/*! 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 19 Two flies sit exactly opposite e... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 1: Problem 19

Two flies sit exactly opposite each other on the surface of a spherical balloon. If the balloon's volume doubles, by what factor does the distance between the flies change?

Short Answer

Expert verified
Answer: The distance between the two flies changes by a factor of (\frac{2}{1})^{1/3}.

Step by step solution

01

Express the volume of a sphere

The volume (V) of a sphere with radius (r) can be written as: V = \frac{4}{3} \pi r^3 Since the volume doubles, we can express the new volume (V') as: V' = 2V
02

Determine the relationship between the new radius and the original radius

Using the formula for the volume of a sphere, we can write the new volume (V') in terms of the new radius (r'): V' = \frac{4}{3} \pi r'^3 Since V' = 2V, we can write: \frac{4}{3} \pi r'^3 = 2(\frac{4}{3} \pi r^3) Now we can solve for the ratio \frac{r'}{r}: \frac{r'^3}{r^3} = 2 \Rightarrow \frac{r'}{r} = (\frac{2}{1})^{1/3}
03

Apply the relationship to the surface distance

Let's denote the initial distance between the flies as d, and the new distance between them as d'. Note that the flies are exactly on opposite sides of the sphere, so the distance between them is equivalent to the spherical distance between the two antipodal points. The initial spherical distance between the flies can be represented as d = \pi r, since it is half the circumference of the sphere. Similarly, the new spherical distance can be represented as d' = \pi r'.
04

Determine the factor by which the distance between the flies changes

Now, we can write the ratio \frac{d'}{d} in terms of the relationship between r' and r: \frac{d'}{d} =\frac{\pi r'}{ \pi r} = \frac{r'}{r} = (\frac{2}{1})^{1/3} Thus, the factor by which the distance between the two flies changes is (\frac{2}{1})^{1/3}.

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

What angle does \(\vec{A}=\left(A_{x}, A_{y}\right)=(30.0 \mathrm{~m},-50.0 \mathrm{~m})\) make with the positive \(x\) -axis? What angle does it make with the negative \(y\) -axis?

According to one mnemonic rhyme, "A pint's a pound, the world around." Investigate this statement of equivalence by calculating the weight of a pint of water, assuming that the density of water is \(1000 . \mathrm{kg} / \mathrm{m}^{3}\) and that the weight of \(1.00 \mathrm{~kg}\) of a substance is 2.21 pounds. The volume of 1.00 fluid ounce is \(29.6 \mathrm{~mL}\).

How many inches are in 30.7484 miles?

Find the magnitude and direction of each of the following vectors, which are given in terms of their \(x\) - and \(y\) -components: \(\vec{A}=(23.0,59.0),\) and \(\vec{B}=(90.0,-150.0)\)

If the radius of a planet is larger than that of Earth by a factor of 5.8 , how much bigger is the volume of the planet than Earth's?
See all solutions

Recommended explanations on Physics Textbooks

Classical Mechanics

Read Explanation

Particle Model of Matter

Read Explanation

Torque and Rotational Motion

Read Explanation

Radiation

Read Explanation

Energy Physics

Read Explanation

Circular Motion and Gravitation

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.