/*! 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} Problem 93 The power expended by a cheetah ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 14: Problem 93

The power expended by a cheetah is \(160 \mathrm{kW}\) while running at $110 \mathrm{km} / \mathrm{h},$ but its body temperature cannot exceed \(41.0^{\circ} \mathrm{C} .\) If \(70.0 \%\) of the energy expended is dissipated within its body, how far can it run before it overheats? Assume that the initial temperature of the cheetah is \(38.0^{\circ} \mathrm{C},\) its specific heat is \(3.5 \mathrm{kJ} /\left(\mathrm{kg} \cdot^{\circ} \mathrm{C}\right),\) and its mass is \(50.0 \mathrm{kg}\)

Short Answer

Expert verified
Answer: The cheetah can run approximately 0.65 km before it overheats.

Step by step solution

01

Calculate energy that can be absorbed by the cheetah

To find out how much energy the cheetah can absorb before overheating, we will use the specific heat formula, which is: \(Q = mc(T_f - T_i)\) Where: Q: energy that can be absorbed m: mass of the object c: specific heat capacity \(T_f\): final temperature \(T_i\): initial temperature Here, \(T_f = 41.0^{\circ} \mathrm{C},\) \(T_i = 38.0^{\circ} \mathrm{C},\) and \(c = 3.5 \mathrm{kJ/ (kg} \cdot^{\circ}\mathrm{C)},\) and \(m = 50.0 \mathrm{kg}\). Let's plug in these values and compute the absorbed energy (Q).
02

Calculate the energy released by the cheetah while running

We know that 70% of the energy is dissipated within the cheetah's body. Thus, to calculate the energy released while running, we multiply the power with the percentage of energy dissipated: \(E = P\cdot t \cdot 0.7\) Where: E: energy released by the cheetah P: power expended (= 160 kW) t: time for which the cheetah runs We need to find t and then the energy released will be equal to the absorbed energy (Q).
03

Find the time for which the cheetah runs

Using the absorbed energy (Q) and energy released (E), we can find the time (t) as: \(t = \frac{Q}{0.7\cdot P}\) Now let's substitute the values for Q and P to find t.
04

Find the distance covered by the cheetah while running

We know the speed of the cheetah is 110 km/h, and we've found the time (t) it takes before overheating. We can now find the distance covered using the formula: Distance = Speed \(\times\) Time Let's plug in the values for speed and time to find the distance covered before the cheetah overheats.

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

If the maximum intensity of radiation for a blackbody is found at $2.65 \mu \mathrm{m},$ what is the temperature of the radiating body?
The thermal resistance of a seal's fur and blubber combined is $0.33 \mathrm{K} / \mathrm{W} .\( If the seal's internal temperature is \)37^{\circ} \mathrm{C}\( and the temperature of the sea is about \)0^{\circ} \mathrm{C},$ what must be the heat output of the seal in order for it to maintain its internal temperature?
A tungsten filament in a lamp is heated to a temperature of $2.6 \times 10^{3} \mathrm{K}\( by an electric current. The tungsten has an emissivity of \)0.32 .$ What is the surface area of the filament if the lamp delivers $40.0 \mathrm{W}$ of power?
A mass of \(1.00 \mathrm{kg}\) of water at temperature \(T\) is poured from a height of \(0.100 \mathrm{km}\) into a vessel containing water of the same temperature \(T,\) and a temperature change of \(0.100^{\circ} \mathrm{C}\) is measured. What mass of water was in the vessel? Ignore heat flow into the vessel, the thermometer, etc.

Consider the leaf of Problem \(70 .\) Assume that the top surface of the leaf absorbs \(70.0 \%\) of \(9.00 \times 10^{2} \mathrm{W} / \mathrm{m}^{2}\) of radiant energy, while the bottom surface absorbs all of the radiant energy incident on it due to its surroundings at \(25.0^{\circ} \mathrm{C} .\) (a) If the only method of heat loss for the leaf were thermal radiation, what would be the temperature of the leaf? (Assume that the leaf radiates like a blackbody.) (b) If the leaf is to remain at a temperature of \(25.0^{\circ} \mathrm{C}\) how much power per unit area must be lost by other methods such as transpiration (evaporative heat loss)?
See all solutions

Recommended explanations on Physics Textbooks

Magnetism and Electromagnetic Induction

Read Explanation

Fields in Physics

Read Explanation

Engineering Physics

Read Explanation

Thermodynamics

Read Explanation

Conservation of Energy and Momentum

Read Explanation

Solid State Physics

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.