SOLUTION: In the video, I show measurements of light intensity (measured in lux) from a flashlight at several distances. With the light off, the ambient room light was 239 lux. At 10 cm, the

Algebra ->  Finance -> SOLUTION: In the video, I show measurements of light intensity (measured in lux) from a flashlight at several distances. With the light off, the ambient room light was 239 lux. At 10 cm, the      Log On


   

Question 1103292: In the video, I show measurements of light intensity (measured in lux) from a flashlight at several distances. With the light off, the ambient room light was 239 lux. At 10 cm, the light sensor read 2513 lux. Unfortunately, the flashlight flickered at 20 cm and the reading wasn't accurate.
L=k/x^2+L ambient
a. Using the data, complete the equation for light in terms of distance, x.
b. If we wanted a light intensity of 1400 lux,at what distances should we place the light .
Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
L=k%2Fx%5E2%2BL%5Bambient%5D
Substituting L%5Bambient%5D=239
and the coordinates system%28x=10%2CL=2513%29 of our data point,
2513=k%2F10%5E2%2B239
2513-239=k%2F100
2274=k%2F100
2274%2A100=k
k=227400
highlight%28L=227400%2Fx%5E2%2B239%29
To get L=1400 ,
1400=227400%2Fx%5E2%2B239%29
means
1400-239=227400%2Fx%5E2
1161=227400%2Fx%5E2
1161x%5E2=227400
x%5E2=227400%2F1161=75800%2F387=100%2A758%2F%289%2A43%29
There are two solutions to that equation,
but we are looking for a positive distance, so
x=sqrt%28227400%2F1161%29=sqrt%2875800%2F758%29
At this point, you could just reach for the calculator,
because it does not simplify much.
The approximate answer from the calculator is
x=13.9952 ,
so we should place the light at highlight%2814cm%29 from whatever point we want with 1400 lux of light intensity.