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

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

where:

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

We are given:

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

We need to substitute these values into the formula to find $D$.

Step-by-step Solution:

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

First, calculate the ratio inside the logarithm:

$\frac{8.5 \times 10^2}{1 \times 10^{-12}} = 8.5 \times 10^{14}$

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

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

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

So:

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

Now multiply by 10:

$D = 10 \times 14.929 = 149.29$

Final Answer:

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

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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 $1 \times 10^{-6}$ 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 $\log(ab) = \log(a) + \log(b)$, which simplifies many calculations involving large numbers.