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What is the wavelength of the radio waves transmitted by an FM station at 90 MHz? (Radio waves travel at \(3.0 \times 10^{8} \mathrm{m} / \mathrm{s} .\)

Short Answer

Expert verified
Answer: The wavelength of the radio waves transmitted by an FM station at 90 MHz is 3.33 meters.

Step by step solution

01

Write down the wave speed formula

The wave speed formula is: speed = frequency × wavelength or v = f × λ Where v is the speed of the radio waves, f is the frequency of the radio waves, and λ is the wavelength of the radio waves.
02

Rearrange the formula to solve for wavelength

To find the wavelength, we need to rearrange the equation: λ = v / f
03

Convert the given frequency to Hz

The frequency is given in MHz, which is megahertz. We need to convert it to hertz (Hz) for our calculations: 90 MHz = 90 × 10^6 Hz
04

Plug in given values and solve for wavelength

Now, we can plug in the values for the speed of radio waves (v = 3.0 × 10^8 m/s) and frequency (f = 90 × 10^6 Hz) into the rearranged equation: λ = (3.0 × 10^8 m/s) / (90 × 10^6 Hz)
05

Calculate the wavelength

Now, we can perform the calculation: λ = (3.0 × 10^8 m/s) / (90 × 10^6 Hz) = (3.0 / 90) × 10^2 m = 0.033 × 10^2 m = 3.33 m The wavelength of the radio waves transmitted by an FM station at 90 MHz is 3.33 meters.

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Most popular questions from this chapter

A guitar's E-string has length \(65 \mathrm{cm}\) and is stretched to a tension of \(82 \mathrm{N}\). It vibrates at a fundamental frequency of $329.63 \mathrm{Hz}$ Determine the mass per unit length of the string.
For a transverse wave on a string described by $y(x, t)=(0.0050 \mathrm{m}) \cos [(4.0 \pi \mathrm{rad} / \mathrm{s}) t-(1.0 \pi \mathrm{rad} / \mathrm{m}) x]$ find the maximum speed and the maximum acceleration of a point on the string. Plot graphs for one cycle of displacement \(y\) versus \(t\), velocity \(v_{y}\) versus \(t\), and acceleration \(a_{y}\) versus \(t\) at the point \(x=0.\)
The formula for the speed of transverse waves on a spring is the same as for a string. (a) A spring is stretched to a length much greater than its relaxed length. Explain why the tension in the spring is approximately proportional to the length. (b) A wave takes 4.00 s to travel from one end of such a spring to the other. Then the length is increased \(10.0 \% .\) Now how long does a wave take to travel the length of the spring? [Hint: Is the mass per unit length constant?]
Using graph paper, sketch two identical sine waves of amplitude $4.0 \mathrm{cm}\( that differ in phase by (a) \)\pi / 3$ rad \(\left(60^{\circ}\right)\) and (b) \(\pi / 2\) rad \(\left(90^{\circ}\right) .\) Find the amplitude of the superposition of the two waves in each case.
A cord of length \(1.5 \mathrm{m}\) is fixed at both ends. Its mass per unit length is \(1.2 \mathrm{g} / \mathrm{m}\) and the tension is \(12 \mathrm{N} .\) (a) What is the frequency of the fundamental oscillation? (b) What tension is required if the \(n=3\) mode has a frequency of \(0.50 \mathrm{kHz} ?\)
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