ttps://ww4.math.msu.edu/webwork2/mth 235 fs1784740/Hw06 1.5-Applications/1/ At l
ID: 2895088 • Letter: T
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ttps://ww4.math.msu.edu/webwork2/mth 235 fs1784740/Hw06 1.5-Applications/1/ At least one of the answers above is NOT correct. This problem has 2 part(s) displayed simultaneously. A similar problem is solved in Example 1.5.1, in Section 1.5, in the MTH 235 Lecture Notes. (10 points) Part 1/2. Remains have been found containing 10 % of the original amount of Carbon-14. Recall that the Carbon-14 half-life is 0-5730 years. Then, answer the following a. (4/10) Find the Carbon-14 decay constant, that is, the positive number k such that N'-k N, where N is the amount of Carbon-14 in a dead organism as function of time, t, with t 0 at the time the organism died. k = | In(1/10/5730 Part 2/2. b. (6/10) Date the remains, that is, find the number t, the number of years since the organism died. 11 = | In(1/10/5730 Preview My Answers Submit AnswersShow me another Your score was recorded. fou have attempted this problem 6 times. ou received a score of 0% for this attempt. Your overall recorded score is 0%. You have 19 attempts remaining.Explanation / Answer
a)
N'=kN
dN/dt =-kN
dN/N=-k dt
dN/N=-k dt
ln(N)=-kt +c
=>N=e-kt+c
=>N=Ce-kt
C =initial amount
halflife =5730 years
=>(C/2)=Ce-k*5730
=>ek*5730=2
=>5730k=ln(2)
=>k=(ln(2))/5730--------------> eact answer
=>k=0.000121 ------------------>approximate answer
=>N=Ce-t*(ln(2))/5730
(b)
remains contain 10% of original amount of carbon-14
=>(10/100)C=Ce-t*(ln(2))/5730
=>(1/10)=e-t*(ln(2))/5730
=>et*(ln(2))/5730=10
=>t*(ln(2))/5730= ln(10)
=>t= 5730*(ln(10))/ln(2)-------------------------------------------> exact answer
=>t=19034.65 years ----------------------------------------------->approximate answer
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