IDNLearn.com offers a comprehensive solution for all your question and answer needs. Join our community to receive prompt, thorough responses from knowledgeable experts.
Sagot :
To determine the increase in intensity level when the intensity of sound increases by a factor of [tex]\(10^5\)[/tex], we follow these steps:
1. Understand the Relationship: The relationship between the intensity increase factor and the increase in intensity level (in decibels, dB) is defined by the formula:
[tex]\[ \text{Increase in dB} = 10 \times \log_{10}(\text{intensity increase factor}) \][/tex]
2. Given Intensity Increase Factor: The intensity increase factor is given as [tex]\(10^5\)[/tex].
3. Substitute and Calculate:
- Substitute the given intensity increase factor into the formula:
[tex]\[ \text{Increase in dB} = 10 \times \log_{10}(10^5) \][/tex]
- Evaluate [tex]\(\log_{10}(10^5)\)[/tex]. Since [tex]\(\log_{10}(10^5)\)[/tex] simplifies to 5 (because the logarithm base 10 of [tex]\(10^5\)[/tex] is the exponent 5):
[tex]\[ \log_{10}(10^5) = 5 \][/tex]
- Now, multiply this result by 10:
[tex]\[ \text{Increase in dB} = 10 \times 5 = 50 \text{ dB} \][/tex]
4. Conclusion: Therefore, the increase in intensity level when the intensity increases by a factor of [tex]\(10^5\)[/tex] is [tex]\(50 \text{ dB}\)[/tex].
So, the correct answer is:
[tex]\[ \boxed{50 \text{ dB}} \][/tex]
1. Understand the Relationship: The relationship between the intensity increase factor and the increase in intensity level (in decibels, dB) is defined by the formula:
[tex]\[ \text{Increase in dB} = 10 \times \log_{10}(\text{intensity increase factor}) \][/tex]
2. Given Intensity Increase Factor: The intensity increase factor is given as [tex]\(10^5\)[/tex].
3. Substitute and Calculate:
- Substitute the given intensity increase factor into the formula:
[tex]\[ \text{Increase in dB} = 10 \times \log_{10}(10^5) \][/tex]
- Evaluate [tex]\(\log_{10}(10^5)\)[/tex]. Since [tex]\(\log_{10}(10^5)\)[/tex] simplifies to 5 (because the logarithm base 10 of [tex]\(10^5\)[/tex] is the exponent 5):
[tex]\[ \log_{10}(10^5) = 5 \][/tex]
- Now, multiply this result by 10:
[tex]\[ \text{Increase in dB} = 10 \times 5 = 50 \text{ dB} \][/tex]
4. Conclusion: Therefore, the increase in intensity level when the intensity increases by a factor of [tex]\(10^5\)[/tex] is [tex]\(50 \text{ dB}\)[/tex].
So, the correct answer is:
[tex]\[ \boxed{50 \text{ dB}} \][/tex]
Thank you for contributing to our discussion. Don't forget to check back for new answers. Keep asking, answering, and sharing useful information. IDNLearn.com provides the answers you need. Thank you for visiting, and see you next time for more valuable insights.