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Mariulka [41]
3 years ago
7

Suppose a ray of light traveling in a material with an index of refraction na reaches an interface with a material having an ind

ex of refraction nb. Which of the following statements must be true for total internal reflection to occur?Check all that apply.a. The angle of incidence must be less than the critical angle.b. na = nbc. na > nbd. The angle of incidence must be greater than the critical angle.e. The angle of incidence must be equal to the critical angle.
Physics
1 answer:
ivolga24 [154]3 years ago
5 0

Answer: Option (b) is the correct answer.

Explanation:

It is known that when a ray of light tends to travel from a denser to rarer medium then there occurs total internal reflection.

For a denser medium the refractive index is greater than that of a rarer medium. This means that for the given situation refractive index of medium n_{a} is greater than medium n_{b}.

this also means that incident angle must be greater than the critical angle of the medium.

Thus, we can conclude that the statement n_{a} > n_{b} must be true for total internal reflection to occur.

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Formula\ for\ period:\\\ T=2 \pi \sqrt{\frac{L}{g}}\\\ g-gravity=9,8 \frac{m}{s^2} ,\ L-pendulum \ length \\\\ \frac{T}{2 \pi } = \sqrt{ \frac{L}{g} }|square\\\\ \frac{T^2}{2 \pi } = \frac{L}{g} \\\\\ \frac{T^2}{2 \pi }*g=L\\\\ L= \frac{2^2}{2*3,14 }*9,8= \frac{39,2}{6,28} =6,24mT=2 \pi   \sqrt{\frac{L}{g}} \\
 \frac{T}{2 \pi } = \sqrt{ \frac{L}{g} }|square\\
 \frac{T^2}{2 \pi }  = \frac{L}{g} \\
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L= \frac{2^2}{2*3,14 }*9,8= \frac{39,2}{6,28} =6,24m

7 0
3 years ago
E asteroid belt circles the sun between the orbits of mars and jupiter. one asteroid has a period of 6.0 earth years. part a wha
anzhelika [568]

To solve the problem, use Kepler's 3rd law : 

T² = 4π²r³ / GM 

Solved for r : 

r = [GMT² / 4π²]⅓ 

but first covert 6.00 years to seconds : 

6.00years = 6.00years(365days/year)(24.0hours/day)(6... 
= 1.89 x 10^8s 

The radius of the orbit then is : 

r = [(6.67 x 10^-11N∙m²/kg²)(1.99 x 10^30kg)(1.89 x 10^8s)² / 4π²]⅓ 
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6 0
3 years ago
Mimas, a moon of Saturn, has an orbital radius of 1.62 × 108 m and an orbital period of about 23.21 h. Use Newton’s version of K
Drupady [299]

Answer:

3.60432\times 10^{26}\ kg

Explanation:

a = Orbital radius = 1.62\times 10^8\ m

T = Orbital period = 23.21 hours

G = Gravitational constant = 6.67 × 10⁻¹¹ m³/kgs²

From Kepler's third law we get

M=\frac{4\pi^2a^3}{GT^2}\\\Rightarrow M=\frac{4\pi^2\times (1.62\times 10^8)^3}{6.67\times 10^{-11}\times (23.21\times 3600)^2}\\\Rightarrow M=3.60432\times 10^{26}\ kg

From the given data the mass of Saturn is 3.60432\times 10^{26}\ kg

8 0
2 years ago
If a car starts at rest and accelerates at 3.4 m/s^2 over a distance of 200 meters, how fast will it be travelling at the end?
CaHeK987 [17]

Answer:

36.9 m/s

Explanation:

From;

v^2 = u^2 + 2as

Where;

v = final velocity =?

a = acceleration = 3.4 m/s^2

u = initial velocity = 0 m/s

s = distance covered = 200 meters

v^2 = 0^2 + 2 * 3.4 * 200

v^2 = 1360

v = √1360

v = 36.9 m/s

8 0
2 years ago
Define what an astronomical unit is and explain why it is used.
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Answer:

AU

Explanation:

Astronomical units are used to measure the  distances within solar system

5 0
3 years ago
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