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allsm [11]
2 years ago
11

Given the following access-control list: access-list 90 deny 10.168.7.0 0.0.0.255 access-list 90 permit host 10.168.7.10 access-

list 90 permit host 10.168.7.11 access-list 90 permit host 10.168.7.12 access-list 90 deny any What would be the end result if the network security engineer applies access-list 90 to the inbound interface of FastEthernet 0/3? Group of answer choices A. The access-list will permit host 10.168.7.10 to pass through the router. B. The access-list will permit host 10.168.7.15 to pass through the router. C. Denial of service for this network because the rules are not placed in the correct order. D. Both hosts 10.168.7.10 and 10.168.7.11 will be permitted. E. The access-list will generate an error message.
Engineering
1 answer:
Yuri [45]2 years ago
7 0

Answer:

D. Both hosts 10.168.7.10 and 10.168.7.11 will be permitted

Explanation:

access-list 90

deny 10.168.7.0 0.0.0.255

permit 10.168.7.10

permit 10.168.7.11

permit 10.168.7.12

deny any

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Compare electromagnets and solenoids
natta225 [31]

An electromagnet is a made coil associated with a ferromagnetic core. This way, the strength of the magnet is controlled by the input current. A solenoid is a simple shape used in magnetostatics or magnetics. ... A solenoid is a cylindrical coil of wire whose diameter is small compared to its length.

8 0
3 years ago
A light bar AD is suspended from a cable BE and supports a 20-kg block at C. The ends A and D of the bar are in contact with fri
babymother [125]

Answer:

Tension in cable BE= 196.2 N

Reactions A and D both are  73.575 N

Explanation:

The free body diagram is as attached sketch. At equilibrium, sum of forces along y axis will be 0 hence

T_{BE}-W=0 hence

T_{BE}=W=20*9.81=196.2 N

Therefore, tension in the cable, T_{BE}=196.2 N

Taking moments about point A, with clockwise moments as positive while anticlockwise moments as negative then

196.2\times 0.125- 196.2\times 0.2+ D_x\times 0.2=0

24.525-39.24+0.2D_x=0

D_x=73.575 N

Similarly,

A_x-D_y=0

A_x=73.575 N

Therefore, both reactions at A and D are 73.575 N

7 0
3 years ago
Consider a very long rectangular fin attached to a flat surface such that the temperature at the end of the fin is essentially t
zepelin [54]

Answer:

attached below

Explanation:

4 0
3 years ago
Identify the phase of the design process illustrated in the following scenario, and justify its importance. Kristin has recently
zaharov [31]

Answer:

The design process is at the verify phase of Design for Six Sigma

Explanation:

In designing for Six Sigma, DFSS, is a product or process design methodology of which the goal is the detailed identification of the customer business needs by using measurements tools such as statistical data, and incorporating the identified need into the created product which in this case is the hydraulic robot Kristin Designed

Implementation of DFSS follows a number of stages that are based on the DMAIC (Define - Measure - Analyze - Improve) projects such as the DMADV which stand for define - measure - analyze - verify

Therefore, since Kristin is currently ensuring that the robot is working correctly and meeting the needs of her client the design process is at the verify phase.

5 0
3 years ago
A rigid tank contains 1 kg of oxygen (O2) at p1 = 35 bar, T1 = 180 K. The gas is cooled until the temperature drops to 150 K. De
andreyandreev [35.5K]

Answer:

a. Volume = 13.36 x 10^-3 m³ Pressure = 29.17 bar  b. Volume = 14.06 x 10^-3 m³ Pressure = 22.5 bar

Explanation:

Mass of O₂ = 1kg, Pressure (P1) = 35bar, T1= 180K, T2= 150k Molecular weight of O₂ = 32kg/Kmol

Volume of tank and final pressure using a)Ideal Gas Equation and b) Redlich - Kwong Equation

a. PV=mRT

V = {1 x (8314/32) x 180}/(35 x 10⁵) = 13.36 x 10^-3

Since it is a rigid tank the volume of the tank must remain constant and hnece we can say

T2/T1 = P2/P1, solving for P2

P2 = (150/180) x 35 = 29.17bar

b. P1 = {RT1/(v1-b)} - {a/v1(v1+b)(√T1)}

where R, a and b are constants with the values of, R = 0.08314bar.m³/kmol.K, a = 17.22(m³/kmol)√k, b = 0.02197m³/kmol

solving for v1

35 = {(0.08314 x 180)/(v1 - 0.02197)} - {17.22/(v1)(v1 + 0.02197)(√180)}

35 = {14.96542/(v1-0.02197)} - {1.2835/v1(v1 + 0.02197)}

Using Trial method to find v1

for v1 = 0.5

Right hand side becomes =  {14.96542/(0.5-0.02197)} - {1.2835/0.5(0.5 + 0.02197)} = 31.30 ≠ Left hand side

for v1 = 0.4

Right hand side becomes =  {14.96542/(0.4-0.02197)} - {1.2835/0.4(0.4 + 0.02197)} = 39.58 ≠ Left hand side

for v1 = 0.45

Right hand side becomes =  {14.96542/(0.45-0.02197)} - {1.2835/0.45(0.45 + 0.02197)} = 34.96 ≅ 35

Specific Volume = 35 m³/kmol

V = m x Vspecific/M = (1 x 0.45)/32 = 14.06 x 10^-3 m³

For Pressure P2, we know that v2= v1

P2 = {RT2/(v2-b)} - {a/v2(v2+b)(√T2)} = {(0.08314 x 150)/(0.45 - 0.02197)} - {17.22/(0.45)(0.45 + 0.02197)(√150)} = 22.5 bar

3 0
2 years ago
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