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Natasha2012 [34]
3 years ago
7

If the index of refraction for a certain glass is 1.5, and the angle of refraction is 15 degrees for a ray of light

Physics
1 answer:
LiRa [457]3 years ago
4 0

Answer:

\theta_i = 23^o

Explanation:

The refractive index of a medium is given by the following formula:

\eta = \frac{Sin\theta_i}{Sin\theta_r}

where,

η = refractive index = 1.5

\theta_i = angle of incidence = ?

\theta_r = angle of refraction = 15°

Therefore,

1.5 = \frac{Sin\theta_i}{Sin15^o}\\\\Sin\theta_i=(1.5)Sin15^o\\\\\theta_i = Sin^{-1}(0.388)\\\\

\theta_i = 23^o

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3 years ago
ou have been called to testify as an expert witness in a trial involving a head-on collision. Car A weighs 660.0 kg and was trav
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Answer:

    vₐ₀ = 29.56 m / s

Explanation:

In this exercise the initial velocity of car A is asked, to solve it we must work in parts

* The first with the conservation of the moment

* the second using energy conservation

let's start with the second part

we must use the relationship between work and kinetic energy

             W = ΔK                             (1)

for this part the mass is

             M = mₐ + m_b

the final velocity is zero, the initial velocity is v

friction force work is

              W = - fr x

the negative sign e because the friction forces always oppose the movement

we write Newton's second law for the y-axis

              N -W = 0

              N = W = Mg

friction forces have the expression

              fr =μ N

              fr = μ M g

we substitute in 1

               -μ M g x = 0 - ½ M v²

             v² = 2 μ g x

let's calculate

              v² = 2  0.750  9.8  6.00

              v = ra 88.5

              v = 9.39 m / s

Now we can work on the conservation of the moment, for this part we define a system formed by the two cars, so that the forces during the collision are internal and therefore the tsunami is preserved.

Initial instant. Before the crash

         p₀ = + mₐ vₐ₀ - m_b v_{bo}

instant fianl. Right after the crash, but the cars are still not moving

         p_f = (mₐ + m_b) v

         p₀ = p_f

         + mₐ vₐ₀ - m_b v_{bo} = (mₐ + m_b) v

           

         mₐ vₐ₀ = (mₐ + m_b) v + m_b v_{bo}

let's reduce to the SI system

          v_{bo} = 64.0 km / h (1000m / 1km) (1h / 3600s) = 17.778 m / s

let's calculate

         660 vₐ₀ = (660 +490) 9.39 + 490 17.778

         vₐ₀ = 19509.72 / 660

         vₐ₀ = 29.56 m / s

we can see that car A goes much faster than vehicle B

5 0
3 years ago
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