Here are the answers to the questions:

1. Wavelength of a radio wave with a frequency of 450 kHz:

The relationship between the speed of a wave ($v$), its frequency ($f$), and wavelength ($\lambda$) is given by:

$\lambda = \frac{v}{f}$

For electromagnetic waves like radio waves, the speed ($v$) is the speed of light in a vacuum, which is approximately $3 \times 10^8 \, \text{m/s}$.

$\lambda = \frac{3 \times 10^8}{450 \times 10^3} = \frac{3 \times 10^8}{4.5 \times 10^5} = 666.67 \, \text{meters}$

So, the wavelength is 666.67 meters.

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2. Frequency of a signal with a period of 0.1057 seconds:

The relationship between frequency ($f$) and period ($T$) is given by:

$f = \frac{1}{T}$

Substituting $T = 0.1057 \, \text{seconds}$:

$f = \frac{1}{0.1057} \approx 9.46 \, \text{Hz}$

So, the frequency is approximately 9.46 Hz.

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3. Signal-to-Noise Ratio (S/N):

The Signal-to-Noise Ratio (S/N) is the ratio of the signal power to the noise power, typically expressed in decibels (dB). The formula for S/N ratio in decibels is:

$S/N \, (\text{dB}) = 20 \log_{10}\left(\frac{V_{\text{signal}}}{V_{\text{noise}}}\right)$

Given:

• $V_{\text{signal}} = 5 \, \text{volts}$
• $V_{\text{noise}} = 7 \, \text{mV} = 7 \times 10^{-3} \, \text{volts}$

$S/N = 20 \log_{10}\left(\frac{5}{7 \times 10^{-3}}\right)$

$S/N = 20 \log_{10}\left(\frac{5}{0.007}\right) = 20 \log_{10}(714.29)$

$S/N \approx 20 \times 2.854 = 57.08 \, \text{dB}$

So, the S/N ratio is approximately 57.08 dB.

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Would you like more details on any of these answers? Here are five related questions to expand your understanding:

1. How does the wavelength of a wave change if the frequency increases?
2. What is the relationship between period and wavelength for a given wave?
3. What are the practical implications of having a high or low Signal-to-Noise Ratio (SNR) in communication systems?
4. How can noise in a circuit affect the quality of a signal?
5. How would the wavelength change if the radio wave traveled through a medium other than a vacuum?

Tip: When working with logarithmic calculations (like S/N ratio in decibels), remember that a 20 dB increase means the signal power is 10 times stronger, while a 40 dB increase means the signal is 100 times stronger!