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I am Lyosha [343]
2 years ago
7

Which statement is true for light passing into a medium that is less optically dense than the first medium through which it pass

ed?
The index of refraction of the second medium is higher.


The index of reflection of the second medium is lower.


The index of reflection of the second medium is higher.


The index of refraction of the second medium is lower.
Physics
2 answers:
BigorU [14]2 years ago
5 0

Answer:

The correct option is;

The index of refraction of the second medium is lower

Explanation:

The index of refraction of a material indicates the magnitude of the optical density of a material. The index of refraction or the refractive index, n, are indices (ratio) of the speed of light through an optically dense medium relative to the speed of light through a vacuum.

The definition of the refractive index is the number of times light travelling through a medium would be slower than light travelling through vacuum

Therefore, the index of refraction of a second medium that is less optically dense than a first medium from which light originates and travels through it would be lower than the index of refraction of the first medium

AleksAgata [21]2 years ago
5 0

Answer:

D

Explanation:

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An electric motor is rated at 900 W. How much force does it apply when moving
brilliants [131]
<h3>Answer:  7.74 newtons</h3>

========================================

Steps Shown:

Work = Force*Displacement

Power = Work/Time

Power = (Force*Displacement)/Time

900 W = (F*1000 m)/(8.6 sec)

900 = (F*1000)/8.6

900*8.6 = 1000F

7740 = 1000F

1000F = 7740

F = 7740/1000

F = 7.74 newtons

6 0
2 years ago
A converging lens of focal length 20 cm is used to form a real image 1.0 m away from the lens. How far from the lens is the obje
Galina-37 [17]

Answer:

0.25 m

Explanation:

We can solve the problem by using the lens equation:

\frac{1}{f}=\frac{1}{p}+\frac{1}{q}

where

f is the focal length

p is the distance of the object from the lens

q is the distance of the image from the lens

In this problem, we have

f = +20 cm=+0.20 m (the focal length is positive for a converging lens)

q = +1.0 m (the image distance is positive for a real image)

Solving the equation for p, we find

\frac{1}{p}=\frac{1}{f}-\frac{1}{q}=\frac{1}{0.20 m}-\frac{1}{1 m}=4 m^{-1}\\p=\frac{1}{4 m^{-1}}=0.25 m

6 0
3 years ago
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
2 years ago
If a change is made to a system in equilibrium, the equilibrium will shift to oppose the change.
Aleksandr [31]

Answer:

No

Explanation:

Because it won’t

5 0
3 years ago
Read 2 more answers
If cells were not able to obtain amino acids, what macromolecule would not be able to be synthesized
Zielflug [23.3K]

Answer:

protein

Explanation:

protein is a very large complex macro-molecule that requires amino acids

5 0
2 years ago
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