Work,
in thermodynamics, is the amount of energy that is transferred from one system
to another system without transfer of entropy. It is equal to the external
pressure multiplied by the change in volume of the system. It is expressed as
follows:<span>
W = PdV
Integrating and assuming that P is not affected
by changes in V or it is constant, then we will have:
W = P (V2 - V1)
Substituting the given values:
P = 1.0 atm = 101325 Pa
(V2 - V1) = 0.50 L =
W = 101325 N/m^3 ( 0.50) (1/1000) m^3
W = 50.66 N-m or 50.66 J
<span>
So, in the expansion process about 50.66 J of work is being done.
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Answer:
I'm not sure if you maybe forgot to add in some information but I can give you the equation you will need to solve this..! (just without the bit of infomration you might need)
In this case you'll be solving for t and I can help you with your first step, you'll divide 2 by 4, getting you to:
If an object changes the speed of <u>direction</u>, its velocity also changes. Any change in <u>velocity</u> results in acceleration.
Hope that helps!
Answer: 2kg
Explanation:
This problem is a textbook conservation of momentum problem. The intial momentum is equal to the final momentum. For the initial state of each block, only the first one was moving. Then they both combine to move together.
Pi = Pf
with p = mv
(6kg)*(4m/s) = (6kg+xkg)(3m/s)
Let x equal the unknown mass of block 2
24 = 18 + 3x
6 = 3x
x = 2kg
