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katen-ka-za [31]
3 years ago
9

What is the length of any side of equilateral triangle TUV? a. 7 b. 15 c. 20 d. 24

Mathematics
1 answer:
NikAS [45]3 years ago
6 0
D
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the half life of c14 is 5730 years. Suppose that wood found at an archeological excavation site contains about 35% as much C14 a
Furkat [3]

Answer:

The wood was cut approximately 8679 years ago.

Step-by-step explanation:

At first we assume that examination occured in 2020. The decay of radioactive isotopes are represented by the following ordinary differential equation:

\frac{dm}{dt} = -\frac{m}{\tau} (Eq. 1)

Where:

\frac{dm}{dt} - First derivative of mass in time, measured in miligrams per year.

\tau - Time constant, measured in years.

m - Mass of the radioactive isotope, measured in miligrams.

Now we obtain the solution of this differential equation:

\int {\frac{dm}{m} } = -\frac{1}{\tau}\int dt

\ln m = -\frac{1}{\tau} + C

m(t) = m_{o}\cdot e^{-\frac{t}{\tau} } (Eq. 2)

Where:

m_{o} - Initial mass of isotope, measured in miligrams.

t - Time, measured in years.

And time is cleared within the equation:

t = -\tau \cdot \ln \left[\frac{m(t)}{m_{o}} \right]

Then, time constant can be found as a function of half-life:

\tau = \frac{t_{1/2}}{\ln 2} (Eq. 3)

If we know that t_{1/2} = 5730\,yr and \frac{m(t)}{m_{o}} = 0.35, then:

\tau = \frac{5730\,yr}{\ln 2}

\tau \approx 8266.643\,yr

t = -(8266.643\,yr)\cdot \ln 0.35

t \approx 8678.505\,yr

The wood was cut approximately 8679 years ago.

5 0
3 years ago
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