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⁻²

The intensity of sound in the busy street is 10⁻⁵ W·m⁻².

the answer is a i belaeve atleast thats what my calculator just gfave me

step-by-step explanation:

answer: idk

step-by-step explanation: