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.
It MUST be either glue or gravity.
From the law of Galileo Galilei :v²=v₀²+2ad we take the speed
v²=0+2*4.90*200=1960=>v=√1960=44.27 m/s
Answer:
Hello your question is incomplete hence I will give you a general answer on how A van de Graaff generator works
answer :
If the electrons falls through a PD of 150mV the electron will gain energy of 150MeV
Explanation:
when a Van de Graff generator is used to accelerate an electron through a PD ( potential difference ) of any value the particle ( electron ) the electron will gain energy ( eV ) which is is equivalent in value of the PD it accelerated through
hence if the electrons falls through a PD of 150mV the electron will gain energy of 150MeV
Answer:
A. fuel mileage and longevity
Explanation:
For a person purchasing a car, car longevity is one of the main concern. They are also interested in many things such as maximum mileage and service life.
By properly monitoring and assessing few measures one can maintain the efficiency and longevity of the car. One such thing is by monitoring the liquid levels of the car. Certain liquids like the coolant or radiator water level should be well maintain in proper level in order to run the car economically.
Thus by doing this, one can optimize the car's longevity and the fuel mileage.
Hence the correct option is (A).
