The energy of a photon is given by:

where h is the Planck constant and f is the photon frequency.
We know the energy of the photon,

, so we can rearrange the equation to calculate the frequency of the photon:

And now we can use the following relationship between frequency f, wavelength

and speed of light c to find the wavelength of the photon:
(a) 0.448
The gravitational potential energy of a satellite in orbit is given by:

where
G is the gravitational constant
M is the Earth's mass
m is the satellite's mass
r is the distance of the satellite from the Earth's centre, which is sum of the Earth's radius (R) and the altitude of the satellite (h):
r = R + h
We can therefore write the ratio between the potentially energy of satellite B to that of satellite A as

and so, substituting:

We find

(b) 0.448
The kinetic energy of a satellite in orbit around the Earth is given by

So, the ratio between the two kinetic energies is

Which is exactly identical to the ratio of the potential energies. Therefore, this ratio is also equal to 0.448.
(c) B
The total energy of a satellite is given by the sum of the potential energy and the kinetic energy:

For satellite A, we have

For satellite B, we have

So, satellite B has the greater total energy (since the energy is negative).
(d) 
The difference between the energy of the two satellites is:

Answer: The main difference between the three is the mode of transmission. The chest pass is straight through the air towards your teammate. While the bounce pass is directed toward the ground and then at your teammate. Finally, the overhead pass is projected high in the air to avoid defenders.
Explanation:
Answer:
the ball didnt hit my face so
Explanation:
Answer:
Theta1 = 12° and theta2 = 168°
The solution procedure can be found in the attachment below.
Explanation:
The Range is the horizontal distance traveled by a projectile. This diatance is given mathematically by Vo cos(theta) t. Where t is the total time of flight of the projectile in air. It is the time taken for the projectile to go from starting point to finish point. This solution assumes the projectile finishes uts motion on the same horizontal level as the starting point and as a result the vertical displacement is zero (no change in height).
In the solution as can be found below, the expression to calculate the range for any launch angle theta was first derived and then the required angles calculated from the equation by substituting the values of the the given quantities.