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stealth61 [152]
2 years ago
14

A glass optical fiber is used to transport a light ray across a long distance. The fiber has an index of refraction of 1.540 and

is submerged in glycerin, which has an index of refraction of 1.473. What is the critical angle (in degrees) for the light ray to remain inside the fiber
Physics
1 answer:
Zinaida [17]2 years ago
4 0

Answer:

73.13°

Explanation:

According to snell's law,

n1sinθi = n2sinθr

n1/n2 = sinθr/sinθi

Critical angle is the angle of incidence at the denser medium when the angle of incidence at the less dense medium is 90°

This means i=C and r = 90°

The Snell's law formula will become

n1/n2 = sinC/sin90°

n2/n1 = 1/sinC

Where n1 is the refractive index of the less dense medium = 1.473

n2 is the refractive index of the denser medium = 1.540

Substituting the values in the formula,

1.540/1.473 = 1/sinC

1.045 = 1/sinC

SinC = 1/1.045

SinC = 0.957

C = sin^-1(0.957)

C = 73.13°

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The statements below are all true. Some of them represent important reasons why the giant impact hypothesis for the Moon’s forma
Molodets [167]

Answer:

the order of importance must be     b e a f c

Explanation:

Modern theories indicate that the moon was formed by the collision of a bad plant with the Earth during its initial cooling period, due to which part of the earth's material was volatilized and as a ring of remains that eventually consolidated in Moon.

Based on the aforementioned, let's analyze the statements in order of importance

b) True. Since the moon is material evaporated from Earth, its compassion is similar

e) True. If the moon is material volatilized from the earth it must train a finite receding speed

a) True. The solar system was full of small bodies in erratic orbits that wander between and with larger bodies

f) False. The moon's rotation and translation are equal has no relation to its formation phase

c) false. The amount of vaporized material on the moon is large

Therefore, the order of importance must be

         b e a f c

5 0
2 years ago
A chair of mass 30.0 kg is at rest on a horizontal floor. The floor is not frictionless. You push on the chair with a force of 8
miv72 [106K]
First make sure you draw a force diagram. You should have Fn going up, Fg going down, Ff going left and another Fn going diagonally down to the right. The angle of the diagonal Fn (we'll call it Fn2) is 35° and Fn2 itself is 80N. Fn2 can be divided into two forces: Fn2x which is horizontal, and Fn2y which is vertical. Right now we only care about Fn2y.

To solve for Fn2y we use what we're given and some trig. Drawing out the actual force of Fn2 along with Fn2x and Fn2y we can see it makes a right triangle, with 80 as the hypotenuse. We want to solve for Fn2y which is the opposite side, so Sin(35)=y/80. Fn2y= 80sin35 = 45.89N

Next we solve for Fg. To do this we use Fg= 9.8 * m. Mass = 30kg, so Fg = 9.8 * 30 = 294N.

Since the chair isn't moving up or down, we can set our equation equal to zero. The net force equation in the vertical direction will be Fn + Fn2y -Fg = 0. If we plug in what we know, we get Fn + 45.89 -294 = 0. Then solve this algebraically.

Fn +45.89 -294 = 0
Fn +45.89 = 294
Fn = 248.11 N

You'll get a more accurate answer if you don't round Fn2y when solving for it, it would be something along the lines of 45.88611 etc
7 0
2 years ago
Read 2 more answers
A 0.600-mm diameter wire stretches 0.500% of its length when it is stretched with a tension of 20.0 n. what is the young's modul
Rashid [163]
The Young modulus is given by:
E= \frac{F /A}{\Delta L / L_0}
where
F is the force applied
L_0 is the initial length of the wire
A is the cross-sectional area of the wire
\Delta L is the stretch of the wire

The wire in the problem stretches by 0.5% of its length, this means 
\frac{\Delta L}{L_0}  = 0.005

We can also calculate the area of the wire; its radius is in fact half the diameter:
r= \frac{d}{2}= \frac{0.600 mm}{2}=0.300 mm=0.3 \cdot 10^{-3} m
and so the area is
A=\pi r^2 = \pi (0.3 \cdot 10^{-3} m)^2 = 2.83 \cdot 10^{-7} m^2

We know the force applied to the wire, F=20 N, so now we have everything to calculate the Young modulus:

E=  \frac{F/A}{\Delta L / L_0} = \frac{20 N/(2.83 \cdot 10^{-7} m^2)}{0.005}=1.42 \cdot 10^{10} N/m^2
3 0
2 years ago
In a charge-free region of space, a closed container is placed in an electric field. Which of the following is a requirement for
galina1969 [7]

Answer:

D. The requirement does not exist -the total electric flux is zero no matter what.

Explanation:

According to Gauss's law , total electric flux over a closed surface is equal to 1 / ε₀ times charge inside.

If charge inside is zero , total electric flux over a closed surface is equal to

zero . It has nothing to do with whether external field is uniform or not. For any external field , lines entering surface will be equal to flux going out.

8 0
2 years ago
A 50-kg platform diver hits the water below with a kinetic energy of 5000 Joules. The height (relative to the water) from which
artcher [175]

Answer:

h = 10 m

Explanation:

given,

mass of platform = 50 Kg

Kinetic energy = 5000 J

height from which the diver dove = ?

taking acceleration due to gravity = 10 m/s²

using conservation of energy

Kinetic energy is converted into mechanical energy

K.E = P.E

K.E =  m g h

5000 = 50 x 10 x h

500 h = 5000

 h = \dfrac{5000}{500}

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The height from which the diver dove is equal to h = 10 m

6 0
2 years ago
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