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Sphinxa [80]
3 years ago
5

What process makes it possible for optical fibers to transmit telephone conversations over long distances?

Physics
1 answer:
11111nata11111 [884]3 years ago
3 0

Light travels down a fiber-optic cable by bouncing repeatedly off the walls. Each tiny photon (particle of light) bounces down the pipe like a bobsleigh going down an ice run. Now you might expect a beam of light, traveling in a clear glass pipe, simply to leak out of the edges. But if light hits glass at a really shallow angle (less than 42 degrees), it reflects back in again—as though the glass were really a mirror. This phenomenon is called total internal reflection. It's one of the things that keeps light inside the pipe.

Artwork: Right: Total internal reflection keeps light rays bouncing down the inside of a fiber-optic cable.

The other thing that keeps light in the pipe is the structure of the cable, which is made up of two separate parts. The main part of the cable—in the middle—is called the core and that's the bit the light travels through. Wrapped around the outside of the core is another layer of glass called the cladding. The cladding's job is to keep the light signals inside the core. It can do this because it is made of a different type of glass to the core. (More technically, the cladding has a lower refractive index.)

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<span>Let F be the force of gravity, G be the gravitational constant, M be the mass of the earth, m your mass and r the radius of the earth, then: 

F = G(Mm / (4(pi)*r^2)) 

The above expression gives the force that you feel on the earth's surface, as it is today! 

Let us now double the mass of the earth and decrease its diameter to half its original size. 

This is the same as replacing M with 2M and r with r/2. 

Now the gravitational force (F' ) on the new earth's surface is given by: 

F' = G(2Mm / (4(pi)(r/2)^2)) = 2G(Mm / ((1/4)*4(pi)*r^2)) = 8G(Mm / (4(pi)*r^2)) = 8F 

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F' = 8F 

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You would be 8 times heavier on this "new" earth!</span>
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You see a marble sign. Its letters are too worn to read. What caused this?
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A track begins at 0 meters and has a total distance of 100 meters. Juliet starts at the 10-meter mark while practicing for a rac
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A car is stopped at a traffic light. When the light turns green at t=0, a truck with a constant speed passes the car with a 20m/
s344n2d4d5 [400]

Answer:

At t = (70 / 3) \; {\rm s} (approximately 23.3 \; {\rm s}.)

Explanation:

Note that the acceleration of the car between t = 0\; {\rm s} and t = 20\; {\rm s} (\Delta t = 20\; {\rm s}) is constant. Initial velocity of the car was v_{0} = 0\; {\rm m\cdot s^{-1}}, whereas v_{1} = 35\; {\rm m\cdot s^{-1}} at t = 20\; {\rm s}\!. Hence, at t = 20\; {\rm s}\!\!, this car would have travelled a distance of:

\begin{aligned}x &= \frac{(v_{1} - v_{0})\, \Delta t}{2} \\ &= \frac{(35\; {\rm m\cdot s^{-1}} - 0\; {\rm m\cdot s^{-1}}) \times (20\; {\rm s})}{2} \\ &= 350\; {\rm m}\end{aligned}.

At t = 20\; {\rm s}, the truck would have travelled a distance of x = v\, t = 20\; {\rm m\cdot s^{-1}} \times 20\; {\rm s} = 400\; {\rm m}.

In other words, at t = 20\; {\rm s}, the truck was 400\; {\rm m} - 350\; {\rm m} = 50\; {\rm m} ahead of the car. The velocity of the car is greater than that of the truck by 35\; {\rm m\cdot s^{-1}} - 20\; {\rm m\cdot s^{-1}} = 15 \; {\rm m\cdot s^{-1}}. It would take another (50\; {\rm m}) / (15\; {\rm m\cdot s^{-1}}) = (10/3)\; {\rm s} before the car catches up with the truck.

Hence, the car would catch up with the truck at t = (20 + (10/3))\; {\rm s} = (70 / 3)\; {\rm s}.

3 0
2 years ago
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