/*! 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 17 Two astronauts are \(1.5 \mathrm... [FREE SOLUTION] | 91Ó°ÊÓ

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Two astronauts are \(1.5 \mathrm{m}\) apart in their spaceship. One speaks to the other. The conversation is transmitted to earth via electromagnetic waves. The time it takes for sound waves to travel at \(343 \mathrm{m} / \mathrm{s}\) through the air between the astronauts equals the time it takes for the electromagnetic waves to travel to the earth. How far away from the earth is the spaceship?

Short Answer

Expert verified
The spaceship is approximately 1311 kilometers away from Earth.

Step by step solution

01

Determine time taken for sound to travel

First, we will calculate the time it takes for the sound to travel from one astronaut to the other. The formula for time based on speed is \( t = \frac{distance}{speed} \). Here the distance is \(1.5\) m and the speed of sound is \(343\, \text{m/s}\). \[ t = \frac{1.5}{343} \approx 0.00437\, \text{seconds} \]
02

Equate times for sound and electromagnetic waves

According to the problem, the time for sound to travel between the astronauts is equal to the time for electromagnetic waves to travel to Earth. This means the time for electromagnetic waves \( t_{em} \) is also approximately \(0.00437\) seconds.
03

Determine distance using electromagnetic speed

Electromagnetic waves (light) travel at a speed of \(3 \times 10^8\, \text{m/s}\). We can use the formula \( distance = speed \times time \) to find how far the spaceship is from the Earth. \[ distance = (3 \times 10^8) \times 0.00437 \approx 1.311 \times 10^6\, \text{meters} \]
04

Convert distance to kilometers

Convert the obtained distance from meters to kilometers by dividing by 1000. \( 1.311 \times 10^6\, \text{meters} = 1311\, \text{kilometers} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding the Speed of Sound
The speed of sound refers to how quickly sound waves move through a medium, such as air. It's important to understand that this speed can change based on factors like temperature, pressure, and the medium itself. For example:
  • In air at 20°C, the speed of sound is approximately 343 meters per second (m/s).
  • In water, sound travels faster due to the denser medium, at about 1482 m/s.
  • Sound cannot travel through a vacuum, as there are no particles to carry the vibrations.
When the exercise mentions a speed of 343 m/s, it assumes a standard scenario typical in air at room temperature. In this problem, the sound travels a short distance of just 1.5 meters, which allows us to quickly calculate how long it takes for sound to reach from one astronaut to another using the formula:\[ t = \frac{\text{distance}}{\text{speed}} \]This gives the astronauts the communicated time as approximately 0.00437 seconds for sound to cover the distance.
Exploring the Speed of Light
The speed of light is vastly higher compared to the speed of sound. Light travels at an astonishing speed of approximately 299,792,458 meters per second, often simplified to \(3 \times 10^8\) m/s for calculations. The properties of electromagnetic waves allow light to move at this speed in a vacuum, which ensures minimal resistance. This constant speed is the cosmic speed limit, meaning nothing can travel faster.In the context of the problem, the signals between the spaceship and Earth are transmitted at the speed of light. We utilized this speed to match the time sound took to travel between the astronauts and calculate the distance the electromagnetic waves traveled. The short time of 0.00437 seconds when multiplied by the speed of light results in a significant distance, highlighting how fast light travels.Keep in mind that, unlike sound, light can travel through a vacuum and does not require a medium. This is why the signal can travel through space from the astronauts back to Earth without any issues.
Basics of Time Calculation in Space
Understanding how to calculate time for various scenarios is crucial in this problem. Time calculation simplifies when you use the basic formulae for speed, distance, and time:\[ t = \frac{\text{distance}}{\text{speed}} \]\[ \text{distance} = \text{speed} \times \text{time} \]In space travel scenarios such as the given exercise, accurately measuring time helps determine distances. Given the brief time sound took to travel between the astronauts, the same duration is applied to determine the distance via the speed of light. This example beautifully exhibits how different speeds in the universe can be used to solve problems related to astronomical distances.By translating the time it takes for light to cover a distance in space, we converted the large measured meters into kilometers to make the distance more comprehensible. This conversion process simplifies our understanding of how far the astronauts are from Earth, aiding in space navigation and communication.

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Most popular questions from this chapter

The electromagnetic wave that delivers a cellular phone call to a car has a magnetic field with an rms value of \(1.5 \times 10^{-10} \mathrm{T}\). The wave passes perpendicularly through an open window, the area of which is \(0.20 \mathrm{m}^{2} .\) How much energy does this wave carry through the window during a \(45-\) s phone call?

The human eye is most sensitive to light with a frequency of about \(5.5 \times 10^{14} \mathrm{Hz},\) which is in the yellow-green region of the electromagnetic spectrum. How many wavelengths of this light can fit across the width of your thumb, a distance of about \(2.0 \mathrm{cm} ?\)

A speeder is pulling directly away and increasing his distance from a police car that is moving at \(25 \mathrm{m} / \mathrm{s}\) with respect to the ground. The radar gun in the police car emits an electromagnetic wave with a frequency of \(7.0 \times 10^{9} \mathrm{Hz} .\) The wave reflects from the speeder's car and returns to the police car, where its frequency is measured to be \(320 \mathrm{Hz}\) less than the emitted frequency. Find the speeder's speed with respect to the ground.

A heat lamp emits infrared radiation whose rms electric field is \(E_{\mathrm{rm}}=2800 \mathrm{N} / \mathrm{C}\) (a) What is the average intensity of the radiation? (b) The radiation is focused on a person's leg over a circular area of radius \(4.0 \mathrm{cm} .\) What is the average power delivered to the leg? (c) The portion of the leg being irradiated has a mass of \(0.28 \mathrm{kg}\) and a specific heat capacity of \(3500 \mathrm{J} /\left(\mathrm{kg} \cdot \mathrm{C}^{\circ}\right)\) How long does it take to raise its temperature by \(2.0 \mathrm{C}^{\circ}\) ? Assume that there is no other heat transfer into or out of the portion of the leg being heated.

A laser emits a narrow beam of light. The radius of the beam is \(1.0 \times\) \(10^{-3} \mathrm{m},\) and the power is \(1.2 \times 10^{-3} \mathrm{W} .\) What is the intensity of the laser beam?

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