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:
An asteroid is a minor planet of the inner Solar System. Historically, these terms have been applied to any astronomical object orbiting the Sun.
300 000 0 squared = 2 x 9.8 distance
KINEMATICS
Uniform or constant motion in a straight line (rectilinear). Speed or velocity constant and/or acceleration constant. If motion is up and down and/or has an up and down component then acceleration omn earth will be g. g is about 10m/s/s.
speed = distance/time
velocity = displacement/time
s=distance ... u=initial speed ... v = final speed ... a = acceleration ... t = time
v=u+at
v^2=u^2+2as
s=ut+1/2at^2
Answer:
v = 0.99 c = 2.99 x 10⁸ m/s
Explanation:
From the special theory of relativity:
where,
v = speed of travel = ?
c = speed of light = 3 x 10⁸ m/s
t = time measured on earth = 90 years
t₀ = time measured in moving frame = 6 months = 0.5 year
Therefore,
<u>v = 0.99 c = 2.99 x 10⁸ m/s</u>
