The decibel level of sound is 50 dB greater on a busy street than in a quiet room where the intensity of sound is 10^-10 watt/m2. The level of sound in the quiet room is (10,20,100) dB, and the intensity of sound in the busy street is (10^-1, 10^-5, 10^-10) watt/m2. Use the formula , β = 10log I/I 0 where β is the sound level in decibels, I is the intensity of sound you are measuring, and Io is the smallest sound intensity that can be heard by the human ear (roughly equal to 1 x 10^-12 watts/m2).

Answers:(a) 20 dB; (b) 10⁻⁵ W·m⁻² Step-by-step explanation:Step-by-step explanation: β = 10log(I/I₀) (a) Quiet room: Data: I =        10⁻¹⁰ W·m⁻² I₀ = 1 × 10⁻¹² W·m⁻² Calculation: β = 10log[(10⁻¹⁰/(1 × 10⁻¹²)] = 10log(10²) = 10 × 2 = 20 dB The level of sound in the quiet room is 20 dB. (b) Street Data: β(street) - β(room) = 50 dB Calculations: Let's rewrite the intensity level equation as β = 10logI - 10 logI₀ Let 1 = the room and 2 = the road. Then, (1) β₂ = 10logI₂ - 10logI₀ (2) β₁ = 10logI₁ - 10log I₀ Subtract (2) from (1)                β₂ - β₁ = 10logI₂  - 10logI₁                                                   50    = 10logI₂  - 10log(10⁻¹⁰) Divide each side by 10               5   =    logI₂   -     log(10⁻¹⁰)                                                      5   =    logI₂   -       (-10)                                                      5   =      logI₂ +       10 Subtract 10 from each side        -5   =      logI₂ Take the antilog of each side       I₂ = 10⁻⁵  W·m⁻²The intensity of sound in the busy street is 10⁻⁵ W·m⁻².