MATH SOLVE

3 months ago

Q:
# 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).

Accepted Solution

A:

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.