To solve this problem it is necessary to apply the concepts given by Malus regarding the Intensity of light.
From the law of Malus intensity can be defined as

Where
Angle From vertical of the axis of the polarizing filter
Intensity of the unpolarized light
The expression for the intensity of the light after passing through the first filter is given by

Replacing we have that


Re-arrange the equation,

Re-arrange to find \theta





The value of the angle from vertical of the axis of the second polarizing filter is equal to 30.2°
Answer:
T = 17649.03 N = 17.65 KN
Explanation:
The tension in the cable must be equal to the apparent weight of the passenger. For upward acceleration:

where,
T = Tension in cable = ?
= Apparent weight
m = mass = 1603 kg
g = acceleration due to gravity = 9.81 m/s²
a = acceleration of elevator = 1.2 m/s²
Therefore,

<u>T = 17649.03 N = 17.65 KN</u>
1. False
2.False
3.False
4.True
5. True
6. False
7.True
8.False
9. True
10. False
We may be positive that an object is in mechanical equilibrium if it is not rotating and experiences no acceleration.
<h3>What is
mechanical equilibrium?</h3>
There are numerous other definitions for mechanical equilibrium that are all mathematically comparable in addition to the definition in terms of force. A system is in equilibrium in terms of momentum if the component motions are all constant. If velocity is constant, the system is in equilibrium in terms of velocity. When an item is in a state of rotational mechanical equilibrium, its angular momentum is preserved and its net torque is zero. More generally, equilibrium is reached in conservative systems at a configuration space location where the gradient of the potential energy concerning the generalized coordinates is zero.
To learn more about mechanical equilibrium, visit:
<u>brainly.com/question/14246949</u>
#SPJ4
Answer:
The handrails must be approximately 10.63 meters long
Explanation:
The given parameters are;
The height of the bleachers, h = 8 m
The depth of the bleachers, d = 7 m
The length of the hand rails to go along the bleachers from bottom to top is given by Pythagoras' Theorem as follows;
The length of the hand rail = √(d² + h²)
∴ The length of the hand rail = √(7² + 8²) = √113 ≈ 10.63
In order for the handrails to go along the bleachers from top to bottom, they must be approximately 10.63 meters long.