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3 years ago

How are project deliverables determined?

Engineering

2 answers:

Greeley [361]3 years ago

7 0

Answer:

The essence including its problem is listed throughout the clarification section following.

Explanation:

Projects build deliverable that seem to be the products of the venture or indeed the implementation of the project. This ensures that perhaps the agile methodology may be as broad as either the goal of the study itself as well as the coverage that would be part of a much larger venture.

For every other production to have been marked as "deliverable" within the same project, this should satisfy a few eligibility requirements:

It should be within the development of the work.
The interested parties-external or internal-must consent to the above. This is perhaps the product of hard effort.

So that the above seems to be the right answer.

Alex17521 [72]3 years ago

7 0

It is determined by:

For any output to be classified as a “deliverable” within a project, it has to meet a few criteria: It must be within the scope of the project. Stakeholders external or internal must agree to it. It must be the result of deliberate work.

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Read the passage.

Archy [21]

The claim being made in in the above passage is that " It makes financial sense to stop using the penny." (Option B)

<h3>What textual evidence backs up the above claim?</h3>

The textual evidence that supports the above claim is "Not only does it make financial sense to take the penny out of circulation, but it also makes environmental sense." [Para. 2]

Textual evidence is evidence related to a text which supports claims made in such a text.

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6 0

2 years ago

A steel rod, which is free to move, has a length of 200 mm and a diameter of 20 mm at a temperature of 15oC. If the rod is heate

kherson [118]

Explanation:

thermal expansion ∝L = (δL/δT)÷L ----(1)

δL = L∝L + δT ----(2)

we have δL = 12.5x10⁻⁶

length l = 200mm

δT = 115°c - 15°c = 100°c

putting these values into equation 1, we have

δL = 200*12.5X10⁻⁶x100

= 0.25 MM

L₂ = L + δ L

= 200 + 0.25

L₂ = 200.25mm

12.5X10⁻⁶ *115-15 * 20

= 0.025

20 +0.025

D₂ = 20.025

as this rod undergoes free expansion at 115°c, the stress on this rod would be = 0

3 0

3 years ago

You have a motor such that if you give it 12 Volt, it will eventually reach a steady state speed of 200 rad/s. If it starts from

Aleksandr [31]

Answer:

a) $\frac{Ws}{Es} = \frac{200}{1+1.2s}$

b) attached below

c) type zero system

d) k > $\frac{g}{200}$

e) The gain K increases above % error as the steady state speed increases

Explanation:

Given data:

Motor voltage = 12 v

steady state speed = 200 rad/s

time taken to reach 63.2% = 1.2 seconds

<u>a) The transfer function of the motor from voltage to speed</u>

let ; $\frac{K1}{1+St}$ be the transfer function of a motor

when i/p = 12v then steady state speed ( k1 ) = 200 rad/s , St ( time constant ) = 1.2 sec

hence the transfer function of the motor from voltage to speed

= $\frac{Ws}{Es} = \frac{200}{1+1.2s}$

<u>b) draw the block diagram of the system with plant controller and the feedback path </u>

attached below is the remaining part of the detailed solution

c) The system is a type-zero system because the pole at the origin is zero

d) ) k > $\frac{g}{200}$

7 0

3 years ago

Basic concepts surrounding electrical circuitry?

Elanso [62]

Hopefully that helps you out and is this for history or science?

3 0

3 years ago

There are two identical oil tanks. The level of oil in Tank A is 12 ft and is drained at the rate of 0.5 ft/min. Tank B contains

Luba_88 [7]

Answer:

16 minutes

Explanation:

This is an example of a class of problems in which two quantities start with different initial values and change at different rates. In such problems, the rates of change are generally ones that cause the values to converge.

The question usually asks when the values will be the same. The generic answer is, "when the difference in rates makes up the difference in initial values."

Here the tanks differ in initial fill height by 12 -8 = 4 ft. The rates of change differ by 0.5 -0.25 = 0.25 ft/min. The more filled tank is draining faster (important), so the fill heights will converge after ...

(4 ft)/(0.25 ft/min) = 16 min

The level in the two tanks will be the same after 16 minutes.

<em>Additional comment</em>

The oil levels at that time will be 4 ft.

You can write two equations for height:

y = 12 -0.5x . . . . . . . height in feet after x minutes (tank A)

y = 8 -0.25x . . . . . . height in feet after x minutes (tank B)

These will be equal when ...

y = y

12 -0.5x = 8 -0.25x

4 = 0.25x . . . . . . . . . . add 0.5x -8

16 = x . . . . . . . . . . . . multiply by 4 . . . . time to equal height

The graph shows when the tanks will have equal heights and when they will be drained.

4 0

2 years ago