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

Answers

Answer 1
Answer:

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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_________ items are similar to the free issue items, but their access is limited. (CLO5) a)-Bin stock items free issue b)-Bin stock controlled issue c)-Critical or insurance spares d)-Rebuildable spares e)-consumables

Answers

Answer:

a)-Bin stock items free issue

Explanation:

Bin stock items free issue items are similar to the free issue items, but their access is limited.

Bin stock items free issue items are similar to the free issue items, but their access is limited.

Transactional Vs Transformational Leadership. Using the Internet, each member of your team should read at least 2 articles each on Transactional Vs Transformational Leadership. Summarize the articles in 300 words or more. Provide appropriate reference. Combine each summarize in one paper but do not change the wording of the original summary. As a term, write a comprehensive summary of the articles. Present a discussion of what your team learned from this exercise?

Answers

Answer: Provided in the explanation section

Explanation:

Transactional Leadership - This leadership style is mainly focused on the transactions between the leader and employees. If the employees work hard, achieve the objectives and deliver the results, they are rewarded through bonuses, hikes, promotions etc. If the employees fail to achieve the desired results, they are punished by awarding lower ratings in the performance appraisal, denying opportunities etc.

In this style, leader lays emphasis on the relation with the followers.

It is a reactive style where the growth of the employee in the organization completely depends on the performance with respect to the activities and deliverables.

It is best suited for regular operations and for a settled environment by developing the existing organizational culture which is not too challenging.

It is a bureaucratic style of leadership where the leader concentrates on planning and execution rather than innovation and creation.

A transactional leader is short-term focused and result oriented. He/she doesn't consider long-term strategic objectives regarding the organization's future.

 

cheers i hope this helped !!

An aquifer has three different formations. Formation A has a thickness of 8.0 m and hydraulic conductivity of 25.0 m/d. Formation B has a thickness of 2.0 m and a conductivity of 142 m/d. Formation C has a thickness of 34 m and a conductivity of 40 m/d. Assume that each formation is isotropic and homogeneous. Compute both the overall horizontal and vertical conductivities.

Answers

Answer:

The horizontal conductivity is 41.9 m/d.

The vertical conductivity is 37.2 m/d.

Explanation:

Given that,

Thickness of A = 8.0 m

Conductivity = 25.0 m/d

Thickness of B = 2.0 m

Conductivity = 142 m/d

Thickness of C = 34 m

Conductivity = 40 m/d

We need to calculate the horizontal conductivity

Using formula of horizontal conductivity

K_(H)=(H_(A)K_(A)+H_(A)K_(A)+H_(A)K_(A))/(H_(A)+H_(B)+H_(C))

Put the value into the formula

K_(H)=(8.0*25+2,0*142+34*40)/(8.0+2.0+34)

K_(H)=41.9\ m/d

We need to calculate the vertical conductivity

Using formula of vertical conductivity

K_(V)=(H_(A)+H_(B)+H_(C))/((H_(A))/(K_(A))+(H_(B))/(K_(B))+(H_(C))/(K_(C)))

Put the value into the formula

K_(V)=(8.0+2.0+34)/((8.0)/(25)+(2.0)/(142)+(34)/(40))

K_(V)=37.2\ m/d

Hence, The horizontal conductivity is 41.9 m/d.

The vertical conductivity is 37.2 m/d.

A new pipeline is installed to convey 2500 gal/min. If the pipeline can not exceed 6 ft/1,000ft of head loss what is the minimum standard diameter to convey the flowrate? Size both PVC (C=130) and steel (C=120) pipelines.

Answers

Answer:

Minimum standard diameter for the PVC pipe = 14.26 inches

Minimum standard diameter for the steel pipe = 14.70 inches

Explanation:

Head loss = 6/1000...................................................(1)

Head loss = hf/l

Head loss = 10.44Q^(1.85) /C^(1.85) D^(4.8655) ............................(2)

Q = 2500 gal/min

a) Minimum standard diameter for PVC

C for PVC = 130

Equating (1) and (2) and putting C = 130

6/1000 = 10.44* 2500^(1.85) /[130^(1.85) * D^(4.8655) ]\nD^(4.8655)  = 10.44* 2500^(1.85) /[0.006*130^(1.85)]\nD = [10.44* 2500^(1.85) /[0.006*130^(1.85)]]^(1/4.8655) \nD = 14.26 inches

b) Minimum standard diameter for steel

C for steel = 120

Equating (1) and (2) and putting C = 120

6/1000 = 10.44* 2500^(1.85) /[120^(1.85) * D^(4.8655) ]\nD^(4.8655)  = 10.44* 2500^(1.85) /[0.006*120^(1.85)]\nD = [10.44* 2500^(1.85) /[0.006*120^(1.85)]]^(1/4.8655) \nD = 14.70 inches

What are two advantages of forging when compared to machining a part from a billet?

Answers

Answer:

Less material waste and time.

Explanation:

Two advantages of forging vs machining would be that with forging there is much less waste of material. With machining you remove a large amount of material turning into not so valuable chips.

There is also a time factor, as machining can be very time intensive. This depends on the speed of the machining, newer machines tend to be very fast, and forging requires a lengthy heating, but for large parts the machining can be excessively long.

A 4-pole, 3-phase induction motor operates from a supply whose frequency is 60 Hz. calculate: 1- the speed at which the magnetic field of the stator is rotating

Answers

Answer:

The answer is below

Explanation:

A 4-pole, 3-phase induction motor operates from a supply whose frequency is 60 Hz. calculate: 1- the speed at which the magnetic field of the stator is rotating. 2- the speed of the rotor when the slip is 0.05. 3- the frequency of the rotor currents when the slip is 0.04. 4- the frequency of the rotor currents at standstill.

Given that:

number of poles (p) = 4, frequency (f) = 60 Hz

1) The synchronous speed of the motor is the speed at which the magnetic field of the stator is rotating. It is given as:

n_s=(120f)/(p)=(120*60)/(4)=1800\ rpm

2) The slip (s) = 0.05

The speed of the motor (n) is the speed of the rotor, it is given as:

n=n_s-sn_s\n\nn=1800-0.05(1800)\n\nn=1800-90\n\nn=1710\ rpm

3) s = 0.04

The rotor frequency is the product of the supply frequency and slip it is given as:

f_r=sf\n\nf_r=0.04*60\n\nf_r=2.4\ Hz

4) At standstill, the motor speed is zero hence the slip = 1:

s=(n_s-n)/(n_s)\n \nn=0\n\ns=(n_s-0)/(n_s)\n\ns=1

The rotor frequency is the product of the supply frequency and slip it is given as:

f_r=sf\n\nf_r=1*60\n\nf_r=60\ Hz