From reliable sources in the internet, the half-live of carbon-14 is given to be 5,730 years. In a span of 10,000 to 12,000 years, there are almost or little more than 2 half-lives. Thus, there should be
A(t) = A(0)(1/2)^t
where t is the number of half-lives, in this case 2. Thus, only about 1/4 of the original amount will be left.
Answer:
The period of rotation is
T=8.025s
Explanation:
The person is undergoing simple harmonic motion on the wheel
Given data
mass of the person =75kg
Radius of wheel r=16m
Velocity =8.25m/s
The oscillating period of simple harmonic motion is given as
T=(2*pi)/2=2*pi √r/g
Assuming that g=9.81m/s
Substituting our data into the expression we have
T=2*3.142 √ 16/9.81
T=6.284*1.277
T=8.025s
A. nucleus
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