An important tool in archeological research is radiocarbon dating, developed by the American chemist Willard F. Libby.3 This is a means of determining the age of certain wood and plant remains, and hence of animal or human bones or artifacts found buried at the same levels. Radiocarbon dating is based on the fact that some wood or plant remains contain residual amounts of carbon-14, a radioactive isotope of carbon. This isotope is accumulated during the lifetime of the plant and begins to decay at its death. Since the half-life of carbon-14 is long (approximately 5730 years),4 measurable amounts of carbon-14 remain after many thousands of years. If even a tiny fraction of the original amount of carbon-14 is still present, then by appropriate laboratory measurements the proportion of the original amount of carbon-14 that remains can be accurately determined. In other words, if Q(t) is the amount of carbon-14 at time t and Q0 is the original amount, then the ratio Q(t)/Q0 can be determined, as long as this quantity is not too small. Present measurement techniques permit the use of this method for time periods of 50,000 years or more. Find an expression for Q(t) at any time t, if Q(0) = Qo.

Q(t) = Q_o*e^(-0.000120968*t)

Step-by-step explanation:

Given:

- The ODE of the life of Carbon-14:

                                       Q' = -r*Q

- The initial conditions Q(0) = Q_o

- Carbon isotope reaches its half life in t = 5730 yrs

Find:

The expression for Q(t).

Solution:

- Assuming Q(t) satisfies:

                                       Q' = -r*Q

- Separate variables:

                                      dQ / Q = -r .dt

- Integrate both sides:

                                       Ln(Q) = -r*t + C

- Make the relation for Q:

                                       Q = C*e^(-r*t)

- Using initial conditions given:

                                       Q(0) = Q_o

                                       Q_o = C*e^(-r*0)

                                      C = Q_o    

- The relation is:

                                       Q(t) = Q_o*e^(-r*t)

- We are also given that the half life of carbon is t = 5730 years:

                                       Q_o / 2 = Q_o*e^(-5730*r)

                                        -Ln(0.5) = 5730*r

                                        r = -Ln(0.5)/5730

                                        r = 0.000120968          

- Hence, our expression for Q(t) would be:

                                       Q(t) = Q_o*e^(-0.000120968*t)                                    

number one is the correct answer

step-by-step explanation: dogs and cats are mentioned, no other animals

1. 3

2. 4

3. 1

hope this ! have a great day.

-wfz