07\).

(c) The athlete鈥檚 percent body fat is \(12\% \).

Step by step solution

01

Concept of law of equilibrium

According to the law of equilibrium, the forces experienced by the object aligned in a direction must be balanced. That suggests the net force on the object must be zero.

02

Given data

The actual mass of the athleteis \({{\rm{m}}_{{\rm{actual}}}} = 70.2\;{\rm{kg}}\).

The apparent mass of the athlete is \({{\rm{m}}_{{\rm{apparant}}}} = 3.4\;{\rm{kg}}\).

The residual volume of air (mainly in the lungs) is\({V_{\rm{R}}} = 1.3 \times {10^{ - 3}}{\rm{ }}{{\rm{m}}^3}\).

The density of water is \({\rho _{{\rm{water}}}} = 1000\;{\rm{kg/}}{{\rm{m}}^3}\).

03

Calculation of total volume of submerged athlete

A free body diagram of the forces acting on athlete is shown below:

Using law of equilibrium of forces, total volume of the submerged athlete is:

\(\begin{aligned}{F_{{\rm{bouyant}}}} + {F_{{\rm{scale}}}} - {F_{{\rm{weight}}}} &= 0\\{\rho _{{\rm{water}}}}Vg + {m_{{\rm{apparant}}}}g - {m_{{\rm{actual}}}}g &= 0\\V = \frac{{{m_{{\rm{actual}}}} - {m_{{\rm{apparant}}}}}}{{{\rho _{{\rm{water}}}}}}\end{aligned}\)

Here, \({F_{{\rm{bouyant}}}}\)is the buoyant force, \({F_{{\rm{scale}}}}\)is the scale force, \({F_{{\rm{weight}}}}\)is the force applied by weight and \(g\) is the acceleration due to gravity.

Substitute the values in the above equation to find volume of submerged athlete.

\(\begin{aligned}V &= \frac{{70.2\;{\rm{kg}} - 3.4\;{\rm{kg}}}}{{1000\;{\rm{kg/}}{{\rm{m}}^3}}}\\ &= 6.68 \times {10^{ - 2}}\;{{\rm{m}}^3}\end{aligned}\)

04

Calculation of specific gravity of athlete’s body

\(\begin{aligned}SG &= \frac{{{\rho _{{\rm{athlete}}}}}}{{{\rho _{\rm{w}}}}}\\ &= \frac{{\frac{{{{\rm{m}}_{{\rm{actual}}}}}}{{V - {V_{\rm{R}}}}}}}{{{\rho _{\rm{w}}}}}\end{aligned}\)

Substitute the known values in the above equation to find the specific gravity of athlete鈥檚 body.

\(\begin{aligned}SG &= \frac{{\frac{{70.2\;{\rm{kg}}}}{{\left( {6.68 \times {{10}^{ - 2}}\;{{\rm{m}}^3} - 1.3 \times {{10}^{ - 3}}\;{{\rm{m}}^3}} \right)}}}}{{1000\;{\rm{kg/}}{{\rm{m}}^3}}}\\ &= 1.07\end{aligned}\)

Hence, specific gravity of athlete鈥檚 body is\(1.07\).

05

Calculation of percent athlete’s body fat

The percent athlete鈥檚 body fat is calculated as:

\(\% {\rm{Body fat}} = \frac{{495}}{{SG}} - 450\)

Substitute the values in the above equation to find athlete鈥檚 body fat.

\(\begin{aligned}\% {\rm{Body fat}} &= \frac{{495}}{{SG}} - 450\\ &= \frac{{495}}{{1.07}} - 450\\ &= 12.62\% \\ &\simeq 12\% \end{aligned}\)

Hence, the athlete鈥檚 body fat percentage is \(12\% \).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Question: You are watering your lawn with a hose when you put your finger over the hose opening to increase the distance the water reaches. If you are holding the hose horizontally, and the distance the water reaches increases by a factor of 4, what fraction of the hose opening did you block?

(a) Show that the flow speed measured by a venturi meter (see Fig. 10-29) is given by the relation

\({{\bf{v}}_{\bf{1}}}{\bf{ = }}{{\bf{A}}_{\bf{2}}}\sqrt {\frac{{{\bf{2}}\left( {{{\bf{P}}_{\bf{1}}}{\bf{ - }}{{\bf{P}}_{\bf{2}}}} \right)}}{{{\bf{\rho }}\left( {{\bf{A}}_{\bf{1}}^{\bf{2}}{\bf{ - A}}_{\bf{2}}^{\bf{2}}} \right)}}} \).

(b) A venturi meter is measuring the flow of water; it has a main diameter of \({\bf{3}}{\bf{.5\;cm}}\) tapering down to a throat diameter of \({\bf{1}}{\bf{.0\;cm}}\). If the pressure difference is measured to be \({\bf{18\;mm - Hg}}\), what is the speed of the water entering the venturi throat?

(II) Calculate the pressure drop per cm along the aorta using the data of Example 10鈥�12 and Table 10鈥�3.

You need to siphon water from a clogged sink. The sink has an area of and is filled to a height of 4.0 cm. Your siphon tube rises 45 cm above the bottom of the sink and then descends 85 cm to a pail as shown in Fig. 10鈥�59. The siphon tube has a diameter of 2.3 cm. (a) Assuming that the water level in the sink has almost zero velocity, use Bernoulli鈥檚 equation to estimate the water velocity when it enters the pail. (b) Estimate how long it will take to empty the sink. Ignore viscosity.

A fire hose exerts a force on the person holding it. This is because the water accelerates as it goes from the hose through the nozzle. How much force is required to hold a \({\bf{7}}{\bf{.0}}\;{\bf{cm}}\)-diameter hose delivering \({\bf{420}}\;{\bf{L/min}}\)through a \({\bf{0}}{\bf{.75}}\;{\bf{cm}}\)-diameter nozzle?
See all solutions

Recommended explanations on Physics Textbooks

Space Physics

Read Explanation

Electric Charge Field and Potential

Read Explanation

Measurements

Read Explanation

Magnetism and Electromagnetic Induction

Read Explanation

Electrostatics

Read Explanation

Geometrical and Physical Optics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.