Math Problem Statement

solve for the indicated value, and graph the situation showing the solution point. The formula for measuring sound intensity in decibels D is defined by the equation D=10log(I/I[0]) using the common (base 10) logarithm. where I is intensity of the sound in watts per square meter and I[0]=1-^-12 is the lowest level of sound that the average person can hear. How many decibels are emitted from a jet plane with a sound in tensity of 8.5*10^2 watts per square meter?

round answer to three decimal places.

Solution

The formula for measuring sound intensity in decibels (D) is given as:

D=10log(II0)D = 10 \log \left( \frac{I}{I_0} \right)

where:

  • II is the intensity of the sound in watts per square meter,
  • I0=1×1012I_0 = 1 \times 10^{-12} is the threshold of hearing, or the lowest intensity of sound the average person can hear.

We are given:

  • I=8.5×102I = 8.5 \times 10^2 watts per square meter.

We need to substitute these values into the formula to find DD.

Step-by-step Solution:

D=10log(8.5×1021×1012)D = 10 \log \left( \frac{8.5 \times 10^2}{1 \times 10^{-12}} \right)

First, calculate the ratio inside the logarithm:

8.5×1021×1012=8.5×1014\frac{8.5 \times 10^2}{1 \times 10^{-12}} = 8.5 \times 10^{14}

Now, take the logarithm (base 10) of 8.5×10148.5 \times 10^{14}:

log(8.5×1014)=log(8.5)+log(1014)\log(8.5 \times 10^{14}) = \log(8.5) + \log(10^{14})

log(8.5)0.929andlog(1014)=14\log(8.5) \approx 0.929 \quad \text{and} \quad \log(10^{14}) = 14

So:

log(8.5×1014)0.929+14=14.929\log(8.5 \times 10^{14}) \approx 0.929 + 14 = 14.929

Now multiply by 10:

D=10×14.929=149.29D = 10 \times 14.929 = 149.29

Final Answer:

The decibels emitted by the jet plane are approximately 149.290 dB.


Would you like me to generate a graph for this situation? Here are 5 related questions for further exploration:

  1. What is the intensity of sound at a decibel level of 100 dB?
  2. How does the sound intensity change as the decibel level increases by 10 dB?
  3. What is the decibel level for a sound with intensity 1×1061 \times 10^{-6} watts per square meter?
  4. If the intensity of a sound increases tenfold, by how many decibels does the sound level increase?
  5. Can the decibel level ever be negative? If so, what does that represent?

Tip: When working with logarithms, recall that log(ab)=log(a)+log(b)\log(ab) = \log(a) + \log(b), which simplifies many calculations involving large numbers.

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Math Problem Analysis

Mathematical Concepts

Logarithms
Exponents
Sound Intensity

Formulas

D = 10 log(I/I0)
log(a * b) = log(a) + log(b)

Theorems

Properties of Logarithms
Sound Intensity to Decibel Conversion

Suitable Grade Level

Grades 11-12