Project Seafarer was an ambitious program to construct an enormous antenna, buried underground on a site about 10000 km2 in area. Its purpose was to transmit signals to submarines while they were deeply submerged. If the effective wavelength were 1.8 × 104 Earth radii, what would be the (a) frequency and (b) period of the radiation emitted? Ordinarily, electromagnetic radiations do not penetrate very far into conductors such as seawater. Take the Earth's radius to be 6370 km. (a) Number Enter your answer for part (a) in accordance to the question statementEntry field with incorrect answer Units Choose the answer for part (a) from the menu in accordance to the question statementEntry field with correct answer

Respuesta :

(a) 0.0026 Hz

The relationship between frequency and wavelength for an electromagnetic wave is

[tex]f=\frac{c}{\lambda}[/tex] (1)

where

[tex]c=3.0\cdot 10^8 m/s[/tex] is the speed of light

f is the frequency

[tex]\lambda[/tex] is the wavelength

For the wave in the problem, the wavelength is [tex]1.8\cdot 10^4[/tex] Earth radii. The Earth radius is

[tex]R=6370 km = 6.37\cdot 10^6 m[/tex]

so the wavelength would be

[tex]\lambda = (1.8\cdot 10^4 )(6.37\cdot 10^6 m)=1.14\cdot 10^{11}m[/tex]

So by using eq.(1) we find the frequency:

[tex]f=\frac{3\cdot 10^8 m/s}{1.14\cdot 10^{11}m}=0.0026 Hz[/tex]

(b) 384.6 s

The period of a wave is given by:

[tex]T=\frac{1}{f}[/tex]

where

T is the period

f is the frequency

For the wave in the problem,

f = 0.0026 Hz

so the period is

[tex]T=\frac{1}{0.0026 Hz}=384.6 s[/tex]