1) n=1 -> n=2 : delta E = -5+11 = 6. The answer is D
<span>2) n=1 -> n=3 : delta E = -2+11 = 9. The answer is B </span>
<span>3) n=1 -> n=4 : delta E = -1+11 = 10. No solution available </span>
<span>4) n=1 -> infinity delta E = 11. The answer is A </span>
<span>5) not absorbed would be C, as there is no transition with delta E of 8. </span>
Military personnel also use periscopes in some gun turrets and in armoured vehicles. More complex periscopes using prisms or advanced fibre optics instead of mirrors and providing magnification operate on submarines and in various fields of science
Car with a mass of 1210 kg moving at a velocity of 51 m/s.
2. What velocity must a 1340 kg car have in order to have the same momentum as a 2680 kg truck traveling at a velocity of 15 m/s to the west? 3.0 X 10^1 m/s to the west.
Hope i helped
Have a good day :)
Its good for a scientist to be skeptical because they dont willingly accept new ideas without thoroughly going over the research themselves and attempting to disprove a theory. skeptical scientists often reveal new information through doing their own examinations of a concept, which is why they are valuable scientists
Answer:
- The work made by the gas is 7475.69 joules
- The heat absorbed is 7475.69 joules
Explanation:
<h3>
Work</h3>
We know that the differential work made by the gas its defined as:

We can solve this by integration:

but, first, we need to find the dependence of Pressure with Volume. For this, we can use the ideal gas law


This give us

As n, R and T are constants

![\Delta W= \ n \ R \ T \left [ ln (V) \right ]^{v_2}_{v_1}](https://tex.z-dn.net/?f=%20%5CDelta%20W%3D%20%5C%20n%20%5C%20R%20%5C%20T%20%20%5Cleft%20%5B%20ln%20%28V%29%20%5Cright%20%5D%5E%7Bv_2%7D_%7Bv_1%7D%20)



But the volume is:



Now, lets use the value from the problem.
The temperature its:

The ideal gas constant:

So:


<h3>Heat</h3>
We know that, for an ideal gas, the energy is:

where
its the internal energy of the gas. As the temperature its constant, we know that the gas must have the energy is constant.
By the first law of thermodynamics, we know

where
is the Work made by the gas (please, be careful with this sign convention, its not always the same.)
So:

