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:
Explanation:
a ) Between r = 0 and r = r₁
Electric field will be zero . It is so because no charge lies in between r = 0 and r = r₁ .
b ) From r = r₁ to r = r₂
At distance r , charge contained in the sphere of radius r
volume charge density x 4/3 π r³
q = Q x r³ / R³
Applying Gauss's law
4πr² E = q / ε₀
4πr² E = Q x r³ / ε₀R³
E= Q x r / (4πε₀R³)
E ∝ r .
c )
Outside of r = r₂
charge contained in the sphere of radius r = Q
Applying Gauss's law
4πr² E = q / ε₀
4πr² E = Q / ε₀
E = Q / 4πε₀r²
E ∝ 1 / r² .
Explanation:
It is given that,
Mass of Millersburg Ferry, m = 13000 kg
Velocity, v = 11 m/s
Applied force, F = 10⁶ N
Time period, t = 20 seconds
(a) Impulse is given by the product of force and time taken i.e.
(b) Impulse is also given by the change in momentum i.e.
(c) For new velocity,
Hence, this is the required solution.
If he keeps that pace he will be at the 34 yard line
A is right because I took the test
