In this scenario we know that carbon14 at a given time A(t) = A0e^(-kt), where A0 is the original carbon present, t is time in years and k is the constant.
As we are working with the half life of carbon being 5730 years, we assume original carbon-14 content, A0 = 1, and carbon-14 at half life 5730 years, A(t) = 0.5.
i.e.
0.5 = 1e^(-5730k)
apply Ln to both sides of equation to cancel e
ln(0.5) = -5730k
k = ln(0.5) / -5730
k = -0.69315 / - 5730 = 1.20968 x 10^-4
9514 1404 393
Answer:
- 4/13 h ≈ 0.308 h
- 22 9/13 km ≈ 22.7 km
Step-by-step explanation:
Mary and Amaruk are starting 35 -15 = 20 km apart. Their closing speed is 40 +25 = 65 km/h. The time it will take to close the gap between them is ...
t = (20 km)/(65 km/h) = 20/65 h = 4/13 h ≈ 0.308 h
They will pass each other after 4/13 hours, about 18 minutes 28 seconds.
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In that time, Mary will have increased her distance from Hall Beach by ...
d = (4/13 h)(25 km/h) = 100/13 km = 7 9/13 km
Her distance from Hall Beach when she passes Amaruk will be ...
15 km + 7 9/13 km = 22 9/13 km ≈ 22.7 km . . . . . distance from Hall Beach