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GREYUIT [131]
3 years ago
14

In this lab, your task is to configure the external vEthernet network adapter with the following IPv6 address: Prefix: 2620:14F0

:45EA:0001 Interface ID: 192:168:0:10 Subnet prefix length: 64 Use ipconfig to verify the information
Engineering
1 answer:
OverLord2011 [107]3 years ago
4 0

Here is the complete question

You work as the IT administrator for a small corporate network. You need to create a separate subnet to use for testing. The test subnet needs access to the rest of the network through a router, but it should not have any local access to production machines.

You have installed Windows Server 2016 on CorpRTR, which you plan to use to isolate the test segment from the rest of the network. You'll use traditional routing or NAT.

In this lab, your task is to add the necessary role and role services to meet the stated requirements. Do not add unnecessary role services.

Answer & Explanation:

Complete this lab as follows:

1.In the notification area, right-click the Network icon and select Open Network and Sharing Center.

2.On the left, select Change adapter settings.

3.Right-click the vEthernet (External)adapter and select Properties.

4.Select Internet Protocol Version 6 (TCP/IPv6).

5.Select Properties.

6.Select Use the following IPv6 address.

7.In the IPv6 address field, enter 2620:14F0:45EA:0001:192:168:0:10as the IPv6 address.

8.In the Subnet prefix length field, enter 64.

9.Click OK.

10.Click Close.

11.In the Search the web and Windows field, enter cmd.

12.Under Best match, right-click Command Prompt and select Run as Administrator.

13.At the command prompt, type ipconfig /all and press Enter to view the IPv6 address.

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A hydrauliic jack is rated at 5000 pound capacity. The area of the large piston on the jack is 4.45 Square inches. Calculate the
sergeinik [125]

Answer:

1123.6 pounds/ square inch.

Explanation:

Fluid pressure is the ratio of force or weight applied by the fluid per unit area.

i.e Fluid pressure = \frac{weight}{area}

The maximum load of the jack is obtained at its maximum capacity = 5000 pounds

Area of the large piston on the jack = 4.45 square inches

Thus,

Fluid pressure = \frac{5000}{4.45}

                        = 1123.5955

Fluid pressure = 1123.6 pounds/ square inch

Thu, the fluid pressure in the jack at maximum load is 1123.6 pounds/ square inch.

7 0
2 years ago
public interface Frac { /** @return the denominator of this fraction */ int getDenom(); /** @return the numerator of this fracti
True [87]

Answer:

The Full details of the answer is attached.

7 0
3 years ago
A vertical piston-cylinder assembly contains water. The piston has a mass of 2 kg and a diameter of 7 cm . Assume atmospheric pr
Step2247 [10]

Answer:

for a) F= 744.97 N

for a) F= 167.85 N

for a) F= 764.57 N

Explanation:

the pressure developed by the piston should be higher than the saturated vapor pressure of water for boiling point at T=120 to ensure boiling.

Then from steam tables

T= 120°C → P required=Pr= 198.67 kPa

then the pressure developed by the piston is

P = (m*g + F)/A

where m= mass of the piston ,g= gravity F= force required and A= area of the piston

then

Pr = P = (m*g + F)/A

F = Pr*A-m*g

since A= π/4*D²

F =π/4* Pr*D²-m*g

replacing values

F =π/4* Pr*D²-m*g = π/4*198.67  *10³Pa*(0.07m)² -2kg* 9.8m/s²

F= 744.97 N

b) for T₂=80°C → Pr₂=47.41 kPa

F₂ =π/4* Pr₂*D²-m*g = π/4*47.41*10³Pa*(0.07m)² -2kg* 9.8m/s²

F₂= 167.85 N

c) for m=0 (mass of the piston neglected) ,the force required is

F₃ =π/4*Pr*D² =  π/4*198.67 *10³Pa*(0.07m)²= 764.57 N

F₃ =764.57 N

4 0
3 years ago
A hair dryer is basically a duct of constant diameter in which a few layers of electric resistors are placed. A small fan pulls
nikitadnepr [17]

Answer:

the percent increase in the velocity of air as it flows through the dryer is 12%

Explanation:

given data

density of air ρ  = 1.18 kg/m³

density of air ρ' = 1.05 kg/m³

solution

we know there is only one inlet and exit

so

ρ × A × v = ρ' × A × v'     ........................1

put here value and we get

\frac{v'}{v} = \frac{\rho }{\rho '}  

\frac{v'}{v} = \frac{1.18}{1.05}  

\frac{v'}{v}  = 1.12

so the percent increase in the velocity of air as it flows through the dryer is 12%

4 0
3 years ago
An aluminum alloy [E = 73 GPa] cylinder (2) is held snugly between a rigid plate and a foundation by two steel bolts (1) [E = 18
V125BC [204]

Answer:

\sigma_A = 58.43 N/mm^2

Explanation:

Given data:

length of Steel bolt L_1 = 335 mm

Length of aluminium cylinder L_2 = 275 mm

Pitch of bolt p = 1mm

Modulus of elasticity of steel E = 215 GPa

Modulus of elasticity of aluminium =  74 GP

Area of bolt = \frac{\pi}{4} 14^2 =  153.93 mm^2

Area of cylinder  = 2300 mm^2

n =1

By equilibrium

\sum F_y = 0

P_A -2P_S = 0

P_A =2P_S

By the compatibility

\delta _s + \delta_A = nP

Displacement in steel is \delta_s = \frac{P_sL_s}{E_sA_s}

Displacement in Aluminium is \delta_A = \frac{P_AL_A}{E_AA_A}

from compatibility equation we have

\frac{P_sL_s}{E_sA_s} +  \frac{P_AL_A}{E_AA_A} = nP

\frac{P_s\times 335}{180\times 10^3\times 153.93} +  \frac{P_A\times 275}{73\times 10^3\times 2300} = 1\times 1

1.20\times 10^[-5} P_s  + 1.44\times 10^{-6}P_A = 1

substituteP_A =2P_S

1.20\times 10^[-5} P_s  + 1.44\times 10^{-6} (2\times P_s) = 1

1.488\times 10^{-5} P_s = 1

P_s = 67204.30 N

P_A = 134,408.60 N

Stress in Aluminium \sigma  = \frac{P_A}{A_A}

                                               = \frac{134,408.60}{2300}

                             \sigma_A = 58.43 N/mm^2

8 0
3 years ago
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