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myrzilka [38]
2 years ago
7

Think about the way optical fibers are designed. why are the ends of the fibers glowing very brightly while the sides of the fib

ers are not?
A. because the light only strikes the ends of the fibers, not the sides
B. because the fibers’ sides absorb any light that strikes it
C. because the ends of the fibers convert the signal into visible light
D. because the light is reflected back into the fiber along its sides
PLEASE HELP ITS MY LAST QUESTION!!
Physics
1 answer:
Effectus [21]2 years ago
6 0

Answer:

D. because the light is reflected back into the fiber along its sides

Explanation:

The fiber is constructed in a way that the light is bent/reflected/refracted toward the center core of glass. So, from the center core, there is a layer above it that has a different propagation than the core, and above that the same thing. To give you a real world visual example, if you look down in a pool of water, then stick a straight stick into it, you see that the straight stick appears to bend. That is what is happening to the light as it travels through a different medium (air to water). This same effect is incorporated in the fiber optic cable construction.

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A 1.5m wire carries a 4 A current when a potential difference of 66 V is applied. What is the resistance of the wire?
kifflom [539]
Resistance = Voltage / current
Resistance = 66/4
=16.5ohms
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3 years ago
_______ was the first person to propose the idea of moving continents as a scientific hypothesis.
yan [13]
The answer is b alfred wegener 


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Imagine how mercury might be different if it had the same mass as earth. Explain.
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How long will it take for a car to travel 200 km if it has an average speed of 55 km/hr? ​
andreyandreev [35.5K]

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4 0
3 years ago
How do I calculate the tension in the horizontal string?
matrenka [14]

ANSWER

T₂ = 10.19N

EXPLANATION

Given:

• The mass of the ball, m = 1.8kg

First, we draw the forces acting on the ball, adding the vertical and horizontal components of each one,

In this position, the ball is at rest, so, by Newton's second law of motion, for each direction we have,

\begin{gathered} T_{1y}-F_g=0_{}_{}_{} \\ T_2-T_{1x}=0 \end{gathered}

The components of the tension of the first string can be found considering that they form a right triangle, where the vector of the tension is the hypotenuse,

\begin{gathered} T_{1y}=T_1\cdot\cos 30\degree \\ T_{1x}=T_1\cdot\sin 30\degree \end{gathered}

We have to find the tension in the horizontal string, T₂, but first, we have to find the tension 1 using the first equation,

T_1\cos 30\degree-m\cdot g=0

Solve for T₁,

T_1=\frac{m\cdot g}{\cos30\degree}=\frac{1.8kg\cdot9.8m/s^2}{\cos 30\degree}\approx20.37N

Now, we use the second equation to find the tension in the horizontal string,

T_2-T_1\sin 30\degree=0

Solve for T₂,

T_2=T_1\sin 30\degree=20.37N\cdot\sin 30\degree\approx10.19N

Hence, the tension in the horizontal string is 10.19N, rounded to the nearest hundredth.

8 0
1 year ago
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