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natima [27]
3 years ago
5

Total internal reflection is the complete reflection of a light ray _____.

Physics
2 answers:
Licemer1 [7]3 years ago
7 0
B- Back into it's original medium.
Hope this helps!
Finger [1]3 years ago
6 0

Answer: (B). Total internal reflection is the complete reflection of a light ray back into its original medium.

Explanation:

Reflection: The reflection will, when the incidence angle will make between 0^{\circ} and 90^{\circ}.

In the reflection, the light ray moves one medium to another medium.

Total internal reflection: The total internal reflection will, when the incidence angle is greater than the critical angle.

In the total reflection, The light ray comes back into the original medium after reflection.

Critical angle: Critical angle permits the ray of light to be reflected back in the medium n₁ and it does not refract into the medium n₂.  

Hence, Total internal reflection is the complete reflection of a light ray back into its original medium.

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A 40kg kid riding their bike down the street at 5m/s reaches the edge of a hill and coast down to the bottom. If the hill was 10
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Which statement best describes a device that has a high energy efficiency? Its ratio of energy absorbed to energy released is hi
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Answer:

It wastes little energy

Explanation:

  • The efficiency of a machine refers to the ratio of work output to the work input as a percentage.
  • The efficiency of most machines is less than 100% because some work is lost  due to friction and heat.
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Attempt 2 You have been called to testify as an expert witness in a trial involving a head-on collision. Car A weighs 15151515 l
Natasha_Volkova [10]

Answer:

v = 28.98 ft / s

Explanation:

For this problem we must solve it in parts, let's start by looking for the speed of the two cars after the collision

In the exercise they indicate the weight of each car

          Wₐ = 1500 lb

          W_b = 1125 lb

Car B's velocity from v_b = 42.0 mph westward, car A travels east

let's find the mass of the vehicles

             W = mg

             m = W / g

             mₐ = Wₐ / g

             m_b = W_b / g

             mₐ = 1500/32 = 46.875 slug

             m_b = 125/32 = 35,156 slug

Let's reduce to the english system

             v_b = 42.0 mph (5280 foot / 1 mile) (1h / 3600s) = 61.6 ft / s

We define a system formed by the two vehicles, so that the forces during the crash have been internal and the moment is preserved

we assume the direction to the east (right) positive

initial instant. Before the crash

           p₀ = mₐ v₀ₐ - m_b v_{ob}

final instant. Right after the crash

           p_f = (mₐ + m_b) v

the moment is preserved

           p₀ = p_f

           mₐ v₀ₐ - m_b v_{ob} = (mₐ + m_b) v

           v = \frac{ m_a \ v_{oa} - m_b \ v_{ob}  }{ m_a +m_b}

we substitute the values

           v = \frac{ 46.875}{82.03} \ v_{oa} -  \frac{35.156}{82.03} \ 61.6

           v = 0.559 v₀ₐ - 26.40                  (1)

Now as the two vehicles united we can use the relationship between work and kinetic energy

the total mass is

              M = mₐ + m_b

              M = 46,875 + 35,156 = 82,031 slug

starting point. Jsto after the crash

              K₀ = ½ M v²

final point. When they stop

             K_f = 0

The work is

             W = - fr x

the negative sign is because the friction forces are always opposite to the displacement

Let's write Newton's second law

Axis y

           N-W = 0

           N = W

the friction force has the expression

            fr = μ N

we substitute

            -μ W x = Kf - Ko

             

            -μ W x = 0 - ½ (W / g) v²

            v² = 2 μ g x  

            v = \sqrt{ 2 \ 0.750 \ 32 \ 17.5}Ra (2 0.750 32 17.5  

            v = 28.98 ft / s

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There are 1000 liters in a cubic meter so this is 81250 liters. Divide by 4.2 to find the number of seconds required to pump out this much water and we get 19345.2 seconds. This equals approximately 5.37 hours.</span>
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