Answer:
Part A
Coriolis effect is used to describe how objects which are not fixed to the ground are deflected as they travel over long distances due to the rotation of the Earth relative to the 'linear' motion of the objects
Due to the Coriolis effect the wind flowing towards the Equator from high pressure belts in the subtropical regions in both the Northern and Southern Hemispheres are deflected towards the western direction because the Earth rotates on its axis towards the east
Part B
In the Northern Hemispheres, the winds are known as northeasterly trade winds and in the Southern Hemisphere, they are known as the southeasterly trade wind. Therefore, Coriolis effect has the same effect on the direction of the Trade Winds in the Southern Hemisphere as it does in the Northern Hemisphere
Explanation:
To solve this problem we will use the definition of the period in a simple pendulum, which warns that it is dependent on its length and gravity as follows:

Here,
L = Length
g = Acceleration due to gravity
We can realize that
is a constant so it is proportional to the square root of its length over its gravity,

Since the body is in constant free fall, that is, a point where gravity tends to be zero:

The value of the period will tend to infinity. This indicates that the pendulum will no longer oscillate because both the pendulum and the point to which it is attached are in free fall.
To develop this problem it is necessary to apply the optical concepts related to the phase difference between two or more materials.
By definition we know that the phase between two light waves that are traveling on different materials (in this case also two) is given by the equation

Where
L = Thickness
n = Index of refraction of each material
Wavelength
Our values are given as





Replacing our values at the previous equation we have




Therefore the thickness of the mica is 6.64μm
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