ਭਾਰਤ ਵਿੱਚ ਸਭ ਤੋਂ ਵੱਧ ਵਰਖਾ ਕਿੱਥੇ ਹੁੰਦੀ ਹੈ।

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

Nikitich [7]

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) 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

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.

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.

4 0

3 years ago

A hockey puck with a mass of 0.18 kg is at rest on the horizontal frictionless surface of the rink. A player applies a horizonta

Leya [2.2K]

Hello!

$\large\boxed{0.018 N}$

Begin by finding the acceleration of the puck. Use the kinematic equation:

$d = v_{i} + \frac{1}{2}at^{2}$

The initial velocity is 0 m/s since the object was at rest, so we can rewrite the equation as:

$d = \frac{1}{2}at^{2}$

Plug in the given distance and time:

$1.25 = \frac{1}{2}a(5^{2})\\\\2.50 = 25a\\\\a = 0.1 m/s^{2}$

Find the net force using the formula F = m · a (Newton's Second Law)

F = 0.18 · 0.1

F = 0.018 N

6 0

3 years ago

Plan an experiment to measure the ideal mechanical advantage of a three-hole punch. (a) What materials would you need? (b) What

Vaselesa [24]

Answer:

A) Three hole punch and either a layered plastic or paper

B) Identify the lengths involved ,

Length of input arm / length of output arm = L1/ L2

Explanation:

a) Materials involved includes :

Three hole punch and either a layered plastic or paper

Identify the forces acting on the three-hole punch which are Input and output forces

Identify the points where they act

B) procedures involved

The mechanical advantage = output force / input force

step one: Identify the lengths involved

assuming no friction or relatively small friction \

mechanical advantage can be calculated as : Length of input arm / length of output arm = L1/ L2

7 0

3 years ago

Compare the surface temperature of a red star with the temperature of a yellow star

inna [77]

The surface temperature of a red star is lower than the surface temperature of a yellow star.

Explanation:

For a red star, the peak of the wavelength of the emitted light is in the red spectrum, therefore the average peak wavelength is:

$\lambda = 680 nm = 680 \cdot 10^{-9} nm$

The corresponding temperature at the surface of the star can be found by using Wien's displacement law:

$T=\frac{b}{\lambda}$

where

$b=2.898\cdot 10^{-3} m\cdot K$ is a constant. Substituting, we find

$T_r=\frac{2.898\cdot 10^{-3}}{680\cdot 10^{-9}}\sim 4300 K$

For a yellow star, the peak of the wavelength of the emitted light is in the yellow spectrum, therefore the average peak wavelength is:

$\lambda = 580 nm = 580 \cdot 10^{-9} nm$

So we can use again the same equation to find the corresponding surface temperature of a yellow star:

$T_y=\frac{2.898\cdot 10^{-3}}{580\cdot 10^{-9}}\sim 5000 K$

Therefore, we see that the surface temperature of a red star is lower than the surface temperature of a yellow star.

Learn more about temperature:

brainly.com/question/1603430

brainly.com/question/4370740

#LearnwithBrainly

6 0

4 years ago

A charge q = 2.00 μC is placed at the origin in a region where there is already a uniform electric field = (100 N/C) î. Calculat

SVETLANKA909090 [29]

Answer:

$\phi = 2.26 \times 10^5 Nm^2/C$

Explanation:

Due to uniform electric field net flux through a closed surface is always zero

so here we have to find the net flux through a sphere due to a point charge placed at origin and due to a uniform electric field

so here we will have

$\phi = \frac{q}{\epsilon_o}$

now we have

$\phi = \frac{2 \times 10^{-6}}{8.85 \times 10^{-12}}$

$\phi = 2.26 \times 10^5 Nm^2/C$

Due to external uniform field the electric flux is zero

so the total flux is only due to charge and it is equal to

$\phi = 2.26 \times 10^5 Nm^2/C$

3 0

3 years ago

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