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9 September, 20:00

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).

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  1. 9 September, 20:10
    0
    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:

    dа ta:

    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

    dа ta:

    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.
  2. 9 September, 20:27
    0
    cant see the answer
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