Answer:
2.83 m
Explanation:
The relationship between frequency and wavelength for an electromagnetic wave is given by
[tex]\lambda=\frac{c}{f}[/tex]
where
[tex]\lambda[/tex] is the wavelength
[tex]c=3.0\cdot 10^8 m/s[/tex] is the speed of light
[tex]f[/tex] is the frequency
For the FM radio waves in this problem, we have:
[tex]f_1=89 MHz=89\cdot 10^6 Hz[/tex] is the minimum frequency, so the maximum wavelength is
[tex]\lambda_2=\frac{c}{f_1}=\frac{3\cdot 10^8}{89\cdot 10^6}=3.37 m[/tex]
The maximum frequency is instead
[tex]f_2=106 MHz=106\cdot 10^6 Hz[/tex]
Therefore, the minimum wavelength is
[tex]\lambda_1=\frac{c}{f_2}=\frac{3\cdot 10^8}{106\cdot 10^6}=2.83 m[/tex]
So, the wavelength at the beginning of the range is 2.83 m.
