SOLUTION: The illumination (in foot-candles) of a light source varies directly as the intensity (in candle-power) of the source and inversely as the square of the distance from the source.

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Question 391349: The illumination (in foot-candles) of a light source varies directly as the intensity (in candle-power) of the source and inversely as the square of the distance from the source. If a certain light source with intensity of 300 candle-power provides an illumination of 2.5 foot-candles at a distance of 40 ft, what is the illumination at a distance of 20 ft?
Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
The illumination (in foot-candles) of a light source varies directly as the intensity (in candle-power) of the source and inversely as the square of the distance from the source.
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L = ki/d^2
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Solve for "k" using "If a certain light source with intensity of 300 candle-power provides an illumination of 2.5 foot-candles at a distance of 40 ft"
2.5 = k*300/40
k = 100/300
k = 3
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Equation for this problem:
L = 3i/d^2
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What is the illumination at a distance of 20 ft?
L = 3*300/20
L = 45 foot-candles
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Cheers,
Stan H.