A sound is recorded at 19 decibels. What is the intensity of the sound? - \(1 \times 10^{-8.7}\) W/m² - \(1 \times 10^{-10.1}\) W/m² - \(1 \times 10^{-11.9}\) W/m² - \(1 \times 10^{-9.4}\) W/m²

To determine the intensity of a sound in watts per square meter (W/m^2) based on its level in decibels (dB), we use the following formula that relates sound intensity to sound level:

L = 10 * log10(I / I_0)

where: - L is the sound level in decibels (dB), - I is the sound intensity in watts per square meter (W/m^2), - I_0 is the reference sound intensity, which is typically 1 x 10^-12 W/m^2 in air.

Rearranging the formula to solve for I:

I = I_0 * 10^(L/10)

Now let's use the given sound level of 19 dB to find the intensity:

I = 1 x 10^-12 W/m^2 * 10^(19/10) I = 1 x 10^-12 W/m^2 * 10^1.9 I = 1 x 10^-12 W/m^2 * 79.43 (since 10^1.9 is approximately 79.43)

Therefore, I ≈ 1 x 10^-12 W/m^2 * 79.43 I ≈ 7.943 x 10^-11 W/m^2

The correct standard form will be 1 x 10^-10.1 W/m^2 (as standard scientific notation requires the coefficient to be between 1 and 10).

Thus, the intensity of the sound is approximately 1 x 10^-10.1 W/m^2.