SOLUTION: The illumination of a small object by a lamp varies directly as the candle-power of the lamp and inversely as the square of the distance between the lamp and the object. If a light

Algebra ->  Test -> SOLUTION: The illumination of a small object by a lamp varies directly as the candle-power of the lamp and inversely as the square of the distance between the lamp and the object. If a light      Log On


   

Question 1204486: The illumination of a small object by a lamp varies directly as the candle-power of the lamp and inversely as the square of the distance between the lamp and the object. If a light-bulb of 8 candle-power, fixed 150 cm above a table, is replaced by a 5 candle-power bulb, how far must the new light be lowered to give the object the same illumination as before?
Answer by mananth(16946) About Me  (Show Source):
You can put this solution on YOUR website!
The illumination can be written as
I = k cp /d^2
I illumination
cp power of lamp
distance of object from lamp
I = k*cp/d^2

k*8/150^2= k*5/d^2
8 d^2 = 5*150^2
d^2 = (5/8)150^2
d = 150 sqrt(5/8)
d = 118.5cm
lower it 150-118.5 = 31.5 cm
The lamp has to be lowered by approx 31.37 cm to provide the same illumination as before.