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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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If a weld is laying into a joint with a concave weld contour with undercutting issues, what might be the cause?
Katarina [22]

Answer:bbbb

Explanation:

4 0
2 years ago
A CL soil is being used for compacted fill on a project. A sample of the compacted soil with a total volume of 1/30 ft3 weighs 4
Genrish500 [490]

Answer:

A. 0.4

B. 1.003

C. 0.83

Explanation:

The void ratio of a mixture is defined as the ratio of the volume of voids to volume of solids.

Total volume of soil = 1/30 ft3

= 1 ft3/30 * 0.0283 m3/1 ft3

= 9.43 x 10^-4 m3

Mass of water is in the soil = 20% * 4.8

= 0.96 pounds of water

= 0.96 * 0.454

= 0.44 kg

SG = density of substance/density of water

= 2.66 * 1 kg/l

Density of the soil = 2.660 kg/l

Mass of solid = 80 %

= 80% * 4.8 * 0.454

= 1.74 kg

Volume of solids = mass/density

= 1.74/2.66

= 6.63 l

= 6.63 x 10^-4 m3.

The volume of voids is found by adding the volume of water and the volume of air.

Total volume of soil = volume of (solids + voids)

9.43 x 10^-4 = 6.63 x 10^-4 + voids

Volume of voids = 2.8 x 10^-4 m3

A.

Void ratio = volume of void : volume of solids

= 2.8 : 6.63

= 0.4

B.

Y = (1 + w) * Gs * Yw * (1 + e)

Y = moist unit weight

Yw = unit weight of water

w = moisture content of the material

Gs = pecific gravity of the solid

e = void ratio

= (1 + 0.2) * 2.66 * 0.44 * (1 + 0.4)

= 1.003.

C.

gd = Y/(1 + w)

Or

= Gs * Yw * 1/(1 + e)

= 0.83.

6 0
3 years ago
For a bronze alloy, the stress at which plastic deformation begins is 275 MPa, and the modulus of elasticity is 115 GPa. (a) Wha
Anton [14]

Answer:

89375 N

Explanation:

Rearrange the formula for normal stress for F:

\sigma=\frac{F}{A}

F=\sigma*A

Convert given values to base units:

275 MPa = 275*10^{6} Pa

325 mm^{2} = 0.000325 m^{2}

Substituting in given values:

F = (275*10^{6})*(0.000325)=89375 N

3 0
2 years ago
Compute the number of kilograms of hydrogen that pass per hour through a 5-mm-thick sheet of palladium having an area of 0.20 m^
Nat2105 [25]

Answer:

The answer is "\bold{ 259.2 \times 10^{11} }".

Explanation:

The amount of kilograms, which travel in a thick sheet of hydrogen:

M= -DAt \frac{\Delta C}{ \Delta x} \\\\

D =1.0 \times 10^{8} \ \ \ \frac{m^2}{s} \\\\ A = 0.20 \ m^2\\\\t = 1\ \ h = 3600 \ \   sec \\\\

calculating the value of \Delta C:

\Delta C =C_A -C_B

  = 2.4 - 0.6 \\\\    = 1.8 \ \ \frac{kg}{m^3}

calculating the value of \Delta X:

\Delta x = x_{A} -x_{B}

     = 0 - (5\ mm) \\\\ = - 5 \ \ mm\\\\= - 5 \times 10^{-3} \ m

M = -(1.0 \times 10^{8}  \times 0.20 \times 3600 \times  (\frac{1.8}{-5 \times 10^{-3}})) \\\\

    = -(1.0 \times 10^{8}  \times 720 \times  (\frac{1.8}{-5 \times 10^{-3}})) \\\\= -(1.0 \times 10^{8}  \times \frac{ 1296}{-5 \times 10^{-3}})) \\\\= (1.0 \times 10^{8}  \times 259.2 \times 10^3)) \\\\= 259.2 \times 10^{11} \\\\

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