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eimsori [14]
3 years ago
11

adiocarbon dating of blackened grains from the site of ancient Jericho provides a date of 1315 BC ± 13 years for the fall of the

city. What is the relative amount of 14C in the old grain vs the new grain in 2007 AD? (A0 = original radioactivity; At = radioactivity in 2007 AD).
Mathematics
1 answer:
Zigmanuir [339]3 years ago
4 0

Answer:

\left(\frac{m(t)}{m_{o}} \right)_{min} \approx 0.659 and \left(\frac{m(t)}{m_{o}} \right)_{max} \approx 0.661

Step-by-step explanation:

The equation of the isotope decay is:

\frac{m(t)}{m_{o}} = e^{-\frac{t}{\tau} }

14-Carbon has a half-life of 5568 years, the time constant of the isotope is:

\tau = \frac{5568\,years}{\ln 2}

\tau \approx 8032.926\,years

The decay time is:

t = 1315\,years + 2007\,years \pm 13\,years (There is no a year 0 in chronology).

t = 3335 \pm 13\,years

Lastly, the relative amount is estimated by direct substitution:

\frac{m(t)}{m_{o}} = e^{-\frac{3335\,years}{8032.926\,years} }\cdot e^{\mp\frac{13\,years}{8032.926\,years} }

\left(\frac{m(t)}{m_{o}} \right)_{min} = e^{-\frac{3335\,years}{8032.926\,years} }\cdot e^{-\frac{13\,years}{8032.926\,years} }

\left(\frac{m(t)}{m_{o}} \right)_{min} \approx 0.659

\left(\frac{m(t)}{m_{o}} \right)_{max} = e^{-\frac{3335\,years}{8032.926\,years} }\cdot e^{\frac{13\,years}{8032.926\,years} }

\left(\frac{m(t)}{m_{o}} \right)_{max} \approx 0.661

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\huge \bf༆ Answer ༄

Here's the solution ~

Area of floor is ;

{ \qquad{ \sf{ \dashrightarrow}}}  \:  \: \sf \:10 \times 12

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{ \qquad{ \sf{ \dashrightarrow}}}  \:  \: \sf \:90 \: ft {}^{2}

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{ \qquad{ \sf{ \dashrightarrow}}}  \:  \: \sf \: \dfrac{90}{120}  \times 100

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{ \qquad{ \sf{ \dashrightarrow}}}  \:  \: \sf \:3 \times 25

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6 0
3 years ago
The area of a rectangle is 56 cm. The length is 2 cm more than x and the width is 5 cm less than twice x. Solve for x. Round to
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 Length = x + 2  (because it is 2 cm more than x)
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Hope I helped and sorry it was really long
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