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Sound waves can be ranked by their intensity, [tex]I[/tex], given in the formula below, where [tex]r[/tex] is the distance from the source of a sound with a power output of [tex]P[/tex].

[tex]\[ P = 4 \pi I r^2 \][/tex]

Solve the formula for the distance from the source, [tex]r[/tex].

A. [tex] r = \frac{1}{2} \sqrt{\frac{P}{\pi I}} [/tex]

B. [tex] r = \frac{1}{4} \sqrt{\frac{P}{\pi I}} [/tex]

C. [tex] r = \frac{1}{2} \sqrt{\frac{\pi I}{P}} [/tex]

D. [tex] r = \frac{1}{4} \sqrt{\frac{\pi I}{P}} [/tex]

Sagot :

GinnyAnswer
Let's start with the given formula:

[tex]\[ P = 4 \pi I r^2 \][/tex]

We need to solve for [tex]\( r \)[/tex]. To do this, we will isolate [tex]\( r \)[/tex] on one side of the equation. Here are the steps:

1. Divide both sides by [tex]\( 4 \pi I \)[/tex] to isolate [tex]\( r^2 \)[/tex]:

[tex]\[ \frac{P}{4 \pi I} = r^2 \][/tex]

2. Take the square root of both sides to solve for [tex]\( r \)[/tex]:

[tex]\[ r = \sqrt{\frac{P}{4 \pi I}} \][/tex]

Notice that [tex]\( \sqrt{\frac{P}{4 \pi I}} \)[/tex] can be rewritten by factoring out the constants under the square root:

[tex]\[ r = \sqrt{\frac{P}{4 \pi I}} = \sqrt{\frac{P}{4 \pi I}} = \frac{\sqrt{P}}{\sqrt{4 \pi I}} \][/tex]

Since [tex]\(\sqrt{4} = 2\)[/tex], we have:

[tex]\[ r = \frac{\sqrt{P}}{2 \sqrt{\pi I}} = \frac{1}{2} \sqrt{\frac{P}{\pi I}} \][/tex]

Therefore, the correct answer is:

A. [tex]\( r = \frac{1}{2} \sqrt{\frac{P}{\pi I}} \)[/tex]
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