To
solve this problem, we assume that the wavelength of the light in air is 500
nanometers.
For this case we
only need the refractive index of the polystyrene. For an antireflective
coating, we need a quarter of wave thickness at the wavelength in the air. Which
means that the antireflective coating needs to be as thick as 1/4 of the
wavelength, divided by the coating’s refractive index. This is expressed
mathematically in the form:
x = λ / (4 * n)
where,
x = thickness
λ = wavelength
of light
n = index of
refraction of polystyrene
Substituting:
x = 500 nm / (4
* 1.49)
x = 500 nm / 5.96
x = 83.90 nm
The correct answer is decreased hearing ability.
To develop this problem, it is necessary to apply the concepts related to Faraday's law and Magnetic Flow, which is defined as the change that the magnetic field has in a given area. In other words
![\Phi = BA Cos\theta](https://tex.z-dn.net/?f=%5CPhi%20%3D%20BA%20Cos%5Ctheta)
Where
B= Magnetic Field
A = Area
Angle between magnetic field lines and normal to the area
The differentiation of this value allows us to obtain in turn the induced emf or electromotive force.
In this case we have that the flat loop of wire is perpendicular to the magnetic field, therefore the angle is 0 degrees, since its magnetic field acts parallel to the area:
0 then our expression can be written as
![\Phi = BA](https://tex.z-dn.net/?f=%5CPhi%20%3D%20BA)
From the same value of the electromotive force we have to
![\epsilon = -\frac{d\Phi}{dt}](https://tex.z-dn.net/?f=%5Cepsilon%20%3D%20-%5Cfrac%7Bd%5CPhi%7D%7Bdt%7D)
Replacing we have
![\epsilon = -A\frac{B}{dt}](https://tex.z-dn.net/?f=%5Cepsilon%20%3D%20-A%5Cfrac%7BB%7D%7Bdt%7D)
Replacing with our values we have that
![\epsilon = -(7.9*10^{-4}m)\frac{(3.5-0.5)}{0.1}](https://tex.z-dn.net/?f=%5Cepsilon%20%3D%20-%287.9%2A10%5E%7B-4%7Dm%29%5Cfrac%7B%283.5-0.5%29%7D%7B0.1%7D)
![\epsilon = -0.0237V](https://tex.z-dn.net/?f=%5Cepsilon%20%3D%20-0.0237V)
Therefore the magnitude of the induced emf in the loop is 0.0237V
On the other hand we have that the current by Ohm's Law can be defined as
![I = \frac{\epsilon}{R}](https://tex.z-dn.net/?f=I%20%3D%20%5Cfrac%7B%5Cepsilon%7D%7BR%7D)
For the given value of the resistance and the previously found potential we have to
![I = \frac{0.237}{1.3}](https://tex.z-dn.net/?f=I%20%3D%20%5Cfrac%7B0.237%7D%7B1.3%7D)
![I= 0.0182A](https://tex.z-dn.net/?f=I%3D%200.0182A)
Answer:
The final velocity,
, is gd.
Explanation:
The condition here is a free falling object. Thus from the third equation of motion under free fall, we have;
=
+ 2gs
where
is the final velocity of the object,
is the initial velocity of the object, g is the gravitational force and s is the height.
Since the object falls from a height of d, then
= 0 m/s, and s = d.
So that;
= 0 + 2gd
= 2gd
= 2gd
When the distance is
,
= 0 m/s.
Then;
= 2g![\frac{d}{2}](https://tex.z-dn.net/?f=%5Cfrac%7Bd%7D%7B2%7D)
= g x d
When the object falls through the height
, then the final velocity is gd.