Algebra 2 Problems-Help!

**ep78** · Jul 30th 2008, 10:54 AM

1) The illumination of a light source is inversely proportional to the square of the distance from the light source. If a light has an illumination of 75 foot-candles at a distance of 4 feet from the light source:

a. Find a model for the illumination of this light

b. What is the illumination of this light at a distance of 8 feet from the light source?

c. How far from the light source would you need to be for the illumination to be 50 foot-candles?

2) The weight of a body varies inversely as the square of the distance from the earth. If a man weighs 200lbs on the top of Mt. McKinley (20320 feet above the surface of the earth), how much would he weigh when he is 3000 miles above the surface of the earth.

**masters** · Jul 30th 2008, 11:11 AM

Originally Posted by ep78

1) The illumination of a light source is inversely proportional to the square of the distance from the light source. If a light has an illumination of 75 foot-candles at a distance of 4 feet from the light source:

a. Find a model for the illumination of this light

$I=\frac{k}{d^2}$

$75=\frac{k}{16}$

$k=1200$ the constant of porportionality

b. What is the illumination of this light at a distance of 8 feet from the light source?

For a distance of 8 feet: $I=\frac{1200}{64}\Longrightarrow I=18.75$

c. How far from the light source would you need to be for the illumination to be 50 foot-candles?

$50=\frac{1200}{d^2}\Longrightarrow d=4\sqrt{6}$

Here's your first one.

**abender** · Jul 30th 2008, 01:24 PM

w = weight
d = distance

$w = \frac{k}{d^2}$

$w_{1} = 200$ lbs
$d_{1} = 20320$ feet

$w_{2} = ???$ lbs
$d_{2} = 15840000$ feet (converted from 3000 miles)

Let's plug in $w_{1}$ and $d_{1}$ in our equation. This will allow us to determine k.

$w_{1} = \frac{k}{d_{1}^2}$

$200 = \frac{k}{20320^2}$

$200 = \frac{k}{412902400}$

$k = 412902400 * 200 = 82580480000$

Now that we know $k$ and are given $d_{2}$ , we can use our equation once again, this time solving for $w_{2}$ , which is unknown.

$w_{2} = \frac{82580480000}{d_{2}^2}$

$w_{2} = \frac{82580480000}{15840000^2} = \frac{82580480000}{250905600000000} = 0.000329129681$

So, the man weighs about 0.00033 lbs when he is 3000 miles above the surface of the earth.

Check my arithmetic.
Good luck!
-Andy

**ep78** · Jul 30th 2008, 03:12 PM

thanks so much!

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