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anonymous

  • one year ago

A hiker in Africa discovers a skull that contains 63% of its original amount of C-14. N=Noe^-kt No=initial amount of C-14(at time t=0) N=amount of C-14 at time t t=time, in years k=0.0001

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  1. anonymous
    • one year ago
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    Ok so we have: \[ N(t)=N_o e^{-kt}\]

  2. anonymous
    • one year ago
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    We are told that their is 63% of C14 remaining in the material hence: \[ \frac{N(t)}{N_o}= 63\% =0.63= e^{-kt}\]

  3. anonymous
    • one year ago
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    yes but it says find the age of the skull to the nearest year.

  4. anonymous
    • one year ago
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    Taking the ln of both sides: \[\ln(0.63)= -(0.0001 yr^{-1})t\]

  5. anonymous
    • one year ago
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    Divide and you have you answer

  6. anonymous
    • one year ago
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    divide 0.63 / -0.0001?

  7. anonymous
    • one year ago
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    Please note it isnt 0.63 but ln(0.63)

  8. anonymous
    • one year ago
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    And then it is a yes

  9. anonymous
    • one year ago
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    Btw, in case your worried about the minus sign... the fact that the argument of the logarithm is <1 means its result will be <0.... so the minus signs cancel and it will yield a positive value for t

  10. anonymous
    • one year ago
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    i got -6300

  11. anonymous
    • one year ago
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    natrual log of 0.63 divided by -10^-4

  12. anonymous
    • one year ago
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    Please note that the CRUCIAL step in this problem was taking the log of both sides. Without that step you cannot reduce the exponential factor.

  13. anonymous
    • one year ago
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    your confusing me

  14. anonymous
    • one year ago
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    What do you find confusing?

  15. anonymous
    • one year ago
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    Please, I cant help if I dont know where your getting confused.

  16. anonymous
    • one year ago
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    Go through the process I did at the top and let me know which step you are uneasy about and I will give a more detailed explanation.

  17. alekos
    • one year ago
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    show him the steps in detail

  18. anonymous
    • one year ago
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    I was under the impression I did alekos but I will go through it again.

  19. anonymous
    • one year ago
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    Ok so we given in the problem that the amount of C14 we find in the skull is 63%

  20. anonymous
    • one year ago
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    This means that of the original amount: \[N_0=100\% (some \ amount \ of \ nuclei)=(some \ amount \ of \ nuclei) \] It decays according to the law \[N(t)=N_0 e^{-kt}\] Where the N's correspond to the number of nuclei of the material in quesion remaining (N(t)) or the original amount (N_0) So that at time T, we have: \[N(T)=63\% (some \ amount \ of \ nuclei)=(some \ amount \ of \ nuclei) e^{-kT}\] Now whatever the total amount of nucleai there are in the sample is irrelevant because I can just divide both sides by that amount leaving just the percent remaining.

  21. anonymous
    • one year ago
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    This results in the equation I gave above: \[63\% = 0.63=e^{-kT}\]

  22. anonymous
    • one year ago
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    Now just like when we have an equation of the form: \[2=x^2\] Which we solve by applying the inverse function of squared (i.e. the square root) to both sides (ignoring the +/- that comes in for simplicity): \[\sqrt{2}=\sqrt{x^2}=x\] When we have an exponential equation of the form: \[2=e^x\] We apply the inverse function to the exponential function (aka the natural logarithm) to both sides in order to simplfy and solve: \[ \ln(2)=\ln(e^x)=x\]

  23. anonymous
    • one year ago
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    Note that an exponential function can NEVER be negative (unless I explicitly multiply it by a -1), which means when I do this inverse function business applying the logarithm, I dont have to worry about a +/- like you do when you apply a square root. So don't let that trouble you.

  24. anonymous
    • one year ago
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    So performing this step (taking the natural log of both sides of the equation) yields: \[\ln{0.63}=\ln{e^{-kT}}=-kT=-(10^{-4}yr^{-1})T\] Then solve for T, by dividing both sides by k: \[T=\frac{\ln{0.63}}{-(10^{-4}yr^{-1})}\]

  25. anonymous
    • one year ago
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    Please calculate this value and express it with the appropriate units here I and I will check your answer.

  26. anonymous
    • one year ago
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    @ix.ty are you still here?

  27. anonymous
    • one year ago
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    im on a diffrent question now

  28. anonymous
    • one year ago
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    A country's population in 1992 was 222 million. In 2001 it was 224 million. Estimate the population in 2004 using the exponential growth formula. Round your answer to the nearest million. P = Aekt

  29. anonymous
    • one year ago
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    So what did you get for the last one?

  30. anonymous
    • one year ago
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    i got it wrong that was my last question but i passed the test

  31. anonymous
    • one year ago
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    Oh so before I even try to start what looks to be a HARDER question on the same material it bears going over the last problem to iron out what confused you. Please tell me what you found confusing.

  32. anonymous
    • one year ago
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    No judgement, I fancy myself like a bit of an auto-mechanic.... I need to know where the problem is in order to try and fix it.

  33. alekos
    • one year ago
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    dont bother plasma, seems like you're wasting you're time

  34. anonymous
    • one year ago
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    Perhaps, but I am stubborn and banging my head against the wall might be a bit of a pastime for me.... But either way I am willing to help @ix.ty but I will only stick around for maybe 10-20 minutes (Ill check back at this tab) then I will move on

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