<span>Speed (rms) = sqrt 3KT/m, where:
K = Boltzmann's costant = 1.38*10^-23 joule/K;
T = temperature in Kelvin degrees = 273 - 63 = 210 K;
m = 44 * 1.672*10^-27 Kg;
1.672*10^-27 Kg (good approximation) is the weight of a single a.m.u. (atomic mass unit);
44 is molecular weight of CO2.
Molecular Speed (CO2) = sqrt 3*1.38*10^-23*210/44*1.672*10^-27 = 344 m/sec.
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Answers:
The acceleration due to gravity on the surface of earth is 9.8 ms^(-2).Time period of a simple pendulum on earth and moon are 3.5 second and 8.4 second respectively. Find the acceleration due to gravity on the moon . <br> Hint : T_(e) = 2pi sqrt((L)/(g_(e))) T_(m)= 2pi sqrt((L)/(g_(m))) <br> (T_(e)^(2))/(T_(m)^(2))= (g_(m))/(g_(e)) <br> g_(m) = (T_(e)^(2))/(T_(m)^(2))g_(e)
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Explanation:
Hope this helps ( not copied and pasted, this answer was done by me so I don't know if it's good or not)
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Explanation:
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A chain position over a certain amount of time