Answered

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The relationship between the intensity of sound((w)/(m^(2)))and the distance from the source of sound is represented by the equation (i)=((k)/(d^(2))).If you sit only one meter away from the stage at a rock concert the intensity of sound is about 0.1(w)/(m^(2)). If vanessa does not want the sound intensity to be more than 0.001(w)/(m^(2)) how close to the stage can she sit?

Sagot :

lublana

Answer:

10 m

Step-by-step explanation:

We are given that the relationship between intensity of sound and distance from the source is given by

Intensity (I)=[tex]\frac{k}{d^2}[/tex]

Where I=Intensity

d=Distance of sound from the source

If distance d=1m

Intensity, I=0.1[tex]w/m^2[/tex]

We have to find how close she can to the stage if vanessa does not want the sound intensity to be more than 0.001(w)/(m^(2)).

Using the formula

[tex]0.1=\frac{k}{1}[/tex]

[tex]k=0.1[/tex]

[tex]\frac{0.1}{d^2}\leq 0.001[/tex]

[tex]\frac{0.1}{0.001d^2}\leq 1[/tex]

[tex]\frac{0.1}{0.001}\leq d^2[/tex]

[tex]100\leq d^2[/tex]

[tex]10\leq d[/tex]

[tex]d=10 m[/tex]

Hence, she can sit 10 m away from the stage.

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