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24 April, 01:01

A student is standing 8 m from a roaring truck engine that is measured at 20 db. The student moves 4 m closer to the engine. What is the measured sound intensity at the new distance?

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  1. 24 April, 01:20
    0
    The answer to the question above is explained below

    Explanation:

    The loudness or sound intensity of the roaring truck engine is inversely proportional to the square of the distance.

    r1^2 / r2^2 = L2/L1

    Where, r1 and r2 = old distance and new distance respectively,

    L1 and L2 = old sound intensity and new sound intensity respectively.

    From the parameters given above,

    r1=8m

    r2=4m

    L2=?

    L1 = 20dB

    8^2 / 4^2 = L2 / 20 dB

    L2 = 80 dB

    The answer is 80 dB. Increased from 20 dB.

    Decibel (dB), is a relative unit of measurement used for expressing the ratio between two physical quantities, usually electric power, or for measuring Sound intensity. Decibels measure the ratio of a given intensity to the threshold of hearing intensity.

    Sound intensity, also known as acoustic intensity, is a vector describing the intensity of a sound is proportional to its amplitude squared.
  2. 24 April, 01:24
    0
    80 dB

    Explanation:

    Sound intensity is inversely proportional to the square of the distance traveled by the sound.

    Thus;

    r1^2 divide by r2^2 = L2divideL1

    where r = distance (m)

    L = sound intensity (decibal dB)

    8squared divide by 4 squared = L2 divide 20 dB

    L2 = 4 * 20

    L2 = 80 dB

    This means that the sound is increased from 20 dB to 80 dB.
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