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Luda [366]
3 years ago
11

• What are common questions, concerns, practices, and experiences across various world religions? Include indigenous religions i

n your discussion.
Physics
1 answer:
faltersainse [42]3 years ago
5 0

<u>Answer:</u>

<em>They are a lot of religions and faiths across the globe. </em>

<u>Explanation:</u>

There are  a lot of monotheistic and polytheistic religions across the world that took origin at different points in the past.

Some of them are:

  • Christianity- It is a monotheistic religion based on the teachings of Jesus Christ believed to be the son of god.  
  • Confucianism it evolved in China and is based on the teachings scholar and philosopher Confucius. It is a largely secular and humanistic.
  • Buddhism it is based on the life and teachings of Gautama Buddha. Buddhism advocates a way of life with meditation and instructs its followers to be compassionate, kind and wise.  
  • Hinduism –It evolved during the period of Indus valley civilisation and is a polytheistic religion. Hinduism beliefs mention rebirths or ‘punarjanma’.
  • Islam- Islam is a monotheistic religion.It considers Muhammad as the messenger of only god Allah.
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3 years ago
A quantum system has three states, with energies 0, 1.6 × 10-21, and 1.6 × 10-21, in Joules. It is coupled to an environment wit
xenn [34]

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}}

Where,

\epsilon_i = energy of that state

k = Boltzmann's constant

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With our values we have that

T= 250K

k = 1.381*10^{23} m^2 kg s^{-2} K^{-1}

\epsilon_1=0J

\epsilon_2=1.6*10^{-21}J

\epsilon_3=1.6*10^{-21}J

To make the calculations easier we can assume that the temperature and Boltzmann constant can be summarized as

\beta = \frac{1}{kT}

\beta = \frac{1}{(1.381*10^{23} m^2)(250)}

\beta = 2.9*10^{20}J

Therefore the average energy would be,

\bar{\epsilon} =\frac{\sum \epsilon_i e^{-\beta \epsilon_i}}{\sum e^{-\beta \epsilon_i}}

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}}}

\bar{\epsilon} = 0.9*10^{-22}J

Therefore the average internal energy is \bar{\epsilon} = 0.9*10^{-22}J

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3 years ago
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Iteru [2.4K]

Answer:

the mass stays the same

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the more mass the faster

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Answer: the amount of work will increase.

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2 years ago
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