If the applied force is in the same direction as the object's displacement, the work done on the object is:
W = Fd
W = work, F = force, d = displacement
Given values:
F = 45N
d = 12m
Plug in and solve for W:
W = 45(12)
W = 540J
Answer:
p(a) * p(b) = .01923
p(b) = .01923 / .07692 = .2500
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:


Answer:
41.16 Joules
Explanation:
Potential energy at a given instant is a function of mass and height of an object. The formula is

Answer:
i really thought that said hater