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timofeeve [1]
2 years ago
9

How many broadcast or vlan is in this switchs and router ? and why? ​

Engineering
1 answer:
Mnenie [13.5K]2 years ago
7 0

Answer:The move from hubs (shared networks) to switched networks was a big improvement. Control over collisions, increased throughput, and the additional features offered by switches all provide ample incentive to upgrade infrastructure. But Layer 2 switched topologies are not without their difficulties. Extensive flat topologies can create congested broadcast domains and can involve compromises with security, redundancy, and load balancing. These issues can be mitigated through the use of virtual local area networks, or VLANs. This chapter provides the structure and operation of VLANs as standardized in IEEE 802.1Q. This discussion will include trunking methods used for interconnecting devices on VLANs.

Problem: Big Broadcast Domains

With any single shared media LAN segment, transmissions propagate through the entire segment. As traffic activity increases, more collisions occur and transmitting nodes must back off and wait before attempting the transmission again. While the collision is cleared, other nodes must also wait, further increasing congestion on the LAN segment.

The left side of Figure 4-1 depicts a small network in which PC 2 and PC 4 attempt transmissions at the same time. The frames propagate away from the computers, eventually colliding with each other somewhere in between the two nodes as shown on the right. The increased voltage and power then propagate away from the scene of the collision. Note that the collision does not continue past the switches on either end. These are the boundaries of the collision domain. This is one of the primary reasons for switches replacing hubs. Hubs (and access points) simply do not scale well as network traffic increases.

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The mathematical relationship between
11Alexandr11 [23.1K]
4) Ohms law thats the answer
7 0
3 years ago
Define a function pyramid_volume with parameters base_length, base_width, and pyramid_height, that returns the volume of a pyram
Maslowich

Hi, you haven't provided the programing language in which you need the code, I'll just explain how to do it using Python, and you can apply a similar method for any programming language.

Answer:

1. def pyramid_volume(base_length, base_width, pyramid_height):

2.     volume = base_length*base_width*pyramid_height/3

3.     return(volume)

Explanation step by step:

  1. In the first line of code, we define the function pyramid_volume and it's input parameters
  2. In the second line, we perform operations with the input values to get the volume of the pyramid with a rectangular base, the formula is V = l*w*h/3
  3. In the last line of code, we return the volume  

In the image below you can see the result of calling the function with input 4.5, 2.1, 3.0.

5 0
3 years ago
A patient has swollen feet, ankles, and legs, and has pain in his lower back. Tests show the level of urea, a waste product, in
myrzilka [38]

Answer: B  Excretory

Explanation: Hope this helps :)

3 0
3 years ago
Tensile testing provides engineers with the ability to verify and establish material properties related to a specific material.
Sedbober [7]

Answer:

True

Explanation:

Tensile testing which is also referred to as tension testing is a process which materials are subjected to so as to know how well it can be stretched before it reaches breaking point. Hence, the statement in the question is true

7 0
2 years ago
Compute the theoretical density of ZnS given that the Zn-S distance and bond angle are 0.234 nm and 109.5o, respectively. The at
andriy [413]

Answer: the theoretical density is 4.1109 g/cm³

Explanation:  

first the image of one set of ZnS bonding in the crystal structure, we calculate the value of angle θ

θ + ∅ + 90° = 180°

θ = 90° - ∅

θ = 90° - ( 109.5° / 2 )

θ = 35.25°

next we calculate the value of x from the geometry

given that;  distance angle d = 0.234

x = dsinθ

= 0.234 × sin35.25°)

= 0.135 nm = 0.135 × 10⁻⁷ cm

next we calculate the length of the unit cell

a = 4x

a = 4(0.135)

a = 0.54 nm = 0.54 × 10⁻⁷ cm

next we calculate number of formula units

n' = (no of corner atoms in unit ell × contribution of each corner atom in unit cell) + ( no of face center atom in a unit cell × contribution of each face center atom in a unit cell)

n' = 8 × 1/8) + ( 6 × 1/2)

= 1 + 3

= 4

next we calculate the theoretical density using  this equation

P = [n'∑(Ac + AA)] / [Vc.NA]

= [n'∑(Ac + AA)] / [(a)³NA]

where the ∑Ac is sum of atomic weights of all cations in the formula unit( 65.41 g/mol)

∑AA is the sum of weights of all anions in the formula unit( 32.06 g/mol)

Na is the Avogadro’s number( 6.023 × 10²³ units/mole)

so we substitute

P = [4( 65.41 + 32.06)] / [ ( 0.54 × 10⁻⁷ )³ × (6.023 × 10²³)]

= 389.88 / 94.84

= 4.1109 g/cm³

therefore the theoretical density is 4.1109 g/cm³

5 0
2 years ago
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