
A radio station broadcasts at \[30\;{\rm{m}}\]. Find the frequency of broadcast.
Answer
586.8k+ views
Hint: The above problem can be resolved using the concepts and fundamentals of the signal broadcast and the mathematical formulas related to this topic. The sound waves travel in a medium and act as a medium to carry the signal from one point to another. The mathematical relation for the broadcast frequency is given by calculating the speed of light and the distance upto which the broadcasting is happening. Moreover, this relation helps calculate the desired variables required to resolve the given problem.
Complete step by step solution:
Given:
The distance of broadcasting is, \[d = 30\;{\rm{m}}\].
The expression for the frequency of broadcast is given as,
\[f = \dfrac{c}{d}\]
Here, c is the speed of light and its value is \[3 \times {10^8}\;{\rm{m/s}}\].
Solve by substituting the value in above equation as,
\[\begin{array}{l}
f = \dfrac{c}{d}\\
f = \dfrac{{3 \times {{10}^8}\;{\rm{m/s}}}}{{30\;{\rm{m}}}}\\
f = {10^7}\;{\rm{Hz}}
\end{array}\]
Therefore, the broadcast frequency is \[{10^7}\;{\rm{Hz}}\] .
Note: To solve the given problem, one must be clear about the concept and applications of the frequency of the sound waves traveling for specific distances. These distances may occur in the space stations, to what is known as the broadcasting. The signals are amplified into the numbers of the channel and the energy waves, the antennas then receive these signals, and specific signals are received. Moreover, this frequency can vary accordingly with the distance for the broadcast. In other sense, the frequency of broadcast varies directly with such distances. The broadcast frequency is much higher for lesser distances.
Complete step by step solution:
Given:
The distance of broadcasting is, \[d = 30\;{\rm{m}}\].
The expression for the frequency of broadcast is given as,
\[f = \dfrac{c}{d}\]
Here, c is the speed of light and its value is \[3 \times {10^8}\;{\rm{m/s}}\].
Solve by substituting the value in above equation as,
\[\begin{array}{l}
f = \dfrac{c}{d}\\
f = \dfrac{{3 \times {{10}^8}\;{\rm{m/s}}}}{{30\;{\rm{m}}}}\\
f = {10^7}\;{\rm{Hz}}
\end{array}\]
Therefore, the broadcast frequency is \[{10^7}\;{\rm{Hz}}\] .
Note: To solve the given problem, one must be clear about the concept and applications of the frequency of the sound waves traveling for specific distances. These distances may occur in the space stations, to what is known as the broadcasting. The signals are amplified into the numbers of the channel and the energy waves, the antennas then receive these signals, and specific signals are received. Moreover, this frequency can vary accordingly with the distance for the broadcast. In other sense, the frequency of broadcast varies directly with such distances. The broadcast frequency is much higher for lesser distances.
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