The refractive index for glycerine is
![n_g=1.473](https://tex.z-dn.net/?f=n_g%3D1.473)
, while for air it is
![n_a = 1.00](https://tex.z-dn.net/?f=n_a%20%3D%201.00)
.
When the light travels from a medium with greater refractive index to a medium with lower refractive index, there is a critical angle over which there is no refraction, but all the light is reflected. This critical angle is given by:
![\theta_c = \arcsin ( \frac{n_2}{n_1} )](https://tex.z-dn.net/?f=%5Ctheta_c%20%3D%20%5Carcsin%20%28%20%5Cfrac%7Bn_2%7D%7Bn_1%7D%20%29)
where n1 and n2 are the refractive indices of the two mediums. If we susbtitute the refractive index of glycerine and air in the formula, we find the critical angle for this case:
An airplane is flying at a constant speed in a positive direction. It slows down when it approaches the airport where it's going to land. this is an example of negative acceleration (D).
The electric field produced by a large flat plate with uniform charge density on its surface can be found by using Gauss law, and it is equal to
![E= \frac{\sigma}{2\epsilon_0}](https://tex.z-dn.net/?f=E%3D%20%5Cfrac%7B%5Csigma%7D%7B2%5Cepsilon_0%7D%20)
where
![\sigma](https://tex.z-dn.net/?f=%5Csigma)
is the charge density
![\epsilon_0](https://tex.z-dn.net/?f=%5Cepsilon_0)
is the vacuum permittivity
We see that the intensity of the electric field does not depend on the distance from the plate. Therefore, the strenght of the electric field at 4 cm from the plate is equal to the strength of the electric field at 2 cm from the plate:
a) 10 m/s
b) 25 m
Explanation:
a)
The body is moving with a constant acceleration, therefore we can solve the problem by using the following suvat equation:
![v=u+at](https://tex.z-dn.net/?f=v%3Du%2Bat)
where
u is the initial velocity
v is the final velocity
a is the acceleration
t is the time
For the body in this problem:
u = 0 (the body starts from rest)
is the acceleration
t = 5 s is the time
So, the final velocity is
![v=0+(2)(5)=10 m/s](https://tex.z-dn.net/?f=v%3D0%2B%282%29%285%29%3D10%20m%2Fs)
b)
In this second part, we want to calculate the distance travelled by the body.
We can do it by using another suvat equation:
![v^2-u^2=2as](https://tex.z-dn.net/?f=v%5E2-u%5E2%3D2as)
where
u is the initial velocity
v is the final velocity
a is the acceleration
s is the distance travelled
Here we have
u = 0 (the body starts from rest)
is the acceleration
v = 10 m/s is the final velocity
Solving for s,
![s=\frac{10^2-0^2}{2(2)}=25 m](https://tex.z-dn.net/?f=s%3D%5Cfrac%7B10%5E2-0%5E2%7D%7B2%282%29%7D%3D25%20m)
The Answer is:
O 3s
Hope you got it right.