/*! 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 73 (II) A space probe enters the th... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 16: Problem 73

(II) A space probe enters the thin atmosphere of a planet where the speed of sound is only about 45 \(\mathrm{m} / \mathrm{s}\) . (a) What is the probe's Mach number if its initial speed is \(15,000 \mathrm{km} / \mathrm{h} ?\) (b) What is the angle of the shock wave relative to the direction of motion?

Short Answer

Expert verified
(a) The Mach number is 92.6. (b) The shock wave angle is \(0.622^{\circ}\).

Step by step solution

01

Convert Speed to m/s

The initial speed of the probe is given as 15,000 km/h. To use the Mach number formula, we first convert this speed to meters per second (m/s): 1 km/h equals \(\frac{1}{3.6}\) m/s. Thus, the probe's initial speed in m/s is:\[15,000 \times \frac{1}{3.6} = 4,166.67 \text{ m/s}\]
02

Calculate the Mach Number

The Mach number is defined as the ratio of the object's speed to the speed of sound. Using the converted speed of the probe and the given speed of sound (45 m/s), the Mach number (M) is:\[M = \frac{4,166.67}{45} \approx 92.6\]
03

Calculate the Shock Wave Angle

The angle \(\theta\) of the shock wave relative to the direction of motion is given by the formula \(\sin(\theta) = \frac{1}{M}\), where \(M\) is the Mach number. Thus:\[\theta = \arcsin\left(\frac{1}{92.6}\right)\]Calculate \(\arcsin(\frac{1}{92.6})\) using a calculator, yielding an angle \(\theta\).

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!

Key Concepts

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

Understanding Shock Waves
When an object travels faster than the speed of sound, it creates a type of wave known as a shock wave. Shock waves form because the object is moving through a medium鈥攍ike air鈥攕o quickly that it compresses the molecules ahead of it. These molecules cannot get out of the way quickly enough, leading to a sudden change in pressure, temperature, and density. This change is what creates the 'boom' associated with supersonic travel.
The shock wave is quite different from normal sound waves, as it travels along with the object. This means it is constantly being generated as long as the object maintains its supersonic speed. One interesting aspect of shock waves is that they create a cone behind the moving object instead of spreading out evenly in all directions like normal sound waves.
The Speed of Sound
The speed of sound is a measure of how quickly sound waves travel through a given medium. This speed varies based on factors such as temperature, pressure, and the type of substance the sound is moving through. In our exercise example, the speed of sound in the planet's atmosphere is 45 m/s.
This is relatively low compared to Earth's atmosphere, where the speed of sound is approximately 343 m/s at sea level. Generally, sound travels faster in solids and liquids than in gases because molecules are more tightly packed in these states, allowing vibrations to transfer more quickly.
Knowing the speed of sound is essential when determining the Mach number of an object, which is a crucial factor in studying aerodynamics and fluid dynamics.
Calculating the Angle of Shock Wave
The angle of the shock wave relative to the direction of an object's movement is an essential consideration in aerodynamics. It helps us understand how the shock wave refracts around the object. This angle is influenced by the Mach number, which is the ratio of the object's speed to the speed of sound.
The mathematical formula for finding the shock wave angle \( \theta \) in terms of the Mach number \( M \) is \( \sin(\theta) = \frac{1}{M} \). A larger Mach number results in a smaller shock wave angle, indicating a more acute formation of the shock wave cone.
In practical terms, knowing this angle assists engineers in designing more efficient aerodynamic structures that can withstand the intense forces encountered at supersonic speeds. This ensures safety and performance in designs like spacecraft, aircraft, and even some high-speed vehicles.

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

a. Calculate the maximum displacement of air molecules when a \(330-\mathrm{Hz}\) sound wave passes whose intensity is at the threshold of pain \((120 \mathrm{~dB}) .\) (b) What is the pressure amplitude in this wave?

(1I) Two piano strings are supposed to be vibrating at 220 \(\mathrm{Hz}\) , but a piano tuner hears three beats every 2.0 \(\mathrm{s}\) when they are played together. \((a)\) If one is vibrating at 220.0 \(\mathrm{Hz}\) what must be the frequency of the other (is there only one answer)? (b) By how much (in percent) must the tension be increased or decreased to bring them in tune?

At a rock concert, a dB meter registered \(130 \mathrm{~dB}\) when placed \(2.2 \mathrm{~m}\) in front of a loudspeaker on the stage. \((a)\) What was the power output of the speaker, assuming uniform spherical spreading of the sound and neglecting absorption in the air? (b) How far away would the sound level be a somewhat reasonable \(85 \mathrm{~dB} ?\)

When a player's finger presses a guitar string down onto a fret, the length of the vibrating portion of the string is shortened, thereby increasing the string's fundamental frequency (see Fig. 16-35). The string's tension and mass per unit length remain unchanged. If the unfingered length of the string is \(\ell=65.0 \mathrm{~cm},\) determine the positions \(x\) of the first six frets, if each fret raises the pitch of the fundamental by one musical note in comparison to the neighboring fret. On the equally tempered chromatic scale, the ratio of frequencies of neighboring notes is \(2^{1 / 12}\)

(I) What is the beat frequency if middle \(\mathrm{C}(262 \mathrm{Hz})\) and \(\mathrm{C}^{4}\) \((277 \mathrm{Hz})\) are played together? What if each is played two octaves lower (each frequency reduced by a factor of 4\() ?\)
See all solutions

Recommended explanations on Physics Textbooks

Geometrical and Physical Optics

Read Explanation

Electric Charge Field and Potential

Read Explanation

Fields in Physics

Read Explanation

Circular Motion and Gravitation

Read Explanation

Rotational Dynamics

Read Explanation

Magnetism

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.