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Chapter 14: Problem 15

A high jumper of mass \(60.0 \mathrm{kg}\) consumes a meal of $3.00 \times 10^{3} \mathrm{kcal}\( prior to a jump. If \)3.3 \%$ of the energy from the food could be converted to gravitational potential energy in a single jump, how high could the athlete jump?

Short Answer

Expert verified
Answer: The high jumper can jump approximately 6.96 meters high.

Step by step solution

01

Convert Kcal to Joules

We need to convert the energy given in Kcal to Joules. We know that 1 Kcal = 4184 Joules. So, we have: Energy in Joules = Energy in Kcal * conversion factor Energy in Joules = 3000 Kcal * 4184 J/Kcal = 12,552,000 J
02

Find the converted gravitational potential energy

We are given that 3.3% of the energy from the food can be converted into GPE. So, we have: Converted GPE = Total energy * percentage converted Converted GPE = 12,552,000 J * 0.033 = 414216 J
03

Find the height the athlete could jump

We use the gravitational potential energy formula, GPE = m * g * h, where m is the mass (60 kg), g is the acceleration due to gravity (approximately 9.81 m/s²), and h is the height in meters. Rearranging the formula for h, we get: h = GPE / (m * g) h = 414216 J / (60 kg * 9.81 m/s²) h ≈ 6.96 m So, the athlete could jump approximately 6.96 meters high.

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Most popular questions from this chapter

For a cheetah, \(70.0 \%\) of the energy expended during exertion is internal work done on the cheetah's system and is dissipated within his body; for a dog only \(5.00 \%\) of the energy expended is dissipated within the dog's body. Assume that both animals expend the same total amount of energy during exertion, both have the same heat capacity, and the cheetah is 2.00 times as heavy as the dog. (a) How much higher is the temperature change of the cheetah compared to the temperature change of the dog? (b) If they both start out at an initial temperature of \(35.0^{\circ} \mathrm{C},\) and the cheetah has a temperature of \(40.0^{\circ} \mathrm{C}\) after the exertion, what is the final temperature of the dog? Which animal probably has more endurance? Explain.
A birch tree loses \(618 \mathrm{mg}\) of water per minute through transpiration (evaporation of water through stomatal pores). What is the rate of heat lost through transpiration?
If a leaf is to maintain a temperature of \(40^{\circ} \mathrm{C}\) (reasonable for a leaf), it must lose \(250 \mathrm{W} / \mathrm{m}^{2}\) by transpiration (evaporative heat loss). Note that the leaf also loses heat by radiation, but we will neglect this. How much water is lost after 1 h through transpiration only? The area of the leaf is \(0.005 \mathrm{m}^{2}\).
The thermal conductivity of the fur (including the skin) of a male Husky dog is \(0.026 \mathrm{W} /(\mathrm{m} \cdot \mathrm{K}) .\) The dog's heat output is measured to be \(51 \mathrm{W}\), its internal temperature is $38^{\circ} \mathrm{C},\( its surface area is \)1.31 \mathrm{m}^{2},$ and the thickness of the fur is \(5.0 \mathrm{cm} .\) How cold can the outside temperature be before the dog must increase its heat output?
A copper bar of thermal conductivity $401 \mathrm{W} /(\mathrm{m} \cdot \mathrm{K})\( has one end at \)104^{\circ} \mathrm{C}$ and the other end at \(24^{\circ} \mathrm{C}\). The length of the bar is \(0.10 \mathrm{m}\) and the cross-sectional area is \(1.0 \times 10^{-6} \mathrm{m}^{2} .\) (a) What is the rate of heat conduction, \(\mathscr{P}\) along the bar? (b) What is the temperature gradient in the bar? (c) If two such bars were placed in series (end to end) between the same temperature baths, what would \(\mathscr{P}\) be? (d) If two such bars were placed in parallel (side by side) with the ends in the same temperature baths, what would \(\mathscr{P}\) be? (e) In the series case, what is the temperature at the junction where the bars meet?
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