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Vlada [557]
4 years ago
12

Currently, system administrators create Ken 7 users in each computer where users need access. In the Active Directory, where wil

l system administrators create Ken 7 users? 2. How will the procedures for making changes to the user accounts, such as password changes, be different in the Active Directory? 3. What action should administrators take for the existing workgroup user accounts after converting to the Active Directory? 4. How will the administrators resolve the differences between the user accounts defined on the different computers? In other words, if user accounts have different settings on different computers, how will the Active Directory address that issue? 5. How will the procedure for defining access controls change after converting to the Active Directory?
Engineering
1 answer:
Sunny_sXe [5.5K]4 years ago
5 0

1. First, you would need to open Active Directory Users and Computers. You click on the folder in which you want to add an account, and point to new, and then user. You would fill in the new user's information, such as name and initials.

2. In Active Directory, you input the user logon name, click on the UPN suffix in the drop-down list. It will prompt you to input password and confirm it.

3. Administrators would need to create new user accounts for all users, then join these to the AD domain manually.

4. Administrators will have to manually change the permissions and privileges of the users in order to meet the new established requirements.

5. After converting to the Active Directory, access control will be administered at the object level by setting different levels of access.

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One - tenth kilogram of air as an ideal gas with k= 1.4 executes a carnot refrigeration cycle as shown i fig. 5,16, the isotherm
maria [59]

Answer:

Hello your question is incomplete attached below is the missing part

a) p1 = 454.83 kPa,  p2 = 283.359 Kpa , p3 = 536.423 kpa , p4 = 860.959kPa

b) W12 = 3.4 kJ, W23 = -3.5875 KJ, W34 = -4.0735 KJ, W41 = 3.5875 KJ

c) 5

Explanation:

Given data:

mass of air ( m ) = 1/10 kg

adiabatic index ( k ) = 1.4

temperature for isothermal expansion = 250K

rate of heat transfer ( Q12 ) = 3.4 KJ

temperature for Isothermal compression ( T4 ) = 300k

final volume ( V4 ) = 0.01m ^3

a)  Calculate the pressure, in Kpa, at each of the four principal states

from an ideal gas equation

P4V4 = mRT4 ( input values above )

hence P4 = 860.959kPa

attached below is the detailed solution

b) Calculate work done for each processes

attached below is the detailed solution

C) Calculate the coefficient of performance

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6 0
3 years ago
Air flows steadily through a variable sized duct in a heat transfer experiment with a speed of u = 20 – 2x, where x is the dista
zysi [14]

Answer:

a) a=-28m/s^{2}

b) \frac{dT}{dx}=-5 ^{o}C/m

Explanation:

a)

In order to solve this problem, we need to start by remembering how the acceleration is related to the velocity of a particle. We have the following relation:

a=\frac{dv}{dt}

in other words, the acceleration is defined to be the derivative of the velocity function with respect to time. So let's take our speed function:

u=20-2x

if we take its derivative we get:

du=-2dx

this is the same as writting:

\frac{du}{dt}=-2\frac{dx}{dt}

we also know that velocity is defined to be:

u=\frac{dx}{dt}

so we get that:

a=-2u

when substituting we get that:

a=-2(20-2x)

when expanding we get:

a=-40+4x

and now we can use this equation to find our acceleration at x=3, so:

a=-40+4(3)

a=-40+12

a=-28 m/s^{2}

b)

the same applies to this problem with the difference that this will be the rate of change of the temperature per m. So we proceed and take the derivative of the temperature function:

T=200-5x

\frac{dT}{dx}=-5

so the rate of change is -5^{o}C/m

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3 years ago
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elena55 [62]

Answer:

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Answer:

Oil is graded by the Society of Automotive Engineers - B.

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1 A power transmission includes a belt drive, a chain drive and a gear drive. Which of the following is the best arrangement bet
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Answer:d

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(d) chain drive belt drive gear drive

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