To develop the problem it is necessary to apply two concepts, the first is related to the calculation of average data and the second is the Boltzmann distribution.
Boltzmann distribution is a probability distribution or probability measure that gives the probability that a system will be in a certain state as a function of that state's energy and the temperature of the system. It is given by
![z = \sum\limit_i e^{-\frac{\epsilon_i}{K_0T}}](https://tex.z-dn.net/?f=z%20%3D%20%5Csum%5Climit_i%20e%5E%7B-%5Cfrac%7B%5Cepsilon_i%7D%7BK_0T%7D%7D)
Where,
energy of that state
k = Boltzmann's constant
T = Temperature
With our values we have that
T= 250K
![k = 1.381*10^{23} m^2 kg s^{-2} K^{-1}](https://tex.z-dn.net/?f=k%20%3D%201.381%2A10%5E%7B23%7D%20m%5E2%20kg%20s%5E%7B-2%7D%20K%5E%7B-1%7D)
![\epsilon_1=0J](https://tex.z-dn.net/?f=%5Cepsilon_1%3D0J)
![\epsilon_2=1.6*10^{-21}J](https://tex.z-dn.net/?f=%5Cepsilon_2%3D1.6%2A10%5E%7B-21%7DJ)
![\epsilon_3=1.6*10^{-21}J](https://tex.z-dn.net/?f=%5Cepsilon_3%3D1.6%2A10%5E%7B-21%7DJ)
To make the calculations easier we can assume that the temperature and Boltzmann constant can be summarized as
![\beta = \frac{1}{kT}](https://tex.z-dn.net/?f=%5Cbeta%20%3D%20%5Cfrac%7B1%7D%7BkT%7D)
![\beta = \frac{1}{(1.381*10^{23} m^2)(250)}](https://tex.z-dn.net/?f=%5Cbeta%20%3D%20%5Cfrac%7B1%7D%7B%281.381%2A10%5E%7B23%7D%20m%5E2%29%28250%29%7D)
![\beta = 2.9*10^{20}J](https://tex.z-dn.net/?f=%5Cbeta%20%3D%202.9%2A10%5E%7B20%7DJ)
Therefore the average energy would be,
![\bar{\epsilon} =\frac{\sum \epsilon_i e^{-\beta \epsilon_i}}{\sum e^{-\beta \epsilon_i}}](https://tex.z-dn.net/?f=%5Cbar%7B%5Cepsilon%7D%20%3D%5Cfrac%7B%5Csum%20%5Cepsilon_i%20e%5E%7B-%5Cbeta%20%5Cepsilon_i%7D%7D%7B%5Csum%20e%5E%7B-%5Cbeta%20%5Cepsilon_i%7D%7D)
Replacing with our values we have
![\bar{\epsilon} = \frac{0e^{-0}+1.6*10^{-21}*e^{-\Beta(1.6*10^{-21})}+1.6*10^{-2-1}*e^{-(2.9*10^{20})(1.6*10^{-21})}}{1+2e^{-2.9*10^{20}*1.6*10^{-21}}}](https://tex.z-dn.net/?f=%5Cbar%7B%5Cepsilon%7D%20%3D%20%5Cfrac%7B0e%5E%7B-0%7D%2B1.6%2A10%5E%7B-21%7D%2Ae%5E%7B-%5CBeta%281.6%2A10%5E%7B-21%7D%29%7D%2B1.6%2A10%5E%7B-2-1%7D%2Ae%5E%7B-%282.9%2A10%5E%7B20%7D%29%281.6%2A10%5E%7B-21%7D%29%7D%7D%7B1%2B2e%5E%7B-2.9%2A10%5E%7B20%7D%2A1.6%2A10%5E%7B-21%7D%7D%7D)
![\bar{\epsilon} = 0.9*10^{-22}J](https://tex.z-dn.net/?f=%5Cbar%7B%5Cepsilon%7D%20%3D%200.9%2A10%5E%7B-22%7DJ)
Therefore the average internal energy is ![\bar{\epsilon} = 0.9*10^{-22}J](https://tex.z-dn.net/?f=%5Cbar%7B%5Cepsilon%7D%20%3D%200.9%2A10%5E%7B-22%7DJ)
Answer:
No.
Explanation:
- According to Faraday's law, the induced emf in the circuit is given by :
, it is proportional to the rate of change of magnetic flux.
- In this case, a short piece of wire that is not attached to anything and move it up and down in a magnetic field. It means that the circuit is not completed here. It is an open circuit. For the induction of current, a circuit must be completed.
- Hence, no current will induce.
Answer:The Women's National Basketball Association,
Explanation:Branliest:)