Question

Question #0a254

Answer 1

Please see the explanation.

Explanation:

The wavelength of `550" kHz"` is:

`(3.0xx10^8" m/s")/(5.5xx10^5" Hz")= 545.5" m"`

The wavelength of `1600" kHz"` is:

`(3.0xx10^8" m/s")/(1.6xx10^6" Hz")= 187.5" m"`

The wavelength of 88 MHz is:

`(3.0xx10^8" m/s")/(8.8xx10^7" Hz")= 3.4" m"`

The wavelength of 108 MHz is:

`(3.0xx10^8" m/s")/(1.08xx10^8" Hz")= 2.8" m"`

Answer 2

AM signals: `188"m"-545"m"`
FM signals: `2.8"m"-3.4"m"`

Explanation:

From the chemistry reference sheet, `C=lambdanu` where `C` is the speed of the wave, `lambda` is the wavelength, and `nu` is the frequency. Solving this formula for wavelength, `lambda=C/nu`.

For AM signals, `C=3.0times10^8"m/s"` and `nu` ranges from `550times10^3"Hz"` to `1600times10^3"Hz"` (since `1"kHz"=10^3"Hz"`). Therefore, the wavelength ranges from `lambda=(3.0times10^8"m/s")/(550times10^3"Hz")=545"m"` to `lambda=(3.0times10^8"m/s")/(1600times10^3"Hz")=188"m"`.

For FM signals, `C=3.0times10^8"m/s"` and `nu` ranges from `88times10^6"Hz"` to `108times10^6"Hz"` (since `1"MHz"=10^6"Hz"`). Therefore, the wavelength ranges from `lambda=(3.0times10^8"m/s")/(88times10^6"Hz")=3.4"m"` to `lambda=(3.0times10^8"m/s")/(108times10^6"Hz")=2.8"m"`.