SOLUTION: The intensity I of light from a light bulb varies inversely as the square of the distance d from the light bulb. Suppose I is 810 w/m^2 when the distance is 3 m. How much farther

Question 1021014: The intensity I of light from a light bulb varies inversely as the square of the distance d from the light bulb. Suppose I is 810 w/m^2 when the distance is 3 m. How much farther would it be to a point where the intensity is 160 w/m^2.
Answer by Theo(13342) (Show Source): You can put this solution on YOUR website!
The intensity I of light from a light bulb varies inversely as the square of the distance d from the light bulb. Suppose I is 810 w/m^2 when the distance is 3 m. How much farther would it be to a point where the intensity is 160 w/m^2.

the formula is i = k / d^2

i equals the intensity of the light in watts per square meter.
d equals the distances from the light bulb in meters.
k is the constant of variation.

i is 810 watts per square meter when d is 3 meters.

the formula if i = k / d^2 becomes 810 = k / 3^2.

this becomes 810 = k / 9.

solve for k to get k = 810 * 9 = 7290.

you are asked to determine how much farther away from the light bulb would a point be if the intensity of the light was 160 watts per square meter.

k is the constant of variation, so it stays the same.

the formula of i = k / d^2 becomes i = 7290 / d^2.

when i = 160, the formula becomes 160 = 7290 / d^2.

solve for d^2 to get d^2 = 7290 / 160 = 45.5625.

solve for d to get d = sqrt(45.5625) = 6.75 meters.

the intensity of light would be 160 watts per square meter when the distance from the light bulb is 6.75 meters.

that would be 6.75 - 3 = 3.75 meters further away.

that's your solution.

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