/*! 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 26 At Zion National Park a loud sho... [FREE SOLUTION] | 91Ó°ÊÓ

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At Zion National Park a loud shout produces an echo \(1.80 \mathrm{s}\) later from a colorful sandstone cliff. How far away is the cliff?

Short Answer

Expert verified
The cliff is 308.7 meters away.

Step by step solution

01

Understand the Problem

The problem involves calculating the distance to a cliff based on the time it takes for an echo to return. This means the sound wave travels to the cliff and back. We know the round trip time and need to find the one-way distance.
02

Identify Known Quantities

We know the time it takes for the echo to return is \(t = 1.80\, \text{s}\). Sound travels at approximately \(v = 343\, \text{m/s}\) in air at room temperature.
03

Calculate the Total Distance

The total distance sound travels is round trip, calculated by \(d = vt\). Substitute the known values into the equation: \[d = 343\, \text{m/s} \times 1.80\, \text{s} = 617.4\, \text{m}.\]
04

Calculate the One-Way Distance

Since \(617.4\, \text{m}\) is the round trip distance for the sound, divide it by 2 to find the one-way distance to the cliff:\[\text{Distance to the cliff} = \frac{617.4}{2} = 308.7\, \text{m}.\]

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Key Concepts

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

Speed of Sound
The speed of sound is a critical factor when calculating distances using echoes. Sound travels as a wave through air, and its speed is influenced by several factors like temperature, humidity, and altitude. At room temperature, which is about 20°C (68°F), the speed of sound is approximately 343 meters per second (m/s).

How does temperature affect speed? As the temperature increases, air molecules move faster, allowing sound to travel faster through the medium. For example, at 0°C, sound travels at about 331 m/s, and at 20°C, it travels at 343 m/s.

Understanding the speed of sound helps in distance calculations, like in echo situations, because it determines how fast the sound wave travels to an object and returns.
Distance Calculation
Distance calculation using echoes involves using the formula \(d = vt\), where \(d\) is the distance, \(v\) is the speed of sound, and \(t\) is the time it takes for the echo to return.
  • The echo takes 1.80 seconds for a round trip.
  • The round trip covers the distance to the object and back, hence the return time is doubled in the equation.
  • To calculate the one-way distance to the object (like the cliff), divide the total distance gained from \(vt\) by 2.
For instance, if the speed of sound is 343 m/s and the echo return time is 1.80 s, the calculation is:
  • Round trip distance: \(d = 343 \text{ m/s} \times 1.80 \text{ s} = 617.4 \text{ m}\)
  • One-way distance: \(\text{Distance to the cliff} = \frac{617.4}{2} = 308.7 \text{ m}\)
In summary, understanding the calculation involves comprehending the concept of round and one-way distances.
Zion National Park
Zion National Park, located in southwestern Utah, is renowned for its stunning sandstone cliffs and diverse landscapes. These remarkable cliffs serve as ideal natural spots for echo-related experiments.

The park's towering sandstone formations, like Zion's iconic red cliffs, not only provide breathtaking views but also unique opportunities to observe how sound behaves in nature. The cliffs at the park reflect sounds effectively, making them perfect for studying concepts like the speed of sound and distance calculation through echoes.

Aside from its geological allure, Zion National Park offers a wealth of recreational activities, such as hiking, rock climbing, and wildlife viewing, attracting millions of visitors each year. These elements make Zion a fascinating place, both scientifically and recreationally.

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Most popular questions from this chapter

Sound 1 has an intensity of \(38.0 \mathrm{W} / \mathrm{m}^{2}\). Sound 2 has an intensity level that is \(2.5 \mathrm{dB}\) greater than the intensity level of sound 1\. What is the intensity of sound \(2 ?\)

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