The intensity of a radio signal from the radio station varies inversely as the square of the distance from the station. Suppose the the intensity is 8000 units at a distance of 2 miles. What will the intensity be at a distance of 6 miles?

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The intensity of a radio signal from the radio station varies inversely as the square of the distance from the station. Suppose the the intensity is 8000 units at a distance of 2 miles. What will the intensity be at a distance of 6 miles?

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Answer 1 · 2016-10-07T20:44:11

$(A p p r .) 888.89 unit.$ $(Appr.) 888.89"unit."$

Explanation:

Let $I, and d$ $I, and d$ resp. denote the intensity of radio signal and the

distance in mile) of the place from the radio station.

We are given that, $I \propto \frac{1}{d^{2}} \Rightarrow I = \frac{k}{d^{2}}, or, I d^{2} = k, k \neq 0 .$ $I prop 1/d^2 rArr I=k/d^2, or, Id^2=k, kne0.$

When $I=8000, d=2 :. k=8000(2)^2=32000.$

Hence, $Id^2=k=32000$

Now, to find $I", when "d=6$

$:. I=32000/d^2=32000/36~~888.89"unit".$

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The intensity of a radio signal from the radio station varies inversely as the square of the distance from the station. Suppose the the intensity is 8000 units at a distance of 2 miles. What will the intensity be at a distance of 6 miles?

1 Answer

Explanation:

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