Question

In an extreme marathon, participants run a total of 100km; world-class athletes maintain a pace of 15 km/h. how many 230 Calorie energy bars would be required to fuel such a run for a 68 kg athlete?

Answer 1

speed of the participants is given as

$v = 15 km/h$

$v = 15 \times \frac{1000 m}{3600s} = 4.2 m/s$

now the kinetic energy of the participant is given as

$KE = \frac{1}{2}mv^2$

$KE = \frac{1}{2}68\times 4.2^2$

$KE = 590.3 J$

As we know that

1 calorie = 4.2 J

$KE = 590.3\times \frac{1 cal}{4.2J} = 141.1 Cal$

so here 230 Calorie is provided by 1 bar

bar required is

$N = \frac{141.1}{230} = 0.61 bars$

Answer 2

1 energy bar would be required to fuel such a run for a 68 kg athlete

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Further explanation

Let's recall Kinetic Energy Formula as follows:

$\large {\boxed{Ek = \frac{1}{2}mv^2} }$

Ek = Kinetic Energy ( Joule )

m = mass of the object ( kg )

v = speed of the object ( m/s )

Let us now tackle the problem !

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Given:

speed of participant = v = 15 km/h = 4¹/₆ m/s

mass of participant = m = 68 kg

Asked:

kinetic energy = Ek = ?

Solution:

Firstly , we will calculate total kinetic energy of the participant:

$Ek = \frac{1}{2}m(v')^2$

$Ek = \frac{1}{2} \times 68 \times (4\frac{1}{6})^2$

$Ek = 590 \frac{5}{18} \texttt{ J}$

$Ek \approx 141 \texttt{ Calories}$

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Next , we could find number of 230 Calorie Energy Bars as follows:

$N = Ek / 230$

$N = 141 / 230$

$N \approx 0.61$

Conclusion:

1 energy bar would be required to fuel such a run for a 68 kg athlete

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Learn more

• Impacts of Gravity : brainly.com/question/5330244
• Effect of Earth’s Gravity on Objects : brainly.com/question/8844454
• The Acceleration Due To Gravity : brainly.com/question/4189441
• Newton's Law of Motion: brainly.com/question/10431582
• Example of Newton's Law: brainly.com/question/498822

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Answer details

Grade: High School

Subject: Physics

Chapter: Dynamics