Physics: For a 225-watt bulb, the intensity I of light in lumens at a distance of x feet is I=225/x^2.
a. What is the intensity of light 5 ft from the bulb?
b. Suppose your distance from the bulb doubles. How does the intensity of the light change? Explain.
a. what is 225/5^2 ?
b. doubles? then it must be .5^2 as much, or 25 percent, or 1/4
That helps, thanks.
a. To find the intensity of light 5 feet from the bulb, we can plug in x = 5 into the equation I = 225/x^2 and calculate it.
Substituting x = 5 into the formula, we get:
I = 225/(5^2)
I = 225/25
I = 9 lumens
Therefore, the intensity of light 5 feet from the bulb is 9 lumens.
b. If the distance from the bulb doubles, we can use the relationship between intensity and distance to determine how the intensity changes.
Let's say the new distance is 2x feet. We can substitute x = 2x into the equation I = 225/x^2 and calculate it.
Substituting x = 2x into the formula, we get:
I = 225/(2x)^2
I = 225/4x^2
I = (225/4)*(1/x^2)
I = (1/4)*(225/x^2)
I = (1/4)*I
Therefore, when the distance from the bulb doubles, the intensity of the light decreases by a factor of 4. In other words, the intensity is reduced to 1/4th of its original value.