Answer:
Q = 47.06 degrees
Explanation:
Given:
- The transmitted intensity I = 0.464 I_o
- Incident Intensity I = I_o
Find:
What angle should the principle axis make with respect to the incident polarization
Solution:
- The relation of transmitted Intensity I to to the incident intensity I_o on a plane paper with its principle axis is given by:
I = I_o * cos^2 (Q)
- Where Q is the angle between the Incident polarized Light and its angle with the principle axis. Hence, Using the relation given above:
Q = cos ^-1 (sqrt (I / I_o))
- Plug the values in:
Q = cos^-1 ( sqrt (0.464))
Q = cos^-1 (0.6811754546)
Q = 47.06 degrees
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Answer:
The branch of applied mathematics dealing with motion and forces producing motion is called mechanics.
Answer:
The acceleration of the satellite is 
Explanation:
The acceleration in a circular motion is defined as:
(1)
Where a is the centripetal acceleration, v the velocity and r is the radius.
The equation of the orbital velocity is defined as
(2)
Where r is the radius and T is the period
For this particular case, the radius will be the sum of the high of the satellite (
) and the Earth radius (
) :


Then, equation 2 can be used:
⇒ 


Finally equation 1 can be used:

Hence, the acceleration of the satellite is 
Answer:
Shortest distance = 0.1914 m
Explanation:
Given that,
Amplitude of the wave is 0.0957 m
Frequency of the wave is 3.75 Hz
Wavelength of the wave is 1.97 m
We need to find the shortest transverse distance between a maximum and a minimum of the wave.
The distance between maximum point in positive axis and the baseline is equal to amplitude.
Shortest distance = 2 A
D = 2 × 0.0957
D = 0.1914 m
So, the shortest transverse distance between a maximum and a minimum of the wave is 0.1914 m.