To solve the problem it is necessary to apply the Malus Law. Malus's law indicates that the intensity of a linearly polarized beam of light, which passes through a perfect analyzer with a vertical optical axis is equivalent to:
Where,
indicates the intensity of the light before passing through the polarizer,
I is the resulting intensity, and
indicates the angle between the axis of the analyzer and the polarization axis of the incident light.
Since we have two objects the law would be,
Replacing the values,
Therefore the intesity of the light after it has passes through both polarizers is 
Solution:
initial sphere mvr = final sphere mvr + Iω
where I = mL²/3 = 2.3g * (2m)² / 3 = 3.07 kg·m²
0.25kg * (12.5 + 9.5)m/s * (4/5)2m = 3.07 kg·m² * ω
where: ω = 2.87 rad/s
So for the rod, initial E = KE = ½Iω² = ½ * 3.07kg·m² * (2.87rad/s)²
E = 12.64 J becomes PE = mgh, so
12.64 J = 2.3 kg * 9.8m/s² * h
h = 0.29 m
h = L(1 - cosΘ) → where here L is the distance to the CM
0.03m = 1m(1 - cosΘ) = 1m - 1m*cosΘ
Θ = arccos((1-0.29)/1) = 44.77 º
Answer:
wen I was in the car toing home from school after a bad day n si si I have crazzyyy
The answer is B
second law
Initially, the experiment has only potential energy (since total energy is the sum of kinetic and potential energy). And at the end, the experment has only kinetic energy.
