the intensity I of light from a light bulb varies inversely as the square of the distance d from the light bulb. Supplose I is 810m/m^2 when the distance is 7m. How much farther would it be to a point where the intensity is 250 w/m^2?

-is the fomula as 250=7^2k??

Question

the intensity I of light from a light bulb varies inversely as the square of the distance d from the light bulb. Supplose I is 810m/m^2 when the distance is 7m. How much farther would it be to a point where the intensity is 250 w/m^2?

-is the fomula as 250=7^2k??

anonymous · Answer

Inverse variation is essentially y = x/ k or yk = x. So in your case I believe you could say l*d^2 = 810. Then plug in 250 for l and solve for d. Not 100% sure this is right but it's my idea.

unklerhaukus · Answer

$$I(d)\propto \frac1{d^2}$$$$I(d)=\frac k{d^2}$$
$$810 \frac{ [\text W]}{[\text m^2]}=\frac{k}{7[\text m]^2}$$$$7[\text m]^2\times810 \frac{ [\text W]}{[\text m^2]}=k$$$$k=7\times810[\text W] $$

$$I(d)=\frac k{d^2}=250 \left[\frac{\text W}{\text{m}^2}\right]$$
solve for $d$

unklerhaukus · Answer

$$\cancel{k=7\times810[\text W]}$$
$$k=7^2\times810[\text W]$$

anonymous · Answer

Since it's an inverse square, you can set it up as:
$$\frac{r_2^2}{r_1^2} = \frac{Intensity_1}{Intensity_2}$$

anonymous · Answer

r is the distance. Then you have to subtract 7 from your answer to get how much further than 7 the new distance is.

anonymous · Answer

i don't get it lol

unklerhaukus · Answer

pick one of the three methods people have offered

anonymous · Answer

You have two equations that have 1/r^2 = the intensity
If you divide the two equations, you end up with the equation I wrote, above