Answer:
8. 2.75·10^-4 s^-1
9. No, too much of the carbon-14 would have decayed for radiation to be detected.
Explanation:
8. The half-life of 42 minutes is 2520 seconds, so you have ...
1/2 = e^(-λt) = e^(-(2520 s)λ)
ln(1/2) = -(2520 s)λ
-ln(1/2)/(2520 s) = λ ≈ 2.75×10^-4 s^-1
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9. Reference material on carbon-14 dating suggests the method is not useful for time periods greater than about 50,000 years. The half-life of C-14 is about 5730 years, so at 65 million years, about ...
6.5·10^7/5.73·10^3 ≈ 11344
half-lives will have passed. Whatever carbon 14 may have existed at the time will have decayed completely to nothing after that many half-lives.
Answer:
chloroplasts
Explanation:
Most plant shoots exhibit positive phototropism, and rearrange their chloroplasts in the leaves to maximize photosynthetic energy and promote growth.
12.00 min = 0.2 hr
8.00 min = 0.15 hr
Total distance:
(10.0 km/hr) (0.2 hr) + (15.0 km/hr) (0.15 hr) + (20.0 km/hr) (0.2 hr)
= 8.25 km
Average speed:
(10.0 km/hr + 15.0 km/hr + 20.0 km/hr) / 3
= 15 km/hr
Change in position:
(10.0 km/hr) (0.2 hr) + (15.0 km/hr) (0.15 hr) - (20.0 km/hr) (0.2 hr)
= 0.25 km
Average velocity:
(10.0 km/hr + 15.0 km/hr - 20.0 km/hr) / 3
≈ 1.67 m/s
