The decibel level of sound is 50 dB greater on a busy street than in a quiet room where the intensity of sound is watt/m2. The level of sound in the quiet room is dB, and the intensity of sound in the busy street is watt/m2. Use the formula , where is the sound level in decibels, I is the intensity of sound, and is the smallest sound intensity that can be heard by the human ear (roughly equal to watts/m2).

Complete question is;

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).

A) The level of sound in the quiet room will be 20 dB

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

Step-by-step explanation:

Formula given is; β = 10log(I/I₀)

(a) For Quiet room:

We are given;

I = 10⁻¹⁰ W·m⁻²

I₀ = 1 × 10⁻¹² W·m⁻²

Plugging these values into the given equation, we have;

β = 10log[(10⁻¹⁰/(1 × 10⁻¹²)]

β = 10log(10²)

β = 10 × 2 = 20 dB

Thus, the level of sound in the quiet room will be 20 dB.

(b) For the Street;

We are given;

β(street) - β(room) = 50 dB

Now, let's rewrite the given intensity level equation;

β = 10logI - 10 logI₀

Now, Let the intensity level for the room be β₁ and let the intensity level for the road be β₂. Thus;

β₁ = 10logI₁ - 10log I₀ - - - - (eq 1)

β₂ = 10logI₂ - 10logI₀ - - - - (eq 2)

Subtract eq 1 from eq 2 to give;

β₂ - β₁ = 10logI₂ - 10logI₁

50 = 10logI₂ - 10log(10⁻¹⁰)

Divide each term by 10 to give:

5 = logI₂ - log(10⁻¹⁰)

5 = logI₂ - (-10)

5 = logI₂ + 10

Subtract 10 from each side to give;

-5 = logI₂

Taking the antilog of both sides to give;

I₂ = 10⁻⁵ W·m⁻²

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