Explanation:
Below is an attachment containing the solution.
Answer: b
Explanation:
When heat is released by the system i.e. system loses heat. So, we take it as negative -Q
When the work is done on the system then it is considered as negative work on the system i.e. -W
In this case, the plunger is pulled out, and work is done on the system. So, we take work as negative work -W
Correct option is b
If the force and the motion are along the same direction (like it is here) then work is force*distance. The time doesn't come into play until you want the power used. So here
W=9.0*3.0=27J
Answer:
The thrown rock will strike the ground
earlier than the dropped rock.
Explanation:
<u>Known Data</u>


, it is negative as is directed downward
<u>Time of the dropped Rock</u>
We can use
, to find the total time of fall, so
, then clearing for
.
![t_{D}=\sqrt[2]{\frac{300m}{4.9m/s^{2}}} =\sqrt[2]{61.22s^{2}} =7.82s](https://tex.z-dn.net/?f=t_%7BD%7D%3D%5Csqrt%5B2%5D%7B%5Cfrac%7B300m%7D%7B4.9m%2Fs%5E%7B2%7D%7D%7D%20%3D%5Csqrt%5B2%5D%7B61.22s%5E%7B2%7D%7D%20%3D7.82s)
<u>Time of the Thrown Rock</u>
We can use
, to find the total time of fall, so
, then,
, as it is a second-grade polynomial, we find that its positive root is
Finally, we can find how much earlier does the thrown rock strike the ground, so 
You should just ask the wave