May I please get some help?

A light ray traveling through a material with an index of refraction of 1.2 is incident on a material that has an index of refra

zzz [600]

Answer:

compared to the incident angle, the refracted angle is 45.56⁰

Explanation:

From Snell's law;

n₁sin(I) = n₂sin(r)

Where;

n₁ is the refractive index of light in medium 1 = 1.2

n₂ is the refractive index of light in medium 2 = 1.4

I is the incident angle

r is the refractive angle

$n = \frac{1}{sin(I)}\\\\sin(I) = \frac{1}{n}\\\\sin(I) =\frac{1}{1.2}\\\\sin(I) =0.8333\\\\I = sin^-{(0.8333)$

I = 56.439⁰

Applying snell's law

$n_1sin(I) = n_2sin(r)\\\\sin(r) = \frac{n_1sin(I) }{n_2}\\\\sin(r) = \frac{1.2*sin(56.439) }{1.4}\\\\sin(r) = 0.714\\\\r = sin^-(0.714)\\\\r = 45.56^o$

Therefore, compared to the incident angle, the refracted angle is 45.56⁰

3 0

3 years ago

Which of the following is not an example of kinetic energy being converted to potential energy?

KengaRu [80]

The list of choices you provided with your question
is utterly devoid of any such examples.

6 0

3 years ago

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What absorb emitted neutrons?

Karolina [17]

Nuclear energy is released during: fission. radioactive decay. man-induced splitting of atoms. Match the basic components of a nuclear reactor with their descriptions. 1. slows down neutrons -> moderator. 2. absorb emitted neutrons -> control rods.

5 0

3 years ago

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Three beads are placed along a thin rod. The first bead, of mass m1 = 23 g, is placed a distance d1 = 1.1 cm from the left end o

Mila [183]

Answer:

a) $x=\frac{m_{1}d_{1}+m_{2}(d_{1}+d_{2})+m_{3}(d_{1}+d_{2}+d_{3} ) }{m_{1}+m_{2}+m_{3} }$

b) x = 4.47 cm

c) $x=\frac{m_{1}d_{2}+m_{2}(0)+m_{3}d_{3} }{m_{1}+m_{2}+m_{3} }$

d) x = 1.48 cm

Explanation:

a) The center of mass is equal to:

$x=\frac{m_{1}x_{1}+m_{2}x_{2}+m_{3}x_{3} }{m_{1}+m_{2} +m_{3}}$

Where m is the mass of beads and x is the distances, if x₁ = d₁, x₂ = d₂ and x₃ = d₃

$x=\frac{m_{1}d_{1}+m_{2}(d_{1}+d_{2})+m_{3}(d_{1}+d_{2}+d_{3} ) }{m_{1}+m_{2}+m_{3} }$

b) If

m₁ = 23g

m₂ = 15 g

m₃ = 58 g

d₁ = 1.1 cm

d₂ = 1.9 cm

d₃ = 3.2 cm

$x=\frac{23*1.1+15*(1.1+1.9)+58(1.1+1.9+3.2) }{23+15+58 } =4.47cm$

c) The center of the mass of the beads realtive to the center of bead is:

$x=\frac{m_{1}d_{2}+m_{2}(0)+m_{3}d_{3} }{m_{1}+m_{2}+m_{3} }$

d) $x=\frac{23*(-1.9)+(15*0)+(58*3.2) }{23+15+58 } =1.48cm$

6 0

3 years ago

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the very high voltage needed to create a spark across the spark plug is produced at the a. transformer's primary winding. b. tra

Karo-lina-s [1.5K]

I think the correct answer from the choices listed above is option B. The very high voltage needed to create a spark across the spark plug is produced at the transformer's secondary winding. <span>The secondary coil is engulfed by a powerful and changing magnetic field. This field induces a current in the coils -- a very high-voltage current.</span>

5 0

3 years ago

Read 2 more answers