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Math Help - exponential word problems

  1. #1
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    exponential word problems

    1.) Art Forgery. A painting attributed to Vermeer (1632 - 1675), a famous dutch painter, w/c should contain no more than 96.2% of its original carbon-14 contains 99.5% instead. About how old is the forgery? ( Carbon dating is a common technique of knowing or estimating the age of an object through its carbon content that undergoes a very slow decaying process)

    2.) Polonium - 210. The half-life of polonium is 140 days, but your sample will not be useful to you after 95% of the radioactive nuclei present on the day the sample arrives has disintegrated. For about how many days after the sample arrives will you be able to use the polonium? ( 1st, research on the meaning and equation of half-life" before proceeding to solve this problem

    3.) A beam of unknown temperature. An aluminum beam was brought into a machine shop where the temperature was held at 65 degrees F. After 10 minutes, the beam's temperature was 35 degrees F and after another 10 minutes it was 50 degrees Farenheight. Use Newton's Law of cooling to estimate the beam's initial temperature.

    (The formula is T - Ts = ( To - Ts) e ^ -kt
    where T is the temperature of the hot object at time t, Ts is the surrounding temperature that is assumed constant, To is the value of T at time zero and k is called the heat transfer coefficient, also a constant. The equation is also applicable to the case of a warming object.)


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  2. #2
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    1rst problem only

    Quote Originally Posted by ^_^Engineer_Adam^_^ View Post
    1.) Art Forgery. A painting attributed to Vermeer (1632 - 1675), a famous dutch painter, w/c should contain no more than 96.2% of its original carbon-14 contains 99.5% instead. About how old is the forgery? ( Carbon dating is a common technique of knowing or estimating the age of an object through its carbon content that undergoes a very slow decaying process)...
    Hi,

    the relative amount p of C14 can be calculated by:

    p = 100*e^(k*t) . t is a variable for the time. In case of Vermeer I take t = 350 years.

    First you have to calculate k:

    96.2 = 100 * e^(k*350), thus k = ln(.962) / 350 ≈ -0.000110688...

    Now you can calculate how old the picture in question is:

    99.5 = 100 * e^(-0.000110688*t). Solve for t:

    t = ln(0.995) / (-0.000110688) ≈ 45 years

    tschüss

    EB
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  3. #3
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    2nd problem only

    Quote Originally Posted by ^_^Engineer_Adam^_^ View Post
    1.) ...

    2.) Polonium - 210. The half-life of polonium is 140 days, but your sample will not be useful to you after 95% of the radioactive nuclei present on the day the sample arrives has disintegrated. For about how many days after the sample arrives will you be able to use the polonium? ( 1st, research on the meaning and equation of half-life" before proceeding to solve this problem
    ...
    Hi,

    with exponential decay use the formula:

    A(t) = A(0) * e^(k * t). A(t) is the amount of the material at the time t, k is a constant, which describes the speed of decay.

    0.5 = 1 * e^(k * 140). Solve for k = ln(0.5) / 140 = -ln(2) / 140

    Now you can calculate the time:

    0.05 = 1 * e^((-ln(2) / 140) * x). Solve for x = ln(0.05) / (-ln(2) / 140) ≈ 605 days

    tschüss

    EB
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  4. #4
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    final lap

    Quote Originally Posted by ^_^Engineer_Adam^_^ View Post
    1.)...
    3.) A beam of unknown temperature. An aluminum beam was brought into a machine shop where the temperature was held at 65 degrees F. After 10 minutes, the beam's temperature was 35 degrees F and after another 10 minutes it was 50 degrees Farenheight. Use Newton's Law of cooling to estimate the beam's initial temperature.

    (The formula is T - Ts = ( To - Ts) e ^ -kt
    where T is the temperature of the hot object at time t, Ts is the surrounding temperature that is assumed constant, To is the value of T at time zero and k is called the heat transfer coefficient, also a constant. The equation is also applicable to the case of a warming object.)
    Hi,

    Use the given formula, you'll get 2 equations:

    35 - 65 = (T(0)-65)*e^(-k*10)
    50 - 65 = (T(0)-65)*e^(-k*20)

    Now divide the sides of th first equation by the corresponding sides of the 2nd equation:

    2 = e^(k*10), thus k = ln(2) / 10 ≈ 0.0693...

    Now you can calculate T(0):

    35 - 65 = (T(0)-65)*e^(-0.0693*10)

    -30 + 65*e^(-0.693) = T(0)*e^(-0.693).

    Solve this equation for T(0). I leave this part for you.

    (You should come up with 37.

    tschüss

    EB
    Quote Originally Posted by ^_^Engineer_Adam^_^ View Post
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    Are you certain?
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  5. #5
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    Thanks very much!!!
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