Answer:
The diagram shows refraction, and medium 1 is less dense than medium 2.
Explanation:
- Reflection occurs when a light ray hits a surface and bounces back into the same medium
- Refraction occurs when a light ray crosses the interface between two different mediums, changing direction
From the diagram, we clearly see that this is a case of refraction, since the light ray crosses the boundary between two mediums.
The direction of a light ray in refraction is given by Snell's Law:
(1)
where
n1 and n2 are the index of refraction of the two mediums: a higher index of refraction means a higher density for the medium
are the angles of the light ray in medium 1 and medium 2, measured with respect to the normal to the interface
We can rewrite eq. (1) as
![\frac{n_1}{n_2}=\frac{sin \theta_2}{sin \theta_1}](https://tex.z-dn.net/?f=%5Cfrac%7Bn_1%7D%7Bn_2%7D%3D%5Cfrac%7Bsin%20%5Ctheta_2%7D%7Bsin%20%5Ctheta_1%7D)
From the diagram, we see that
![\theta_1 > \theta_2](https://tex.z-dn.net/?f=%5Ctheta_1%20%3E%20%5Ctheta_2)
so
![sin \theta_1 > sin \theta_2](https://tex.z-dn.net/?f=sin%20%5Ctheta_1%20%3E%20sin%20%5Ctheta_2)
and so
![\frac{sin \theta_2}{sin \theta_1}](https://tex.z-dn.net/?f=%5Cfrac%7Bsin%20%5Ctheta_2%7D%7Bsin%20%5Ctheta_1%7D%3C1)
which means
![\frac{n_1}{n_2}](https://tex.z-dn.net/?f=%5Cfrac%7Bn_1%7D%7Bn_2%7D%3C1%5C%5Cn_1%20%3C%20n_2)
so, medium 2 is denser than medium 1, and the correct answer is
The diagram shows refraction, and medium 1 is less dense than medium 2.
There are a few different formulas for electrical power. They involve
different combinations of voltage, current, and resistance.
==> Power = (current)² x (resistance)
==> Power = (voltage)² / (resistance)
==> Power = (voltage) x (current) .
In this case right now, we know the voltage and current, so maybe
it's a good idea to use the formula that involves voltage and current.
Power = (voltage) x (current)
= (120 volts) x (2.3 Amperes)
= 276 watts .
Kinetic, thermal and electrical. There is more then one form of energy
Let us say that:
1 = 1st player notation
2 = 2nd player notation (the opponent)
a. First let us establish the distance travelled by the 2nd
player:
d2 = 13 m/s * (t + 1.5)
d2 = 13 t + 19.5
Then the distance of the 1st player:
d1 = v0 t + 0.5 a t^2 (v0
initial velocity = 0 since he started from rest)
d1 = 0.5 * 4 m/s^2 * t^2
d1 = 2 t^2
The two distances must be equal, d1 = d2:
2 t^2 = 13 t + 19.5
t^2 – 6.5 t = 9.75
Completing the square:
(t – 3.25)^2 = 9.75 + (- 3.25)^2
t – 3.25 = ±4.5
t = -1.25, 7.75
Since time cannot be negative, therefore:
t = 7.75 seconds
So he catches his opponent after 7.75 seconds.
b. Using the equation:
d1 = 2 t^2
d1 = 2 * (7.75)^2
d1 = 120.125 m
So he travelled about 120.125 meters when he catches up
to his opponent.
Answer:
node jaer le cotenidonv
Explanation:
escirbloa em cuietw n bl********