Book edition 1st Edition

Author(s) Paul Peter Urone

Pages 1272 pages

ISBN 9781938168000

Answers without the blur.

Just sign up for free and you're in.

Short Answer

Kanellos Kanellopoulos flew 119 km from Crete to Santorini, Greece, on April 23, 1988, in the Daedalus 88, an aircraft powered by a bicycle-type drive mechanism (see Figure 7.42). His useful power output for the 234-min trip was about 350 W. Using the efficiency for cycling from Table 7.2, calculate the food energy in kilojoules he metabolized during the flight.

The energy used in metabolism during flight is \(5872.37{\rm{ kcal}}\).

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: Efficiency

The efficiency is the ratio of the total useful output by the total input. It is a measure of how efficiently a device uses energy.

Mathematically,

\(\eta \% = \frac{{{P_o}}}{{{P_i}}} \times 100\% \)

Here, \(\eta \) is the efficiency, \({P_o}\) is the useful power output, and \(P\) is the total power input.

Step 2: Food energy in kilojoules he metabolized during the flight

The total power input of the person can be calculated using equation (1.1).

Rearranging equation (1.1) in order to get an expression for the total power input.

\({P_i} = \frac{{{P_o}}}{{\eta \% }} \times 100\% \)

Here, \({P_o}\) is the useful power output \(\left( {{P_o} = 350{\rm{ W}}} \right)\), and \(\eta \) is the efficiency of cycling\(\left( {\eta = 20\% } \right)\).

Putting all known values,

\(\begin{aligned} {P_i} &= \frac{{\left( {350{\rm{ W}}} \right)}}{{20\% }} \times 100\% \\ &= 1750{\rm{ W}}\end{aligned}\)

The energy consumed is,

\(E = {P_i}t\)

Here, \({P_i}\) is the input power \(\left( {{P_i} = 1750{\rm{ W}}} \right)\), and t is the time of operation \(\left( {t = 234{\rm{ min}}} \right)\).

Putting all known values,

\(\begin{aligned} E &= \left( {1750{\rm{ W}}} \right) \times \left( {234{\rm{ min}}} \right)\\ &= \left( {1750{\rm{ W}}} \right) \times \left( {234{\rm{ min}}} \right) \times \left( {\frac{{60{\rm{ sec}}}}{{1{\rm{ min}}}}} \right)\\ &= 24570000{\rm{ J}} \times \left( {\frac{{1{\rm{ kcal}}}}{{4184{\rm{ J}}}}} \right)\\ &= 5872.37{\rm{ kcal}}\end{aligned}\)

Therefore, the required energy used in metabolism during flight is \(5872.37{\rm{ kcal}}\).

Find Study Materials for

Create Study Materials

Select your language

College Physics (Urone)

Answers without the blur.

Short Answer

Step by Step Solution

Step 1: Efficiency

Step 2: Food energy in kilojoules he metabolized during the flight

Most popular questions for Physics Textbooks

Want to see more solutions like these?

Recommended explanations on Physics Textbooks

Select your language

College Physics (Urone)

Answers without the blur.

Short Answer

Step by Step Solution

Step 1: Efficiency

Step 2: Food energy in kilojoules he metabolized during the flight

Most popular questions for Physics Textbooks

Want to see more solutions like these?

Recommended explanations on Physics Textbooks

Work Energy and Power

Radiation

Astrophysics

Physics of Motion

Scientific Method Physics

Fluids

Torque and Rotational Motion

Nuclear Physics

Fields in Physics

Measurements

Space Physics

Electricity

Physical Quantities and Units

Medical Physics

Electrostatics

Further Mechanics and Thermal Physics

Mechanics and Materials

Engineering Physics

Energy Physics

Force

Particle Model of Matter

Waves Physics

Magnetism

Modern Physics

Atoms and Radioactivity

Dynamics

Electricity and Magnetism

Conservation of Energy and Momentum

Fundamentals of Physics

Kinematics Physics

Circular Motion and Gravitation

Linear Momentum

Rotational Dynamics

Oscillations

Translational Dynamics

Geometrical and Physical Optics

Thermodynamics

Magnetism and Electromagnetic Induction

Electromagnetism

Electric Charge Field and Potential

Turning Points in Physics