Answer:
0.71121 km/s
Explanation:
= Velocity of planet initially = 54 km/s
= Distance from star = 0.54 AU
= Final velocity of planet
= Final distance from star = 41 AU
As the angular momentum of the system is conserved

When the exoplanet is at its farthest distance from the star the speed is 0.71121 km/s.
A parallel plate capacitor can store electric charge and
electrical energy, and if the plates are far enough apart,
you can store your lunch in there too.
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)
Answer:
Option C
Explanation:
Given
Pressure = Force /Area
For any gas sample, force is defined as basically the force exerted by the gas molecules when they strike the surface (with in which they are contained) and bounce back.
Hence, Option C is the correct