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Alexxx [7]
3 years ago
13

Each of the following activities are commonly performed during the implementation of the Database Life Cycle (DBLC). Fill in the

blank, before each activity, with the phase number of the DBLS that this activity would normally be performed.
DBLC Task Numbers:

1. Database initial study
2. Database design
3. Implementation and loading
4. Testing and evaluation
5. Operation
6. Maintenance and evolution

1.______Load the initial values into the tables
2.______Finish user documentation
3.______Adding tables, attributes, and indexes
4._______Attempt to gain unauthorized access to the data
5.______Interview management
6.______Convert existing data
7._______Study the competition's database
8._______Plan how to grant different levels of access to different user groups
9._______Install the database
10.________Train users
11._______Changing constraints to match changes in business rules
12._______Define budget and scope
13.________Select a DBMS software solution
14._______Draw a logical ERD
15.________Performing software patches to the DBMS
16._______Create the database
17.________Understand how this database will connect to other databases in the organization
18.________Develop a Conceptual Model
19.________Make sure application software updates the database
20._______Regular security audits
21.__________Define objectives
22._______Create a detailed model that can be physically implemented
Engineering
1 answer:
kicyunya [14]3 years ago
4 0
Yessiree I agree with yu cause yu are right
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A viscous fluid flows in a 0.10-m-diameter pipe such that its velocity measured 0.012 m away from the pipe wall is 0.8 m/s. If t
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Answer:

A) centerline velocity = 1.894 m/s

B) flow rate = 7.44 x 10^(-3) m³/s

Explanation:

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U(r) = (Δp•D²/16μL) [1 - (2r/D)²]

Velocity at the centre of the tube can be expressed as;

V_c = (Δp•D²/16μL)

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But we are to find the velocity at the centre of the tube, thus;

We will use the radius across the horizontal distance which will be;

0.05 - 0.012 = 0.038m

Thus, let's put 0.038 for r in the velocity intensity equation and put other relevant values to get the velocity at the centre.

Thus;

U(r) = (V_c)[1 - (2r/D)²]

0.8 = (V_c)[1 - {(2 * 0.038)/0.1}²]

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B) flow rate is given by;

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Answer:

0.1788 ,180°,90°,60°

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  \frac{\pi}{3}= \frac{\pi}{3}×\frac{180}{\pi}

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