Westonci.ca is your go-to source for answers, with a community ready to provide accurate and timely information. Discover the answers you need from a community of experts ready to help you with their knowledge and experience in various fields. Join our platform to connect with experts ready to provide precise answers to your questions in different areas.
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]
We appreciate your time. Please come back anytime for the latest information and answers to your questions. Thanks for using our service. We're always here to provide accurate and up-to-date answers to all your queries. We're here to help at Westonci.ca. Keep visiting for the best answers to your questions.