Light intensity (I) is proportional (a) to the inverse square of distance (d) of a subject from a lightsource. The relationship in intensities for subjects at different distances from the same source canbe likewise seen as a ratio or proportional relationship.I x 1 1A proportional relationship can be mathematically expressed as follows for light intensity (I) inlumens when a subject is at the light source.=Lumens at origin or sourceSo, if a light intensity (I) is 569 lumens at the source, what is the light intensity (I) at 9 distancefrom the source,?Round the value to the nearest tenth if necessary. You do not need to include a label for lumens.Only the number, rounded to the tenth, will be necessary.

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ANSWER

The light intensity is 7.0

STEP-BY-STEP EXPLANATION:

What to find? The value of the proportionality constant.

Given parameters

• Light intensity = 569 lumens

,

• Distance = 9

According to the question, Light intensity (I) is proportional to the inverse square of the distance.

This can be expressed mathematically as

Let the intensity of light be represented as I

Let the distance be represented as d

[tex]I\text{ }\propto\text{ }\frac{1}{d^2}[/tex]

The next thing is to introduce a constant k

[tex]\begin{gathered} I\text{ = }\frac{K\cdot\text{ 1}}{d^2} \\ I\text{ = }\frac{K}{d^2} \end{gathered}[/tex]

Recall that,

I = 569 lumens

d = 9

The next thing is to substitute the parameters into the above formula

[tex]\begin{gathered} \frac{Lumens\text{ at current distance}}{\text{Lumens at origin or source}}\text{ = }\frac{1}{d^2} \\ \frac{\text{Lumens at current distance}}{\text{5}69}\text{ = }\frac{1}{(9)^2} \\ \frac{\text{Lumens at current distance}}{\text{5}69}\text{ = }\frac{1}{81} \\ \text{Cross multiply} \\ 569\cdot\text{ }1\text{ = Lumens at current distance }\cdot\text{ 81} \\ \text{Divide both sides by 81} \\ \frac{569}{81}\text{ = }\frac{Lumens\text{ at current distance }\cdot\text{ 81}}{81} \\ \text{Lumens at current distance = 7.0} \end{gathered}[/tex]