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Thread: Algebra 2 Problems-Help!

  1. #1
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    Algebra 2 Problems-Help!

    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.

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  2. #2
    A riddle wrapped in an enigma
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    Quote Originally Posted by ep78 View Post
    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

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

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

    $\displaystyle k=1200$ 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: $\displaystyle I=\frac{1200}{64}\Longrightarrow I=18.75$

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

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



    Here's your first one.
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  3. #3
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    w = weight
    d = distance

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

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

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

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

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

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

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

    $\displaystyle k = 412902400 * 200 = 82580480000 $

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

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

    $\displaystyle 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
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  4. #4
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    thanks so much!
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