1 milliliter = 1 cubic centimeter (cm^3)
Answer:
See the answers below.
Explanation:
In order to solve this problem we must use the principle of energy conservation. Which tells us that the energy of a body will always be the same regardless of where it is located. For this case we have two points, point A and point B. Point A is located at the top at 120 [m] and point B is in the middle of the cliff at 60 [m].

The important thing about this problem is to identify the types of energy at each point. Let's take the reference level of potential energy at a height of zero meters. That is, at this point the potential energy is zero.
So at point A we have potential energy and since a velocity of 18 [m/s] is printed, we additionally have kinetic energy.

At Point B the rock is still moving downward, therefore we have kinetic energy and since it is 60 [m] with respect to the reference level we have potential energy.

Therefore we will have the following equation:
![(6.5*9.81*120)+(0.5*6.5*18^{2} )=(6.5*9.81*60)+(0.5*6.5*v_{B}^{2} )\\3.25*v_{B}^{2} =4878.9\\v_{B}=\sqrt{1501.2}\\v_{B}=38.75[m/s]](https://tex.z-dn.net/?f=%286.5%2A9.81%2A120%29%2B%280.5%2A6.5%2A18%5E%7B2%7D%20%29%3D%286.5%2A9.81%2A60%29%2B%280.5%2A6.5%2Av_%7BB%7D%5E%7B2%7D%20%29%5C%5C3.25%2Av_%7BB%7D%5E%7B2%7D%20%3D4878.9%5C%5Cv_%7BB%7D%3D%5Csqrt%7B1501.2%7D%5C%5Cv_%7BB%7D%3D38.75%5Bm%2Fs%5D)
The kinetic energy can be easily calculated by means of the kinetic energy equation.
![KE_{B}=\frac{1}{2} *m*v_{B}^{2}\\KE_{B}=0.5*6.5*(38.75)^{2}\\KE_{B}=4878.9[J]](https://tex.z-dn.net/?f=KE_%7BB%7D%3D%5Cfrac%7B1%7D%7B2%7D%20%2Am%2Av_%7BB%7D%5E%7B2%7D%5C%5CKE_%7BB%7D%3D0.5%2A6.5%2A%2838.75%29%5E%7B2%7D%5C%5CKE_%7BB%7D%3D4878.9%5BJ%5D)
In order to calculate the velocity at the bottom of the cliff where the reference level of potential energy (potential energy equal to zero) is located, we must pose the same equation, with the exception that at the new point there is only kinetic energy.
![E_{A}=E_{C}\\6.5*9.81*120+(0.5*9.81*18^{2} )=0.5*6.5*v_{C}^{2} \\v_{c}^{2} =\sqrt{2843.39}\\v_{c}=53.32[m/s]](https://tex.z-dn.net/?f=E_%7BA%7D%3DE_%7BC%7D%5C%5C6.5%2A9.81%2A120%2B%280.5%2A9.81%2A18%5E%7B2%7D%20%29%3D0.5%2A6.5%2Av_%7BC%7D%5E%7B2%7D%20%5C%5Cv_%7Bc%7D%5E%7B2%7D%20%3D%5Csqrt%7B2843.39%7D%5C%5Cv_%7Bc%7D%3D53.32%5Bm%2Fs%5D)
I'd answer that but I can't text graphs and tables...
The answer is D
<em>"</em><em>a property of matter by which it continues in its existing state of rest or uniform motion in a straight line, unless that state is changed by an external force.</em><em>"</em>
It will land in your lap because there's different frames of motion relative to yourself. For example, if you're running at a speed of 6 mph, it doesn't mean you'll run as fast as the Earth spins. Also, since you're on the interior of the plane, any kind of wind or weather on the outside will not affect the coin. A law to back up this claim is Einsteins Special Law of Relativity.