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3 years ago

How light is channelled down an optical fibre

Physics

1 answer:

coldgirl [10]3 years ago

3 0

Explanation:

Suppose you want to shine a flashlight beam down a long, straight hallway. Just point the beam straight down the hallway -- light travels in straight lines, so it is no problem. What if the hallway has a bend in it? You could place a mirror at the bend to reflect the light beam around the corner. What if the hallway is very winding with multiple bends? You might line the walls with mirrors and angle the beam so that it bounces from side-to-side all along the hallway. This is exactly what happens in an optical fiber.

The light in a fiber-optic cable travels through the core (hallway) by constantly bouncing from the cladding (mirror-lined walls), a principle called total internal reflection. Because the cladding does not absorb any light from the core, the light wave can travel great distances.

However, some of the light signal degrades within the fiber, mostly due to impurities in the glass. The extent that the signal degrades depends on the purity of the glass and the wavelength of the transmitted light (for example, 850 nm = 60 to 75 percent/km; 1,300 nm = 50 to 60 percent/km; 1,550 nm is greater than 50 percent/km). Some premium optical fibers show much less signal degradation -- less than 10 percent/km at 1,550 nm.

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A squirrel jumps into the air with a velocity of 4 m/s at an angel of 50 degrees. What is the maximum height reached by the squi

Debora [2.8K]

Answer:

Explanation:

Assuming the squirrel is jumping off the ground, here's what we know but don't really know...

v₀ = 4.0 at 50.0°

So that's not really the velocity we are looking for. We are dealing with a max height problem, which is a y-dimension thing. Therefore, we need the squirrel's upward velocity, which is NOT 4.0 m/s. We find it in the following way:

$v_{0y}=4.0sin(50.0)$ which gives us that the upward velocity is

v₀ = 3.1 m/s

Moving on here's what we also know:

a = -9.8 m/s/s and

v = 0

Remember that at the very top of the parabolic path, the final velocity is 0. In order to find the max height of the squirrel, we need to know how long it took him to get there. We are using 2 of our 3 one-dimensional equations in this problem. To find time:

v = v₀ + at and filling in:

0 = 3.1 - 9.8t and

-3.1 = -9.8t so

t = .32 seconds.

Now that we know how long it took him to get to the max height, we use that in our next one-dimensional equation:

Δx = $v_0t+\frac{1}{2}at^2$ and filling in:

Δx = $3.1(.32)+\frac{1}{2}(-9.8)(.32)^2$ and using the rules for adding and subtracting sig fig's correctly, we can begin to simplify this:

Δx = .99 - .50 so

Δx = .49 meters

7 0

3 years ago

8. A car travels at a constant velocity of 70 mph for one hour. By the end of the second hour, the car’s velocity was 60 mph. At

Mrac [35]

<u>Answer:</u>

Positive acceleration is in third hour and negative acceleration is in second hour.

<u>Explanation:</u>

Velocity of car in first hour = 70 mph

Velocity of car in second hour = 60 mph

Velocity of car in third hour = 80 mph

Acceleration = Change in velocity / Time

Acceleration in second hour = (60 - 70)/1 = -10 mph²

Acceleration in third hour = (80 - 60)/1 = 20 mph²

So positive acceleration is in third hour and negative acceleration is in second hour.

8 0

3 years ago

A 10- kilogram block is pushed across a horizontal surface with a horizontal force of 20 N against a friction force of 10 N. The

Temka [501]

Answer:

$1m/s^2$

Explanation:

Mass of block=10 kg

Applied horizontal force =F=20 N

Friction force=f=10 N

We have to find the acceleration of block.

Net force=Applied horizontal force-friction force

$ma=F-f$

Where F= Horizontal force

f=Friction force

m=Mass of object

a=Acceleration of object

$10a=20-10=10$

$a=\frac{10}{10}=1 m/s^2$

Hence, the acceleration of the block= $1m/s^2$

4 0

3 years ago

What Kinetic energy is exactly equal to Gravitational Potential Energy why is height halfway between the maximum height?

prohojiy [21]

Explanation:

Let us calculate the work done in lifting an object of mass m through a height h, such as in Figure 1. If the object is lifted straight up at constant speed, then the force needed to lift it is equal to its weight mg. The work done on the mass is then W = Fd = mgh. We define this to be the gravitational potential energy (PEg) put into (or gained by) the object-Earth system. This energy is associated with the state of separation between two objects that attract each other by the gravitational force

Potential energy is a property of a system rather than of a single object—due to its physical position. An object’s gravitational potential is due to its position relative to the surroundings within the Earth-object system. The force applied to the object is an external force, from outside the system. When it does positive work it increases the gravitational potential energy of the system. Because gravitational potential energy depends on relative position, we need a reference level at which to set the potential energy equal to 0. We usually choose this point to be Earth’s surface, but this point is arbitrary; what is important is the difference in gravitational potential energy, because this difference is what relates to the work done. The difference in gravitational potential energy of an object (in the Earth-object system) between two rungs of a ladder will be the same for the first two rungs as for the last two rungs.

5 0

2 years ago

As a laudably skeptical physics student, you want to test Coulomb's law. For this purpose, you set up a measurement in which a p

monitta

Answer:

force F = 1.66 × $10^{-13}$ N