SOLUTION: the intensity at which an object is illuminated varies inversely as the square of its distance from the light source at 1.5 feet from a light bulb the intensity of the light is 60

Algebra ->  Rate-of-work-word-problems -> SOLUTION: the intensity at which an object is illuminated varies inversely as the square of its distance from the light source at 1.5 feet from a light bulb the intensity of the light is 60       Log On


   

Question 1183768: the intensity at which an object is illuminated varies inversely as the square of its distance from the light source at 1.5 feet from a light bulb the intensity of the light is 60 units what is the intensity at 1 feet
Found 2 solutions by Boreal, greenestamps:
Answer by Boreal(15235) About Me  (Show Source):
You can put this solution on YOUR website!
I=k/d^2 is the basic formula
60=k/1.5^2
k=135
so
I=135/1^2
=135 units at one foot.

Answer by greenestamps(13200) About Me  (Show Source):
You can put this solution on YOUR website!


The ratio of the new to the old distance is 1/1.5 = 2/3, so the intensity is changed by a factor of %281%2F%282%2F3%29%29%5E2+=+%283%2F2%29%5E2+=+9%2F4

ANSWER: 60%2A%289%2F4%29+=+9%2A15+=+135