Find answers to your most challenging questions with the help of IDNLearn.com's experts. Join our interactive Q&A community and get reliable, detailed answers from experienced professionals across a variety of topics.
Sagot :
Let's find the wavelength of a radio wave given its frequency and speed. We'll use the formula that relates wavelength ([tex]\(\lambda\)[/tex]), speed (v), and frequency (f). The formula is:
[tex]\[ \lambda = \frac{v}{f} \][/tex]
Here are the given values:
- Frequency ([tex]\(f\)[/tex]) = [tex]\(5.5 \times 10^4\)[/tex] hertz
- Speed ([tex]\(v\)[/tex]) = [tex]\(3.0 \times 10^8\)[/tex] meters/second
Plugging in these values into the formula:
[tex]\[ \lambda = \frac{3.0 \times 10^8 \, \text{m/s}}{5.5 \times 10^4 \, \text{Hz}} \][/tex]
Performing the division:
[tex]\[ \lambda = \frac{3.0 \times 10^8}{5.5 \times 10^4} \approx 5454.545454545455 \, \text{meters} \][/tex]
When you round this to 4 significant figures, [tex]\( \lambda \approx 5455 \, \text{meters} \)[/tex]
Based on the given choices, the closest correct answer is:
B. [tex]$5.0 \times 10^3$[/tex] meters
[tex]\[ \lambda = \frac{v}{f} \][/tex]
Here are the given values:
- Frequency ([tex]\(f\)[/tex]) = [tex]\(5.5 \times 10^4\)[/tex] hertz
- Speed ([tex]\(v\)[/tex]) = [tex]\(3.0 \times 10^8\)[/tex] meters/second
Plugging in these values into the formula:
[tex]\[ \lambda = \frac{3.0 \times 10^8 \, \text{m/s}}{5.5 \times 10^4 \, \text{Hz}} \][/tex]
Performing the division:
[tex]\[ \lambda = \frac{3.0 \times 10^8}{5.5 \times 10^4} \approx 5454.545454545455 \, \text{meters} \][/tex]
When you round this to 4 significant figures, [tex]\( \lambda \approx 5455 \, \text{meters} \)[/tex]
Based on the given choices, the closest correct answer is:
B. [tex]$5.0 \times 10^3$[/tex] meters
We are delighted to have you as part of our community. Keep asking, answering, and sharing your insights. Together, we can create a valuable knowledge resource. IDNLearn.com is committed to providing accurate answers. Thanks for stopping by, and see you next time for more solutions.