Maybe friction..? my best answer.
<span>The water is clearer, has higher oxygen levels, and freshwater fish such as trout and heterotrophs can be found. In the middle of the river, the width increases, so do the species of aquatic green plants and algae. Toward the mouth of the river, the water becomes murky from all the sediments that it has picked up upstream.<span>
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Answer:
he wavelength is different (greater) than the wavelength of the incident photon
Explanation:
The Compton effect is the scattering of a photon by an electron, this process is analyzed using the conservation of momentum, in which we assume that initially the electron is at rest and after the collision it recedes, therefore the energy of the incident photon decreases and consequently its wavelength changes
To complete the sentence we use the wavelength is different (greater) than the wavelength of the incident photon
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)
Explanation:
Kepler's third law gives the relationship between the orbital radius and the orbital period of the planet. Its mathematical form is given by :

Here,
G is gravitational constant
M is mass of sun
It means that the mass of Sun is constant for all planets orbiting the sun, assuming circular orbits.