The final volume of the gas is 144.25 L
Explanation:
For an ideal gas kept at constant pressure, the work done by the gas on the surroundings is given by

where
p is the pressure of the gas
is the initial volume
is the final volume
For the gas in the cylinder in this problem,
p = 2.00 atm

And we also know the work done,
W = 288 J
So we can solve the equation for
, the final volume:

Learn more about ideal gases:
brainly.com/question/9321544
brainly.com/question/7316997
brainly.com/question/3658563
#LearnwithBrainly
D.) Vertical relationships involve unequal status, while horizontal relationships represent equal status.
HOPE THIS HELPS!
i believe it's C but i'm not completely sure
Answer:
Yes , it is possible for two different atoms of carbon to have different numbers of neutrons in their nuclei .
Explanation:
Isotopes -
Atoms of the element with same number of the electrons and protons , but differ in the number of neutrons , are called as the isotopes .
The isotopes of the element have the same number of protons and electrons , hence have the same chemical and physical properties .
Many isotopes occurs naturally .
In case of Carbon ,
Carbon too have isotopes ,
i.e.
Carbon - 13 and Carbon - 14
carbon - 13 , have seven neutrons
Carbon - 14 , have eight neutrons .
Explanation:
At point B, the velocity speed of the train is as follows.

= 
= 26.34 m/s
Now, we will calculate the first derivative of the equation of train.
y = 

Now, second derivative of the train is calculated as follows.
Radius of curvature of the train is as follows.
![\rho = \frac{[1 + (\frac{dy}{dx})^{2}]^{\frac{3}{2}}}{\frac{d^{2}y}{dx^{2}}}](https://tex.z-dn.net/?f=%5Crho%20%3D%20%5Cfrac%7B%5B1%20%2B%20%28%5Cfrac%7Bdy%7D%7Bdx%7D%29%5E%7B2%7D%5D%5E%7B%5Cfrac%7B3%7D%7B2%7D%7D%7D%7B%5Cfrac%7Bd%5E%7B2%7Dy%7D%7Bdx%5E%7B2%7D%7D%7D)
= ![\frac{[1 + 0.2e^{\frac{400}{1000}}^{2}]^{\frac{3}{2}}}{0.2(10^{-3})e^{\frac{400}{1000}}}](https://tex.z-dn.net/?f=%5Cfrac%7B%5B1%20%2B%200.2e%5E%7B%5Cfrac%7B400%7D%7B1000%7D%7D%5E%7B2%7D%5D%5E%7B%5Cfrac%7B3%7D%7B2%7D%7D%7D%7B0.2%2810%5E%7B-3%7D%29e%5E%7B%5Cfrac%7B400%7D%7B1000%7D%7D%7D)
= 3808.96 m
Now, we will calculate the normal component of the train as follows.

= 
= 0.1822 
The magnitude of acceleration of train is calculated as follows.
a = 
= 
= 
Thus, we can conclude that magnitude of the acceleration of the train when it reaches point B, where sAB = 412 m is
.