Explanation Given data: A hunter is standing on flat ground between two vertical cliffs that are directly opposite one another. The hunter is closer to one cliff than the other cliff. He fires a gun and, after a while, hears three echoes. The second echo arrives 1.6 s after the first, and the third echo arrives 1.1 s after the second. The speed of sound ( v s ) is 343 m s . There are no reflections of sound from the ground. Formula used: The sound spreads uniformly in all the directions. Consider only the sound that travels in the straight line parallel to the ground and reflects from the vertical walls of the cliffs. Consider the distance between the hunter and the closer cliff as x 1 , and the distance between the hunter and the farther cliff as x 2 . From the given data, draw the figure as shown in Figure 1. Consider the echo which moves towards the near cliff as first echo and the echo which moves towards the far cliff as second echo. From the Figure 1, write the expression for the distance between the cliffs as follows. d = x 1 + x 2 (1) Here, x 1 is the distance between hunter and the near cliff, and x 2 is the distance between hunter and the far cliff. The first echo returns to the hunter (firing point) when it travels twice between the hunter and the near cliff. That is, the first echo initially starts at the firing point and moves to near cliff and again returns from the near cliff to the hunter. Consider the time taken by the first echo to return to the hunter as t 1 , and write the expression for the time taken by the first echo to return to the hunter as follows. t 1 = 2 x 1 v s (2) Here, v s is the speed of the echo (since the echo travels at the speed of sound). Similarly, consider the time taken by the second echo to returns to the hunter as t 2 , and write the expression for the time taken by the second echo to returns to the hunter as follows. t 2 = 2 x 2 v s (3) The third echo arrives after the second echo from the near cliff. Consider the time taken by the third echo after the second echo (time between second and third echo) as Δ t 32 , which is given as 1.1 s. As the third echo returns from the near cliff wall, write the expression for the time taken by the third echo after the second echo as follows. Δ t 32 = 2 x 1 v s Rearrange the expression for the distance between hunter and the near cliff. x 1 = v s Δ t 32 2 Substitute 343 m s for v s , and 1

11th Edition

ISBN: 9781119460176

Author: CUTNELL, John D., Johnson, Kenneth W., YOUNG, David, Stadler, Shane

Publisher: WILEY

See similar textbooks

Concept explainers

Properties of sound

A sound wave is a mechanical wave (or mechanical vibration) that transit through media such as gas (air), liquid (water), and solid (wood).

Quality Of Sound

A sound or a sound wave is defined as the energy produced due to the vibrations of particles in a medium. When any medium produces a disturbance or vibrations, it causes a movement in the air particles which produces sound waves. Molecules in the air vib…

Categories of Sound Wave

People perceive sound in different ways, like a medico student takes sound as vibration produced by objects reaching the human eardrum. A physicist perceives sound as vibration produced by an object, which produces disturbances in nearby air molecules th…

Videos

Question

Chapter 16, Problem 43P

To determine

To find: The distance between the cliffs.

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Chapter 16, Problem 42P

Chapter 16, Problem 44P

Students have asked these similar questions

Can i get answer and solution for this question and can you teach me What we use to get the answer.

Can i get answer and solution and can you teach me how to get it.

Consider a image that is located 30 cm in front of a lens. It forms an upright image 7.5 cm from the lens. Theillumination is so bright that that a faint inverted image, due to reflection off the front of the lens, is observedat 6.0 cm on the incident side of the lens. The lens is then turned around. Then it is observed that the faint,inverted image is now 10 cm on the incident side of the lens.What is the index of refraction of the lens?

Chapter 16 Solutions

Physics, Volume 2

Ch. 16.1 - Prob. 2CYU Ch. 16.2 - 3. A sound wave (a periodic longitudinal wave)...Ch. 16.3 - Prob. 4CYU Ch. 16.3 - Prob. 5CYU Ch. 16.3 - Prob. 6CYU Ch. 16.3 - Prob. 7CYU Ch. 16.3 - Prob. 8CYU Ch. 16.3 - Prob. 9CYU Ch. 16.5 - 10. In a traveling sound wave, are there any...Ch. 16.6 - 11. Do you expect an echo to return to you more...

