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toxicsugar22

  • 5 years ago

Dichloro- is a peticide that was used in the middle decades of the twenith century to control maleria. After 1945, It was widely used on crops in the US, AND AS MUCH AS one ton might be sprayed on a single field. However, after the toxic effects of Dichloro on the environment began to appear, the chemical was banned in 1972. A common estimate for the half-life of Dichloro in the soil is 15 years. Write a decay law for Dichloro in the soil?

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  1. anonymous
    • 5 years ago
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    Proportional growth and decay are governed by the differential equation,\[\frac{dN}{dt}=k(N-P)\]where k and P are constants. It leads to an expression for N of the form,\[N(t)=N(0)e^{kt}\]where P is zero here. You'll see that, at time t=0, the amount of substance hanging around is N(0). If you're given initial data, you can substitute it in here. Also, you're told that 15 years is the half life. This means, if you start with N(0) amount of dichloro at time t=0, then at time t=15, you'll have N(15)=N(0)/2. But that means, looking at the formula,\[\frac{N(0)}{2}=N(0)e^{15k}\rightarrow \frac{1}{2}=e^{15k}\rightarrow \log \frac{1}{2} =15k\]that is,\[k=-\frac{\log 2}{15}\]So your equation is\[N(t)=N(0)e^{-\frac{\log 2}{15}t}\]

  2. anonymous
    • 5 years ago
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    I use log for the natural logarithm, ln.

  3. anonymous
    • 5 years ago
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    You can also note, using exponent laws and definition of the log, that\[e^{-\frac{\log 2}{15}t}=\left( e^{\log 2} \right)^{-\frac{t}{15}}=2^{\frac{-t}{15}}\]so that\[N(t)=N(0)2^{-t/15}\]

  4. toxicsugar22
    • 5 years ago
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    OK nOW AS A PART OF THIS QUESTION In 1970, many soils samples in the US contained about 0.5 mg of Dichloro per kg of soil. The Mational safe level for Dichloro in the soil is 0.008 mg/kg. when will Dichloro content in the soil be reduced to a safe level?

  5. toxicsugar22
    • 5 years ago
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    hurry up

  6. anonymous
    • 5 years ago
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    Take a basis of 1kg of soil. Then N(0)=0.5mg. You need t such that N(t)=0.008mg. \[N(t)=N(0)2^{-t/15} \rightarrow \log N(t)=\log [N(0)2^{-t/15}]=\log N(0)+\log2^{-t/15}\]\[=\log N(0)-\frac{t}{15}\log 2\]so\[\log N(t)=\log N(0)-\frac{t}{15}\log 2 \rightarrow t=\frac{15(\log N(0)-\log N(t))}{\log 2}=\frac{15(\log 0.5-\log 0.008)}{\log 2}\]\[\approx89.5 yrs\]

  7. anonymous
    • 5 years ago
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    and that's f***ing rude.

  8. toxicsugar22
    • 5 years ago
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    srry

  9. toxicsugar22
    • 5 years ago
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    do u forgive

  10. anonymous
    • 5 years ago
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    ok

  11. toxicsugar22
    • 5 years ago
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    the number of websites is growing by 15% each month. In June 2005, there were 4000 websites. Write a formula for the number of websites as a function of months since june 2005.

  12. anonymous
    • 5 years ago
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    \[N(t)=4000(1.15)^t\]

  13. toxicsugar22
    • 5 years ago
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    Now with that question how long will it take fr the number of sites to reach 8000? Give the answer in exact value and decimal form

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