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Solved: (a) What is the efficiency of an out-of-condition

College Physics | 1st Edition | ISBN: 9781938168000 | Authors: Paul Peter Urone, Roger Hinrichs ISBN: 9781938168000 42

Solution for problem 47PE Chapter 7

College Physics | 1st Edition

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College Physics | 1st Edition | ISBN: 9781938168000 | Authors: Paul Peter Urone, Roger Hinrichs

College Physics | 1st Edition

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Problem 47PE

(a) What is the efficiency of an out-of-condition professor who does 2.10×105 J of useful work while metabolizing 500 kcal of food energy? (b) How many food calories would a well-conditioned athlete metabolize in doing the same work with an efficiency of 20%?

Step-by-Step Solution:

Solution 47PE

a.)

Step 1 of 4</p>

The efficiency of the out of condition professor can be calculated by using the definition of the efficiency which is defined as the ratio of the work output to work input.

Efficiency (=

 Workoutput is the amount of useful work done by the professor is 2.10105 J .

Workoutput = 2.10105 J

Workinput is the amount of energy in 500 kcal of food .

1 kcal = 4184 J

So,  500 kcal = 500 kcal

                    = 2092000 J

Workinput = 2092000 J

Step 2 of 4</p>

The efficiency () can be calculated as

Efficiency(=

 =

                = 0.10

           or= 10%

Therefore, the efficiency of an out-of -condition professor who does 2.10×105 J of useful work while metabolizing 500 kcal of food energy is 10 %.

b.)

Step 3 of 4

Chapter 7, Problem 47PE is Solved
Step 4 of 4

Textbook: College Physics
Edition: 1
Author: Paul Peter Urone, Roger Hinrichs
ISBN: 9781938168000

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Solved: (a) What is the efficiency of an out-of-condition

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