How does the angle of incoming light compare to the angle of light reflecting off of the surface?
- <em>The measurements of the angles are equal.</em>
<u>For</u><u> </u><u>a</u><u> </u><u>smooth</u><u> </u><u>surface</u><u>,</u><u> </u><u>The</u><u> </u><u>angle</u><u> </u><u>of</u><u> </u><u>reflection</u><u> </u><u>and</u><u> </u><u>the</u><u> </u><u>angle</u><u> </u><u>of</u><u> </u><u>incidence</u><u> </u><u>are</u><u> </u><u>equal</u><u>.</u>
Answer:
0.012 m
Explanation:
m = mass of the marble = 0.0265 kg
M = mass of the pendulum = 0.250 kg
v = initial velocity of the marble before collision = 5.05 m/s
V = final velocity of marble-pendulum combination after the collision = ?
using conservation of momentum
m v = (m + M) V
(0.0265) (5.05) = (0.0265 + 0.250) V
V = 0.484 m/s
h = height gained by the marble-pendulum combination
Using conservation of energy
Potential energy gained by the combination = Kinetic energy of the combination just after collision
(m + M) gh = (0.5) (m + M) V²
gh = (0.5) V²
(9.8) h = (0.5) (0.484)²
h = 0.012 m
Answer:
The transmitted intensity through all polarizers is ![I_3 =41.31 W/m^2](https://tex.z-dn.net/?f=I_3%20%3D41.31%20W%2Fm%5E2)
Explanation:
According to Malu's law the intensity of a polarized light having an initial intensity
is mathematically represented as
![I = I_0cos^2 \theta](https://tex.z-dn.net/?f=I%20%3D%20I_0cos%5E2%20%5Ctheta)
Now considering the polarizer(The polarizing disk) the equation above becomes
![I = I_0 (cos^2 \theta)^n](https://tex.z-dn.net/?f=I%20%3D%20I_0%20%28cos%5E2%20%5Ctheta%29%5En)
Where n is the number of polarizers
Substituting
for the initial intensity 3 for the n and 20° for the angle of rotation
![I_3 = 60 (cos^220)^3](https://tex.z-dn.net/?f=I_3%20%3D%2060%20%28cos%5E220%29%5E3)
![=41.31 W/m^2](https://tex.z-dn.net/?f=%3D41.31%20W%2Fm%5E2)
3rd nearest planet.
Neighborhood of Sun.
Orion arm.
Milky Way galaxy.
Local cluster.
13.8 billion years.
Universe.
Mind of God.