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Digiron [165]
3 years ago
7

My questions are in the picture: Please help.

Physics
1 answer:
frutty [35]3 years ago
8 0

Answer:

a.)

I.) Open system

II.) Closed system

b.)

The total momentum of the object and earth system stay the same as the object fall toward the earth

The kinetic energy of the object increases as the object falls toward the earth.

Explanation:

a.)

I.) The system containing only the object is an open system because of the influence of external forces and presence of matters. External forces in this scenario means gravitational force acting on the object. And air is the matter that has influence on the object.

II.) The system containing only the object and earth is a closed system. Because in a closed system, there are no external dissipative forces acting on it. Universe is a closed system. And the mechanical energy of a close system is conserved. The mechanical energy will remain constant. In other words, it will not change (become more or less). This is called the Law of Conservation of Mechanical Energy.

b.) The total momentum of the object and earth system stay the same as the object fall toward the earth. Because momentum will always be conserved.

The kinetic energy of the object increases as the object falls toward the earth. When the object was initially at rest, kinetic energy equals zero

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jek_recluse [69]

Answer:

a)    I= I₀ (cos²θ - cos⁴θ)    b) 75.5º

Explanation:

a) For this exercise we must use Malus's law

         I = I₀ cos² θ

where tea is the angle between the two polarizers.

We apply this expression to our case

* Polarizer 1 suppose that it is vertical and polarizer 2 (intermediate) is at an angle θ with respect to the vertical

         I₁ = I₀ cos² θ

* We analyze for the polarity 2 and the last polarizer 3 which indicate that it must be at 90º from the first one, therefore it must be horizontal.

The angle of polarizers 2 and 3 is θ' measured from the horizontal, if we measure with respect to the vertical

              θ₂ = 90- θ’ = θ

fiate that in the exercise we must take a reference system and measure everything with respect to this system.

          I = I₁ cos² θ'

       

we substitute

         I = (I₀ cos² tea) cos² (θ - 90)

        cos (θ -90) = cos θ cos 90 + sin θ sin 90 = sin θ

         I = Io cos² θ sin² θ

        1= cos²θ+ sin²θ

       sin²θ = 1 - cos²θ

        I= I₀ (cos²θ - cos⁴θ)

b) to find when the intensity is maximum,

we can use that we have an extreme point when the drift is zero

          \frac{dI}{d \theta} = 0

          \frac{dI}{d \theta}= Io (2 cos θ - 4 cos³θ) = 0

whereby

            cos θ - 2 cos³ θ = 0

            cos θ ( 1 - 2 cos² θ) = 0  

The zeros of this function are in

           θ = 90º

           1-2cos²θ =0       cos θ = 0.25  θ =  75.5º

Let's analyze this two results for the angle of 90º the intnesidd is zero with respect to the first polarizer, so it is not an acceptable solution.

Consequently, the angle that allows the maximum intensity to pass is 75.5º

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