Answer:
1) At the highest point of the building.
2) The same amount of energy.
3) The kinetic energy is the greatest.
4) Potential energy = 784.8[J]
5) True
Explanation:
Question 1
The moment when it has more potential energy is when the ball is at the highest point in the building, that is when the ball is at a height of 40 meters from the ground. It is taken as a point of reference of potential energy, the level of the soil, at this point of reference the potential energy is zero.
![E_{p} = m*g*h\\E_{p} = 2*9.81*40\\E_{p} = 784.8[J]](https://tex.z-dn.net/?f=E_%7Bp%7D%20%3D%20m%2Ag%2Ah%5C%5CE_%7Bp%7D%20%3D%202%2A9.81%2A40%5C%5CE_%7Bp%7D%20%3D%20784.8%5BJ%5D)
Question 2)
The potential energy as the ball falls becomes kinetic energy, in order to be able to check this question we can calculate both energies with the input data.
![E_{p}=m*g*h\\ E_{p} = 2*9.81*20\\ E_{p} = 392.4[J]\\](https://tex.z-dn.net/?f=E_%7Bp%7D%3Dm%2Ag%2Ah%5C%5C%20E_%7Bp%7D%20%3D%202%2A9.81%2A20%5C%5C%20E_%7Bp%7D%20%3D%20392.4%5BJ%5D%5C%5C)
And the kinetic energy will be:
![E_{k}=0.5*m*v^{2}\\ where:\\v = velocity = 19.8[m/s]\\E_{k}=0.5*2*(19.8)^{2}\\ E_{k}=392.04[J]](https://tex.z-dn.net/?f=E_%7Bk%7D%3D0.5%2Am%2Av%5E%7B2%7D%5C%5C%20%20where%3A%5C%5Cv%20%3D%20%20velocity%20%3D%2019.8%5Bm%2Fs%5D%5C%5CE_%7Bk%7D%3D0.5%2A2%2A%2819.8%29%5E%7B2%7D%5C%5C%20%20E_%7Bk%7D%3D392.04%5BJ%5D)
Therefore it is the ball has the same potential energy and kinetic energy as it is half way through its fall.
Question 3)
As the ball drops all potential energy is transformed into kinetic energy, therefore being close to the ground, the ball will have its maximum kinetic energy.
![E_{k}=E_{p}=m*g*h = 2*9.81*40\\ E_{k} = 784.8[J]\\ E_{k} = 0.5*2*(28)^{2}\\ E_{k} = 784 [J]](https://tex.z-dn.net/?f=E_%7Bk%7D%3DE_%7Bp%7D%3Dm%2Ag%2Ah%20%3D%202%2A9.81%2A40%5C%5C%20%20E_%7Bk%7D%20%3D%20784.8%5BJ%5D%5C%5C%20E_%7Bk%7D%20%3D%200.5%2A2%2A%2828%29%5E%7B2%7D%5C%5C%20E_%7Bk%7D%20%3D%20784%20%5BJ%5D)
Question 4)
It can be easily calculated using the following equation
![E_{p} =m*g*h\\E_{p}=2*9.81*40\\E_{p} =784.8[J]](https://tex.z-dn.net/?f=E_%7Bp%7D%20%3Dm%2Ag%2Ah%5C%5CE_%7Bp%7D%3D2%2A9.81%2A40%5C%5CE_%7Bp%7D%20%3D784.8%5BJ%5D)
Question 5)
True
The potential energy at 20[m] is:
![E_{p}=2*9.81*20\\ E_{p}= 392.4[J]\\The kinetic energy is:\\E_{k}=0.5*2*(19.8)^{2} \\E_{k}=392[J]](https://tex.z-dn.net/?f=E_%7Bp%7D%3D2%2A9.81%2A20%5C%5C%20E_%7Bp%7D%3D%20392.4%5BJ%5D%5C%5CThe%20kinetic%20energy%20is%3A%5C%5CE_%7Bk%7D%3D0.5%2A2%2A%2819.8%29%5E%7B2%7D%20%5C%5CE_%7Bk%7D%3D392%5BJ%5D)
If the object is not at rest how?
A law stating that electric current is proportional to voltage and inversely proportional to resistance.
Explanation:
Yes, the law of conservation of energy still applies even if there is waste energy.
The waste energy are the transformation products of energy from one form to another.
According to the law of conservation of energy "energy is neither created nor destroyed by transformed from one form to another in a system".
But of then times, energy is lost as heat or sound within a system.
- If we take into account these waste energy, we can see that energy is indeed conserved.
- The sum total of the energy generated and those produced will be the same if we factor in other forms in which the energy has been transformed into.
In this case, you need the formula below where:
F = force
k = coulombs constant 8.99 x10^{9} N.m^{2} . C^{-2}
q1 = electric charge 1
q2 = electric charge 2
r = the distance between the charges
pls note: make sure your units are correct (in meters etc, not fm (<em>femto-meters</em>)).
Curiously, this question doesn't tell you what atom you are next to the nucleus of. Different numbers of protons in the nucleus of the atom will make for vastly different forces in your answer...
