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Elanso [62]
3 years ago
7

What’s the difference between fiber optic cable, twisted pair cable and coaxial cable?

Physics
1 answer:
Nikitich [7]3 years ago
4 0
Twisted pair cable consists of a pair of insulated wires twisted together, which is adapted in the field of telecommunication for a long time. With the cable twisting together, it helps to reduce noise from outside sources and crosstalk on multi-pair cables. Basically, twisted pair cable can be divided into two types: unshielded twisted-pair (UTP<span>) and shielded twisted-pair (STP). The former serves as the most commonly used one with merely two insulated wires twisted together. Any data communication cables and normal telephone cables belong to this category. However, shielded twisted pair distinguishes itself from UTP in that it consists of a foil jacket which helps to prevent crosstalk and noise from outside source. It is typically used to eliminate inductive and capacitive coupling, so it can be applied between equipment, racks and buildings. There exist following several different types of 




</span><span>Coaxial cable acts as a high-frequency transmission cable which contains a single solid-copper core. A coaxial cable has over 80 times the transmission capability of the twisted-pair. It is commonly used to deliver television signals and to connect computers in a network as well, so people may get more familiar with this kind of cable. There are two coaxial cables: 75 Ohm and 50 Ohm.
</span>


omputing and data communications are fast-moving technologies. There comes a new generation of transmission media—fiber optic cable. It refers to the complete assembly of fibers, which contain one or more optical fibers that are used to transmit data. Each of the optical fiber elements is individually coated by plastic layers and contained in a protective tube. Fiber optic cable transmits data as pulses of light go through tiny tubes of glass, the transmission capacity of which is 26,000 times higher than that of twisted-pair cable. When comparing with coaxial cables, fiber optic cables are lighter and reliable for transmitting data. They transmit information using beams of light at light speed rather than pulses of electricity.

Nowadays, two types of fiber optic cables are widely adopted in the field of data transfer—single-mode fiber optic cables and multimode fiber optic cables. A single-mode optical fiber is a fiber that has a small core, and only allows one mode of light to propagate at a time. So it is generally adapted to high speed, long-distance applications. While a multimode optical fiber is a type of optical fiber with a core diameter larger than the wavelength of light transmitted and it is designed to carry multiple light rays, or modes at the same time. It is mostly used for communication over short distances because of its high capacity and reliability, serving as a backbone applications in buildings.

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2 years ago
I need help, please answer
Burka [1]

This being a perfect collision means no energy is lost during the collision. Because this question asks for speed and not velocity, the speed will be the same because the final energy is the same. The speed after the collision would therefore be 1.27 m/s.

7 0
3 years ago
A force of 40 N is required to hold a spring that has been stretched from its natural length of 10 cm to a length of 15 cm. How
Sati [7]

0.36 J of work is done in stretching the spring from 15 cm to 18 cm.

To find the correct answer, we need to know about the work done to strech a string.

<h3>What is the work required to strech a string?</h3>
  • Mathematically, the work done to strech a string is given as 1/2 ×K×x².
  • K is the spring constant.
<h3>What will be the spring constant, if 40N force is required to hold a 10 cm to 15 cm streched spring?</h3>
  • The force experienced by a streched spring is given as Kx. x is the length of the spring streched from its natural length.
  • Then K = Force / x.
  • Here x = 15 - 10 = 5 cm = 0.05 m
  • K = 40/0.05 = 800N/m.
<h3>What will be the work required to strech that spring from 15 cm to 18 cm?</h3>
  • Work done = 1/2×k×x²
  • Here x= 18-15=3cm or 0.03 m
  • So, W= 1/2×800×0.03² = 0.36 J.

Thus, we can conclude that the work done is 0.36 J.

Learn more about the spring force here:

brainly.com/question/14970750

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6 0
2 years ago
What must her minimum speed be just as she leaves the top of the cliff so that she will miss the ledge at the bottom, which is w
Yanka [14]
<span>1.37 m/s Assuming her initial velocity is totally horizontal and her vertical velocity is only affected by gravity, let's first calculate how much time she has until she reaches the ledge 8.00 m below her. d = 1/2AT^2 8.00m = 1/2 * 9.8 m/s^2 * T^2 Solve for T 8.00 m = 4.9 m/s^2 * T^2 Divide both sides by 4.9 m/s^2 1.632653061 s^2 = T^2 Take square root of both sides 1.277753 s = T So we now know that she has 1.277753 seconds in which to reach a horizontal distance of 1.75 m. So how fast does she need to be going? 1.75 m / 1.277753 s = 1.369592 m/s Since we only have 3 significant figures in our data, round the result to 3 figures giving 1.37 m/s</span>
8 0
3 years ago
What is the kinetic energy of a 150kg object that is moving with a speed of 15 m/s?
otez555 [7]

ke =  \frac{1}{2}  \times m {v}^{2}

so you :

\frac{1}{2}  \times 150 \times  {15}^{2}

= 16875 ( J )

8 0
3 years ago
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