Ch. 16.6 - 12. Carbon monoxide (CO), hydrogen (H2). and...Ch. 16.6 - Prob. 13CYU Ch. 16.7 - 14. Sonic animals rely on an acute sense of...Ch. 16.7 - Prob. 15CYU Ch. 16.8 - Prob. 16CYU Ch. 16.8 - Prob. 17CYU Ch. 16.9 - Prob. 18CYU Ch. 16.9 - 19. When a car is at rest, its horn emits a...Ch. 16.9 - Prob. 20CYU Ch. 16.9 - Prob. 21CYU Ch. 16.9 - Prob. 22CYU Ch. 16 - Prob. 2FC Ch. 16 - Prob. 3FC Ch. 16 - Prob. 6FC Ch. 16 - Prob. 7FC Ch. 16 - Prob. 10FC Ch. 16 - 11. As the amplitude of a sound wave in air...Ch. 16 - Prob. 12FC Ch. 16 - Prob. 13FC Ch. 16 - Prob. 15FC Ch. 16 - Prob. 17FC Ch. 16 - Prob. 18FC Ch. 16 - Prob. 19FC Ch. 16 - Prob. 1P Ch. 16 - Prob. 2P Ch. 16 - Prob. 3P Ch. 16 - Prob. 4P Ch. 16 - Prob. 5P Ch. 16 - Prob. 6P Ch. 16 - Prob. 7P Ch. 16 - Prob. 8P Ch. 16 - Prob. 9P Ch. 16 - Prob. 10P Ch. 16 - Prob. 12P Ch. 16 - Prob. 13P Ch. 16 - Prob. 14P Ch. 16 - Prob. 15P Ch. 16 - Prob. 16P Ch. 16 - Prob. 17P Ch. 16 - Prob. 18P Ch. 16 - Prob. 19P Ch. 16 - Prob. 20P Ch. 16 - Prob. 21P Ch. 16 - Prob. 23P Ch. 16 - Prob. 24P Ch. 16 - Prob. 25P Ch. 16 - Prob. 26P Ch. 16 - Prob. 27P Ch. 16 - Prob. 28P Ch. 16 - Prob. 29P Ch. 16 - Prob. 30P Ch. 16 - Prob. 31P Ch. 16 - Prob. 32P Ch. 16 - Prob. 33P Ch. 16 - Prob. 34P Ch. 16 - Prob. 35P Ch. 16 - Prob. 36P Ch. 16 - Prob. 37P Ch. 16 - Prob. 38P Ch. 16 - Prob. 39P Ch. 16 - Prob. 40P Ch. 16 - 41. An explosion occurs at the end of a pier. The...Ch. 16 - Prob. 42P Ch. 16 - Prob. 43P Ch. 16 - Prob. 44P Ch. 16 - Prob. 45P Ch. 16 - Prob. 46P Ch. 16 - Prob. 47P Ch. 16 - Prob. 48P Ch. 16 - Prob. 52P Ch. 16 - Prob. 53P Ch. 16 - 54. A source of sound is located at the center of...Ch. 16 - Prob. 56P Ch. 16 - 57. A loudspeaker has a circular opening with a...Ch. 16 - Prob. 58P Ch. 16 - Prob. 59P Ch. 16 - Prob. 60P Ch. 16 - Prob. 61P Ch. 16 - Prob. 63P Ch. 16 - Prob. 64P Ch. 16 - Prob. 65P Ch. 16 - Prob. 66P Ch. 16 - Prob. 67P Ch. 16 - Prob. 68P Ch. 16 - Prob. 69P Ch. 16 - Prob. 70P Ch. 16 - Prob. 71P Ch. 16 - Prob. 72P Ch. 16 - 73. A portable radio is sitting at the edge of a...Ch. 16 - Prob. 76P Ch. 16 - Prob. 77P Ch. 16 - Prob. 78P Ch. 16 - Prob. 79P Ch. 16 - Prob. 80P Ch. 16 - 81. a car is parked 20.0 m directly south of a...Ch. 16 - 82. A loudspeaker in a parked car is producing...Ch. 16 - Prob. 83P Ch. 16 - Prob. 84P Ch. 16 - Prob. 85P Ch. 16 - Prob. 86P Ch. 16 - Prob. 87P Ch. 16 - 89. A recording engineer works in a soundproofed...Ch. 16 - Prob. 92AP Ch. 16 - Prob. 93AP Ch. 16 - Prob. 94AP Ch. 16 - Prob. 95AP Ch. 16 - Prob. 96AP Ch. 16 - Prob. 97AP Ch. 16 - Prob. 98AP Ch. 16 - Prob. 99AP Ch. 16 - Prob. 100AP Ch. 16 - Prob. 101AP Ch. 16 - Prob. 102AP Ch. 16 - Prob. 103AP Ch. 16 - Prob. 104AP Ch. 16 - Prob. 105AP Ch. 16 - 106. spider hangs from a strand of silk whose...Ch. 16 - Prob. 108AP Ch. 16 - Prob. 111AP Ch. 16 - Prob. 112CCP Ch. 16 - Prob. 113CCP Ch. 16 - Prob. 114TP Ch. 16 - Prob. 115TP

Knowledge Booster

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, physics and related others by exploring similar questions and additional content below.