/*! 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 7 A truck driver is broadcasting a... [FREE SOLUTION] | 91Ó°ÊÓ

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A truck driver is broadcasting at a frequency of \(26.965 \mathrm{MHz}\) with a CB (citizen's band) radio. Determine the wavelength of the electromagnetic wave being used. The speed of light is \(c=2.9979 \times 10^{8} \mathrm{m} / \mathrm{s}\).

Short Answer

Expert verified
The wavelength is approximately 11.119 meters.

Step by step solution

01

Identify the Formula

The wavelength of a wave is related to its speed and frequency by the formula \( \lambda = \frac{c}{f} \), where \( \lambda \) is the wavelength, \( c \) is the speed of light, and \( f \) is the frequency.
02

Assign Known Values

We know from the problem statement that the frequency \( f = 26.965 \) MHz and the speed of light \( c = 2.9979 \times 10^8 \text{ m/s} \). First, convert the frequency from MHz to Hz: \( 26.965 \text{ MHz} = 26.965 \times 10^6 \text{ Hz} \).
03

Calculate the Wavelength

Insert the known values into the formula: \( \lambda = \frac{2.9979 \times 10^8 \text{ m/s}}{26.965 \times 10^6 \text{ Hz}} \). Evaluating this expression gives \( \lambda \approx 11.119 \text{ meters} \).

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Key Concepts

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

Wavelength Calculation
Wavelength calculation is a fundamental concept in understanding electromagnetic waves. The wavelength (\( \lambda \) ) is the distance between consecutive peaks of a wave. To find it, we use the formula \( \lambda = \frac{c}{f} \).
This formula involves three components:
  • \( \lambda \): Wavelength in meters (m)
  • \( c \): Speed of light, approximately \( 2.9979 \times 10^8 \text{ m/s} \)
  • \( f \): Frequency, the number of peaks passing a point per second in Hertz (Hz)
For example, if you are given a frequency, you can substitute it into the equation along with the speed of light. This will give you the wavelength. In our exercise, the given frequency is \( 26.965 \text{ MHz} \), or \( 26.965 \times 10^6 \text{ Hz} \). By placing these values into the formula, you can easily solve for the wavelength: \( \lambda \approx 11.119 \text{ meters} \).
Learning to calculate wavelength is crucial because it helps in analyzing wave behavior, communication signals, and many other applications.
Frequency and Wavelength Relationship
The relationship between frequency and wavelength is inversely proportional. This means that as the frequency increases, the wavelength decreases, and vice versa. In the formula \( \lambda = \frac{c}{f} \), you can see how these two variables relate. Substituting different values will show how they affect one another.
- **Frequency (\( f \))**: Measured in Hertz (Hz), it represents how often the wave cycles occur per second.- **Wavelength (\( \lambda \))**: Measured in meters (m), it describes the distance over which the wave's shape repeats.For higher frequency waves, such as those used in citizen's band radios like in the exercise, expect shorter wavelengths. Conversely, low frequency waves, like radio waves, have longer wavelengths.Understanding this relationship is critical in fields like telecommunications, where specific frequencies are chosen based on their transmission properties.
Speed of Light
The speed of light in a vacuum is an incredibly important constant in physics, approximately \( 2.9979 \times 10^8 \, \text{m/s} \). It represents the fastest speed at which electromagnetic waves, such as light and radio waves, can travel.
This constant is crucial for calculating wave properties. For example, the speed of light enables the calculation of wavelengths and frequencies using the formula \( c = \lambda \cdot f \). Here's why it's important:
  • Sets the foundation for understanding how far waves travel over time.
  • Provides a reference for comparing other speeds in physics.
  • Supports the theory of relativity, which impacts our understanding of time and space.
In practical applications, this constant is vital for communication technology and understanding natural phenomena.

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Most popular questions from this chapter

The team monitoring a space probe exploring the outer solar system finds that radio transmissions from the probe take 2.53 hours to reach earth. How distant (in meters) is the probe?

Two radio waves are used in the operation of a cellular telephone. To receive a call, the phone detects the wave emitted at one frequency by the transmitter station or base unit. To send your message to the base unit, your phone emits its own wave at a different frequency. The difference between these two frequencies is fixed for all channels of cell phone operation. Suppose that the wavelength of the wave emitted by the base unit is \(0.34339 \mathrm{m}\) and the wavelength of the wave emitted by the phone is \(0.36205 \mathrm{m}\). Using a value of \(2.9979 \times 10^{8} \mathrm{m} / \mathrm{s}\) for the speed of light, determine the difference between the two frequencies used in the operation of a cell phone.

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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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