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Math Help - -HELP-

  1. #1
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    -HELP-

    After the release of radioactive material into the atmosphere from a nuclear power plant at Chernobyl (Ukraine) in 1986, the hay in Austria was contaminated by iodine 131 (half-life 8 days). If it is alright to feed the hay to cows when 10% of the iodine 131 remains, how long do the farmers need to wait to use this hay?
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  2. #2
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    by using

    N_t = N_0 (1/2)^{t/t_{1/2}}

    we get

    .1 = 1(1/2)^{t/8}

    now just solve for t, using logarithims

    i think thats how you set it up, if not change the 1 and 0.1 to 100 and 10
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  3. #3
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    Quote Originally Posted by treetheta View Post
    by using

    N_t = N_0 (1/2)^{t/t_{1/2}}

    we get

    .1 = 1(1/2)^{t/8}

    now just solve for t, using logarithims

    i think thats how you set it up, if not change the 1 and 0.1 to 100 and 10
    Um...I don't get it.
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  4. #4
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    No= Original
    So we know that the orginal has to be 100% so we have to let it equal 1, or we can let it equal 100, it doesn't matter

    Nt = Amount leftover
    So we want 10% to be leftover so it is usable to feed the cows thats why we let it equal .1 or 10, it doesnt matter,

    t1/2 = constant half-life decay
    Since it tells us in the questions that radiocative decay in the hay is 8 days, we can let this be 8 then when we solve for t

    We can find out how many days the hay hasn't to not be eaten for it to decay to 10% so it can be eatable for the cows

    I hope that clears it up

    you can always use the beloved,

    Half-life - Wikipedia, the free encyclopedia

    to clear things up


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  5. #5
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    Quote Originally Posted by treetheta View Post
    No= Original
    So we know that the orginal has to be 100% so we have to let it equal 1, or we can let it equal 100, it doesn't matter

    Nt = Amount leftover
    So we want 10% to be leftover so it is usable to feed the cows thats why we let it equal .1 or 10, it doesnt matter,

    t1/2 = constant half-life decay
    Since it tells us in the questions that radiocative decay in the hay is 8 days, we can let this be 8 then when we solve for t

    We can find out how many days the hay hasn't to not be eaten for it to decay to 10% so it can be eatable for the cows

    I hope that clears it up

    you can always use the beloved,

    Half-life - Wikipedia, the free encyclopedia

    to clear things up


    o k... i got this so far.
    ln .1 = ln (1/2)^(t/8)
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  6. #6
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    you forgot the 1, and u can bring the exponent down, by properties of logarithms, then isolate for t,

    if you need to know the logarithms:

    Logarithm - Wikipedia, the free encyclopedia
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  7. #7
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    Quote Originally Posted by treetheta View Post
    you forgot the 1, and u can bring the exponent down, by properties of logarithms, then isolate for t,

    if you need to know the logarithms:

    Logarithm - Wikipedia, the free encyclopedia
    o k...this is what I have now.
    log .1 = log 1(1/2)^(t/8)
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  8. #8
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    er...
    did you look at the wikipedia site?
    <br />
.1 = 1(1/2)^{t/8}<br />

    take ln of both sides

     ln.1 = ln(1/2)^{t/8}
    ln.1 = t/8[ln0.5]
    8(ln.1)/(ln0.5) = t
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