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vesna_86 [32]
4 years ago
7

What impact has the use of optical fibre cables had on the way we receive and communicate information? Discuss both positive AND

negative effects.
Physics
1 answer:
frozen [14]4 years ago
5 0

Due to rising demand for higher bandwidth and faster speed connections, fiber optic transmission is becoming more and more common in the developing world. Optical fiber has had a great impact on the way we receive and transmit signals. They have largely replaced copper wire communications in core networks in the developed world, because of its advantages over electrical transmission. But they brought some difficulties as well.

Following are the advantages and disadvantages of optical fiber:

Advantages

  • They provide extremely High Bandwidth.
  • Optical fiber is capable of low power loss at long distances.
  • Optical fiber is highly resistant to electromagnetic disturbances. They are virtually error free.
  • The data is transferred through light that is why there is no security risk of data being detected.
  • Optical fiber has lesser weight and size as compared to other cables.

Disadvantages

  • They are made of glass that is why they are very fragile and they have to be handled carefully.
  • They are difficult to install. If you bend them too much, they will break. And fiber cable is highly susceptible to damage during installation or construction activities.
  • Extra optical components have to be added to handle dispersion of light at longer distances.
  • Optical fiber costs a lot more than copper cables. The installation itself is very costly.

Keywords: fiber optics, bandwidth, transmission, impact, advantages, disadvantages

Learn more about Fiber Optics from brainly.com/question/1373678

#learnwithBrainly

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A cyclist rides at 6.20 m/s through a intersection. A stationary car begins to
Xelga [282]

Answer:

The width of the intersection is 20 meters

Explanation:

The speed with which the cyclist is riding, v₁ = 6.20 m/s

The rate at which the car starts to accelerate, a = 3.844 m/s²

The initial velocity of the car = The car is stationary at the start = 0 m/s

The time at which the cyclist and the car reach the other side of the intersection = The same time;

Let 't' represent the time at which the cyclist and the car both reach the other side of the intersection, we have;

The distance travelled by the cyclist = The distance traveled by the car

∴ v₁ × t = 1/2 × a × t²

Plugging in the values for 'v₁', and 'a' in the above equation, we get;

6.20 × t = 1/2 × 3.844 × t²

∴ 1.922·t² - 6.20·t = 0

∴ t·(1.922·t - 6.20) = 0

t = 0, or t = 6.20/1.922 = 100/31

The time at which the cyclist and the car both reach the other side of the intersection, t = 100/31 seconds

The with of the intersection, w = v₁ × t

∴ w = 6.20 × 100/31 = 100/5 = 20

The width of the intersection, w = 20 meters.

8 0
3 years ago
3. How is using a model to study cells helpful?
4vir4ik [10]

Answer:

Yes, it is very helpful.

Explanation:

It's helpful since in a cell, plant or animal, there are a lot of different things. It's hard to memorize everything and know what they look like. Using a model can help you memorize everything better and even understand it better. If someone asked me where or what something was in a cell I think I would be able to recognize it better.

I hope this helps!

6 0
3 years ago
1. Two forces act on a box as follows: F= 100 N at 02 = 170. and F2= 75 N at
Setler [38]

Answer:

yfv.

Explanation:

7 0
3 years ago
A bicyclist starting at rest produces a constant angular acceleration of 1.30 rad/s2 for wheels that are 35.5 cm in radius.
Debora [2.8K]

Answer:

a) 0.462 m/s^2

b) 31.5 rad/s

c) 381 rad

d) 135m

Explanation:

the linear acceleration is given by:

a=\alpha *r\\a=1.30rad/s^2*(35.5*10^{-2}m)\\a=0.462m/s^2

the angular speed is given by:

\omega=\frac{v}{r}\\\\\omega=\frac{11.2m/s}{35.5*10^{-2}m}\\\\\omega=31.5rad/s

to calculate how many radians have the wheel turned we need the apply the following formula:

\theta=\frac{1}{2}\alpha*t^2\\\\t=\frac{\omega}{\alpha}\\\\t=\frac{31.5rad/s}{1.30rad/s^2}\\\\t=24.2s\\\\\theta=\frac{1}{2}*1.30rad/s^2*(24.2s)^2\\\\\theta=381rad

the distance is given by:

d=\theta*r

d=381rad*(35.5*10^{-2}m)\\d=135m

4 0
4 years ago
A body weights 50 N in air and 45 N when wholly immersed in water calculate (i) the loss in weight of the body in water (ii) the
Lelechka [254]

Answer:

difference  \: in \: weight = 150n - 100n = 50n

Now,buyantant force

difference \: in \: weight \: = volume(body) \times density \: of \: water \:  \times g

so;

50 =  {v}^{b}  \times 1 \times  {10}^{3}  \times 9.8m {s}^{2}

{v}^{b}  =  \frac{50}{1000 } \times 9.8

=  \frac{50}{9800}

= 0.0051

Now,

mass \: in \: air \:  = 150n =  \frac{150}{9.8kg}

density =  \frac{weght}{volume}

=  \frac{150}{0.0051}  \times 9.8 \\ x = 3000

And now,

specific \: density \:  =  \frac{density of \: the \: body}{density \: of \: water}

=  \frac{3000}{1000}

= 3

Hence that,specific density of a given body is 3

please mark me as brainliest, please

3 0
3 years ago
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