Answer:
Option A applies.
A. Greater than its escape speed from the mass within the volume
Explanation:
Here it is mentioned that the spherical volume is large enough for the space to be considered as homogeneous. Also, the pressure within the volume is negligible, so that will not result into the re collapse of the Universe. Now as per our knowing, Hubble's Law relates the average speed of the particle to the distance R between the Earth and the particle. So, if the particle's speed is greater than it's escape speed from the mass within the volume, then the Universe is bound to re collapse back again. Option A applies.
Answer:
The object will travel at a constant rate in along a straight line.
Explanation:
In the given situation, it is mentioned that there is no external force acting on the given object. Thus, it will retain its initial velocity along a straight path.
Heat energy and thermal energy are the same because heat energy is thermal energy. Also thermal energy and temperature are the same because temperature is measuring heat in degrees Celsius or degrees Fahrenheit. Hope this helps!
To solve this problem we will use the work theorem, for which we have that the Force applied on the object multiplied by the distance traveled by it, is equivalent to the total work. From the measurements obtained we have that the width and the top are 14ft and 7ft respectively. In turn, the bottom of the tank is 15ft. Although the weight of the liquid is not given we will assume this value of 
(Whose variable will remain modifiable until the end of the equations subsequently presented to facilitate the change of this, in case be different). Now the general expression for the integral of work would be given as
Basically under this expression we are making it difficult for the weight of the liquid multiplied by the area (Top and widht) under the integral of the liquid path to be equivalent to the total work done, then replacing
![W = (14*7*62)\big [15y-\frac{y^2}{2}\big ]^{15}_0](https://tex.z-dn.net/?f=W%20%3D%20%2814%2A7%2A62%29%5Cbig%20%5B15y-%5Cfrac%7By%5E2%7D%7B2%7D%5Cbig%20%5D%5E%7B15%7D_0)
![W = (14*7*62)[15(15)-\frac{(15)^2}{2}]](https://tex.z-dn.net/?f=W%20%3D%20%2814%2A7%2A62%29%5B15%2815%29-%5Cfrac%7B%2815%29%5E2%7D%7B2%7D%5D)
Therefore the total work in the system is 
