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Sever21 [200]
2 years ago
7

When viewing a piece of art that is behind glass, one often is affected by the light that is reflected off the front of the glas

s (called glare), which can make it difficult to see the art clearly. One solution is to coat the outer surface of the glass with a thin film to cancel part of the glare. Part A If the glass has a refractive index of 1.58 and you use TiO2, which has an index of refraction of 2.62, as the coating, what is the minimum film thickness that will cancel light of wavelength 480 nm
Physics
1 answer:
Elza [17]2 years ago
5 0

Answer:

  t = 9.16 10⁻⁸ m

Explanation:

In this exercise we fear a case of interference by reflection,

We have two aspects

* when the light ray passes from the air to the TiO2 that has a higher refraction start, the reflected ray has a phase change of 180º

* when the beam is inside the material its wavelength changes

            λₙ = λ₀ / n

with these two done the equation for destructive interference remains

           2t = m  λₙ

           2nt = m λ₀

They ask us for the thickness t of the film for m = 1

           t = m λ₀ / 2n

let's calculate

          t = \frac{1 \ 480 1 10^9 }{2 \ 2.62}

          t = 9.16 10⁻⁸ m

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The air pressure is most likely lower.
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3 years ago
What is the net force on a cart that is pulled to the right with 100 pounds of force and to the left with 30 pounds of force?
Arlecino [84]

Answer:

The answer to your question is: F = 70 pounds

Explanation:

To calculate the net force, only substract the left value to the right value.

                                F = F right  -  F left

                                F = 100 p - 30 p

                                F = 70 pounds

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3 years ago
Finish the sentence
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The first blank is character
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8 0
3 years ago
Block A of mass M is at rest and attached to the top of a spring. The block compresses the spring a distance d from its uncompre
Anni [7]

Answer:

a)  k = Mg / d , b)   v = √2gh , c)  v_{f} = \frac{2}{3} \ \sqrt{2gh},  d)   x² + 6d x - \frac{8}{3} dh = 0

e)the spring must compress a greater distance.

Explanation:

a) when the block of mass M is placed on the spring, we have an equilibrium condition,

             ∑ F  = 0

             F_{e}- W = 0

             k d = Mg

             k = Mg / d

b) let's use the concepts of energy to find the velocity of the block just before the collision

starting point. Position when released

          Em₀ = U = m g h

lowest point. Right at the point of shock

          Em_{f} = K = ½ m v²2

as there is no friction, energy is conserved

          Em₀ = Em_{f}

          mg h = ½ m v²

          v = √2gh

         

c) The velocity of the two blocks after the collision, we define a system formed by the two blocks, in such a way that the forces during the collision are internal and the moment is conserved

initial instant. Just before the crash

          p₀ = 2M v + M 0

final instant. Just after the shock, before the spring compression begins

         p_{f} = (2M + M) v_{f}

 the moment is preserved

          p₀ = p_{f}

          2M v = 3M v_{f}

          v_{f} = ⅔ v

          v_{f} = \frac{2}{3} \ \sqrt{2gh}

d) now we work with the joined system after the collision, let's use the concepts of energy

starting point. After shock, before beginning spring compression

        Em₀ = K = ½ (3M) v_{f}^2

        Em₀ = 3/2 M (\frac{2}{3} \ \sqrt{2gh})²

        Em₀ = 4/3 M gh

final point. With the spring fully compressed

       Em_f = K_e + U = ½ k x² + (3M) g x

in this case we have taken the zero of gravitational potential energy at the point where the blocks collide, as there is no friction, the energy is conserved

         Em₀ = Em_f

        4/3 M g h = ½ k x² + 3M g x

        ½ k x² + 3Mg x - 4/3 Mgh = 0

we substitute the expression for k

         \frac{1}{2} (\frac{Mg}{d}) x² + 3Mg x - \frac{4}{3} Mgh = 0

          \frac{x^{2} }{2d} + 3 x - \frac{4}{3}h = 0

to find the value of the spring compression, the second degree equation must be solved

          x² + 6d x - \frac{8}{3} dh = 0

         x = [-6d ±\sqrt{(36 d^{2} - 4 \frac{8}{3} dh)  } ] / 2

         x = [-6d ± 6d \sqrt{ 1 -  \frac{32}{3 \ 36}  \ \frac{h}{d}    }  ]/2

         x = 3d ( -1±  \sqrt{ 1 - 0.296 \frac{h}{d}   }  )

e) If the collision elastic force would not lose any part of the kinetic energy during the collision, therefore the speed of the block of mass M would be much higher and therefore the spring must compress a greater distance.

8 0
3 years ago
In the Rutherford model of hydrogen the electron (mass=9.11x10-31 kg) is in a planetary type orbit around the proton (mass=1.67x
Andrei [34K]

Answer:

(a) F_g=1.62*10^{-48}N

(b) F_e=3.68*10^{-9}N

Explanation:

(a) We use Newton's law of universal gravitation, in order to calculate the gravitational force between electron and proton:

F_g=-G\frac{m_1m_2}{r^2}

Where G is the Cavendish gravitational constant, m_1 and m_2 are the masses of the electron and the proton respectively and r is the distance between them:

F_g=-6.67*10^{-11}\frac{N\cdot m^2}{kg^2}\frac{(9.11*10^{-31}kg)(1.67*10^{-27}kg)}{(2.5*10^{-10}m)^2}\\F_g=-1.62*10^{-48}N

The minus sing indicates that the force is repulsive. Thus, its magnitude is:

F_g=1.62*10^{-48}N

(b) We use Coulomb's law, in order to calculate the electric force between electron and proton, here k is the Coulomb constant and e is the elementary charge:

F_e=-k\frac{e^2}{r^2}\\F_e=-8.99*10^{9}\frac{N\cdot m^2}{C^2}\frac{(1.6*10^{-19}C)^2}{(2.5*10^{-10}m)^2}\\F_e=-3.68*10^{-9}N

Its magnitude is:

F_e=3.68*10^{-9}N

6 0
3 years ago
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