To solve the problem it is necessary to identify the equation in the manner given above.
This equation corresponds to the displacement of a body under the principle of simple harmonic movement.
Where,
PART A) Our equation corresponds to
Therefore the value of omega is equivalent to that of
From the definition we know that the period as a function of angular velocity is equivalent to
This same point is the equivalent of the maximum point of the speed that the body can reach, since the internal expression of the 
Is equivalent to . So the maximum speed that the body can reach is,
Therefore the maximum felocity will be 5ft / s
PART B) The period of graph is the time taken to reach from one maximum point to next point maximum point, then
Answer:
The separation distance between the parallel planes of an atom is hc/2sinθ(EK - EL)
Explanation:
The relationship between energy and wavelength is expressed below:
E = hc/λ
λ = hc/EK - EL
Considering the condition of Bragg's law:
2dsinθ = mλ
For the first order Bragg's law of reflection:
2dsinθ = (1)λ
2dsinθ = hc/EK - EL
d = hc/2sinθ(EK - EL)
Where 'd' is the separation distance between the parallel planes of an atom, 'h' is the Planck's constant, 'c' is the velocity of light, θ is the angle of reflection, 'EK' is the energy of the K shell and 'EL' is the energy of the K shell.
Therefore, the separation distance between the parallel planes of an atom is hc/2sinθ(EK - EL)
1) 0.0011 rad/s
2) 7667 m/s
Explanation:
1)
The angular velocity of an object in circular motion is equal to the rate of change of its angular position. Mathematically:
where
is the angular displacement of the object
t is the time elapsed
is the angular velocity
In this problem, the Hubble telescope completes an entire orbit in 95 minutes. The angle covered in one entire orbit is
rad
And the time taken is
Therefore, the angular velocity of the telescope is
2)
For an object in circular motion, the relationship between angular velocity and linear velocity is given by the equation
where
v is the linear velocity
is the angular velocity
r is the radius of the circular orbit
In this problem:
is the angular velocity of the Hubble telescope
The telescope is at an altitude of
h = 600 km
over the Earth's surface, which has a radius of
R = 6370 km
So the actual radius of the Hubble's orbit is
Therefore, the linear velocity of the telescope is:
Is their a multiple choice to choose from I'm not sure the answer I got is even right.
That would be very helpful.
I am not so sure about this it is too difficult
