They retrieved two ice cores at the bottom of the glacier, more than 132 meters (433 feet) deep.Trapped within the cores were insects and bark fragments from local trees.A 1998 study provided evidence that the tropics were much colder during the last glacial maximum than previously thought.Prior understanding had been that tropical regions were mostly unaffected by past ice ages.Since a radioactive substance such as carbon 14 has a known half life the amount of carbon 14 remaining in an object can be used to date that object. In 1991, hikers in the Tyrolean Alps of Europe made a remarkable discovery.Carbon from organic material near the bottom of the cores dated to the coldest period of the last ice age.
Ötzi’s fate was matched by a variety of well-preserved plant and animal species that were found close by.
Only round off the final answers.) Now that we know k, we can calculate the carbon 14 decimal amount remaining after 5300 years. N(t) = N(0)e^kt N(5300) = N(0)e^kt since we are dealing with percentages, we know that 100% is the original amount, and that is decimal 1.0. N(5300) = (1.0) * e^(-0.000120968 * 5300) N(5300) = e^(-0.000120968 * 5300) N(5300) = e^(-0.6411309) N(5300) = e^(-0.6411309) N(5300) = 0.526696446 = 52.67% remaining 2) N(t) = N(0)e^kt Since 5.5% = 0.055 in decimal, then N(t) = (0.055)N(0).
(0.055)N(0) = N(0)e^(-0.000120968 * t) cancel out the N(0) on both sides and replace with 1's (0.055) * 1 = 1 * e^(-0.000120968 * t) 0.055 = e^(-0.000120968 * t) take natural log of both sides ln(0.055) = ln(e^(-0.000120968 * t)) -2.900422094 = -0.000120968 * t t = 23976.75279 = 23,977 years ago 3) Since we know that 8% of the carbon has been lost, then 92% remains (100 - 8 = 92).
Since we know the time and the amounts, we can calculate the constant k.
N(5730) = [N(0)e^(k*5730)] Since after 5730 years, 1/2 will remain, then N(t) = N(0)/2.