Final answer:
The question asks for the decibel level produced by a sound wave with pressure peaks of 1/10 of atmospheric pressure, but it requires the sound pressure in Pascals to compute the level. Without knowing the specific pressure in Pascals or a proper reference pressure, it's not possible to give a definitive answer. Moreover, the decibel level directly calculated from the pressure ratio given in the question without further conversion could be incorrect.
Step-by-step explanation:
To determine the decibel level (dB) produced by a sound wave with pressure peaks of 1/10 of normal atmospheric pressure, we can use the formula for sound pressure level in decibels:
L_p = 20 × log10(p/p_ref)
where L_p is the sound pressure level in decibels, p is the sound pressure, and p_ref is the reference sound pressure.
Nominal atmospheric pressure is approximately 101,325 Pascals (Pa), so 1/10 of this is 10,132.5 Pa. However, when measuring sound pressure in decibels, a common reference pressure p_ref of 20 μPa is used, which corresponds to the threshold of human hearing at a frequency of 1000 Hz. This reference pressure is equivalent to 2 × 10⁻µ atm, which is much less than the atmospheric pressure. Using the reference pressure of 20 μPa for the dB scale, we would first need to convert the pressure peaks from atm to Pa to make use of the formula correctly. As the peaks are 1/10 of normal atmospheric pressure:
p = (1/10) × 101,325 Pa = 10,132.5 Pa
Now, if we apply the formula:
L_p = 20 × log10(10,132.5 / 20 μPa)
Calculating the decibel level directly from the pressure ratio provided without further conversion could produce an incorrect answer. Therefore, it is unclear which decibel level the sound wave produces without additional information regarding the sound wave's pressure in Pascals or its comparison to the reference pressure used in sound level calculations.