/*! 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 2 A sailor strikes the side of his... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A sailor strikes the side of his ship just below the waterline. He hears the echo of the sound reflected from the ocean floor directly below 2.5 s later. How deep is the ocean at this point? Assume the speed of sound in sea water is \(1560 \mathrm{~m} / \mathrm{s}\) (Table \(16-1\) ) and does not vary significantly with depth.

Short Answer

Expert verified
The ocean is 1950 meters deep at that point.

Step by step solution

01

Understand the Problem

The sailor strikes the side of his ship and hears the echo 2.5 seconds later. We need to find out how deep the ocean is. The speed of sound in seawater is given as 1560 m/s.
02

Calculate the Total Distance Sound Travels

The sound has to travel from the ship to the ocean floor and back to the ship. This means the total distance covered is twice the depth of the ocean, as it goes down to the ocean floor and then comes back up.
03

Use the Speed-Time-Distance Relationship

The formula to use is the basic distance formula: \(\text{Distance} = \text{Speed} \times \text{Time}\). Here, the total distance traveled by the sound is \(2 \times \text{Depth}\). Set up the equation: \[ 2 \times \text{Depth} = 1560 \, \text{m/s} \times 2.5 \, \text{s} \]
04

Solve for the Depth

From the equation in Step 3, solve for Depth:\[ 2 \times \text{Depth} = 1560 \times 2.5 \]\[ 2 \times \text{Depth} = 3900 \, \text{m} \]Now, divide both sides of the equation by 2 to find the depth:\[ \text{Depth} = \frac{3900}{2} = 1950 \, \text{m} \]
05

Conclusion

The depth of the ocean at the point where the sailor struck the ship is 1950 meters.

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 91Ó°ÊÓ!

Key Concepts

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

distance formula
The distance formula is a fundamental concept in physics and mathematics that relates the three key components of motion: distance, speed, and time.
The formula can be expressed as:
  • \( \text{Distance} = \text{Speed} \times \text{Time} \)
In the context of this exercise, the sailor uses the distance formula to determine how far sound travels in water.
Once the speed of sound is known, which is given as 1560 meters per second in seawater, the total time taken by the sound to travel is the crucial next data point.
The sound's journey doesn’t just stop when it reaches the ocean floor, it also includes the return trip back to the sailor's ears.
This doubles the total distance traveled. To solve the original problem, you calculate the total distance sound travels and adjust for its round trip, effectively doubling the ocean depth measured in meters.
In practical applications, always ensure you adjust for any extra trips sound makes. This is a common pitfall for those initially learning these calculations.
echo
An echo is a sound that is reflected back to its source after bouncing off a surface.
It occurs when sound waves strike a surface, like the ocean floor, and are redirected back towards the original source, such as the sailor's location on the ship.
This is a simple, yet powerful natural phenomenon that allows us to gather information about our surroundings through sound. In the sailor's scenario, he hears the echo 2.5 seconds after striking the ship.
This time interval includes the duration it takes for sound to travel to the ocean floor and come back up.
Using echoes, scientists can determine the depth and characteristics of underwater environments.
It’s important to note that while the time for the echo to return is crucial, this must be halved to find the one-way time, which corresponds to just the depth of the ocean.
This makes calculating underwater distances with sound an efficient and effective method.
seawater acoustics
Seawater acoustics is a fascinating field that examines how sound behaves underwater.
The unique properties of seawater significantly affect sound speed and behavior, differentiating it from air travel. Several factors affect seawater acoustics:
  • Temperature
  • Salinity
  • Pressure (depth)
These factors can change the speed of sound in seawater, but for simplicity, our problem assumes a constant speed of 1560 m/s regardless of depth.
This assumption is practical for certain standard calculations but may vary in complex real-world scenarios.
Sound generally travels faster in water than in air due to water’s higher density.
Understanding this can help in different fields, including navigation, scientific research, and military strategies.
By experimenting with how sound reflects and refracts in the ocean, we can gain insights into marine life habitats, ocean floor mapping, and more.

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

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 \(85 \mathrm{~dB} ?\)

A tuning fork is set into vibration above a vertical open tube filled with water (Fig. 16-40). The water level is allowed to drop slowly. As it does so, the air in the tube above the water level is heard to resonate with the tuning fork when the distance from the tube opening to the water level is \(0.125 \mathrm{~m}\) and again at \(0.395 \mathrm{~m}\). What is the frequency of the tuning fork?

(II) A police car sounding a siren with a frequency of 1280 \(\mathrm{Hz}\) is traveling at 120.0 \(\mathrm{km} / \mathrm{h}\) . (a) What frequencies does an observer standing next to the road hear as the car approaches and as it recedes? (b) What frequencies are heard in a car traveling at 90.0 \(\mathrm{km} / \mathrm{h}\) in the opposite direction before and after passing the police car? (c) The police car passes a car traveling in the same direction at 80.0 \(\mathrm{km} / \mathrm{h}\) . What two frequencies are heard in this car?

A bat flies toward a wall at a speed of \(7.0 \mathrm{~m} / \mathrm{s}\). As it flies, the bat emits an ultrasonic sound wave with frequency \(30.0 \mathrm{kHz}\). What frequency does the bat hear in the reflected wave?

(1) An organ pipe is 124 \(\mathrm{cm}\) long. Determine the fundamental and first three audible overtones if the pipe is \((a)\) closed at one end, and \((b)\) open at both ends.

See all solutions

Recommended explanations on Physics Textbooks

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.