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Candis · Answer 1 · 2023-08-11T21:27:20+0000

ANSWER:

1000 times

Explanation:

Sound intensity level train = 90 dB

Sound intensity level machine = 120 dB

We can calculate the sound intensity in each case using the following formula:

$\begin{gathered} \beta =10\log _(10)\left((I)/(I_0)\right) \\ \\ I_0=10^(-12)\text{ W/m}^2 \end{gathered}$

We substitute each value and calculate I in each case:

$\begin{gathered} \text{ For the train:} \\ \\ 90=10\cdot\log_(10)\left((I_1)/(10^(-12))\right) \\ \\ \log_(10)\left((I_1)/(10^(-12))\right)=(90)/(10) \\ \\ \log_(10)\left((I_1)/(10^(-12))\right)=9 \\ \\ (I_1)/(10^(-12))=10^9 \\ \\ I_1=10^9\cdot\:10^(-12) \\ \\ I_1=0.001\text{ W/m}^2 \\ \\ \\ \text{ For the machine:} \\ \\ 120=10\cdot\log_(10)\left((I_2)/(10^(-12))\right) \\ \\ \log_(10)\left((I_2)/(10^(-12))\right)=(120)/(10) \\ \\ \log_(10)\left((I_2)/(10^(-12))\right)=12 \\ \\ (I_2)/(10^(-12))=10^(12) \\ \\ I_2=10^(12)\cdot\:10^(-12) \\ \\ I_2=1\text{ W/m}^2 \\ \\ \text{ Therefore:} \\ \\ (I_2)/(I_1)=(1)/(0.001)=1000 \\ \\ \text{ It is 1000 times greater is the sound intensity of the machine than that of the train} \end{gathered}$

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