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Roman55 [17]
3 years ago
14

The amount of time an activity can be delayed and yet not delay the project is termed:_________

Engineering
1 answer:
Shalnov [3]3 years ago
6 0

The correct answer is A. Total slack

Explanation:

Projects include multiple tasks or activities; moreover, each activity and the project itself have a specific due date. Besides this, activities can be delayed and this might affect or not the due date of the project. Indeed, some activities can have extra time without any effect on the date of the project, which is known as total slack or total float (delay in activities that does not delay the project.) Additionally, total slack is possible if activities or tasks are flexible, for example, if each task in a project should take 1 day and one of the tasks takes one day and a half, this can be compensated if another task takes less time. According to this, the correct answer is A.

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What is an example of a product made of textile?
Otrada [13]

beach \: towel \\  \\ hope \: it \: helps

4 0
2 years ago
For some metal alloy, a true stress of 345 MPa (50040 psi) produces a plastic true strain of 0.02. How much will a specimen of t
Strike441 [17]

Answer:

the elongation of the metal alloy is 21.998 mm

Explanation:

Given the data in the question;

K = σT/ (εT)ⁿ

given that metal alloy true stress σT = 345 Mpa, plastic true strain εT = 0.02,

strain-hardening exponent n = 0.22

we substitute

K = 345 / 0.02^{0.22

K = 815.8165 Mpa

next, we determine the true strain

(εT) = (σT/ K)^1/n

given that σT = 412 MPa

we substitute

(εT) = (412 / 815.8165 )^(1/0.22)

(εT) = 0.04481 mm

Now, we calculate the instantaneous length

l_i = l_0e^{ET

given that l_0 = 480 mm

we substitute

l_i =480mm × e^{0.04481

l_i =  501.998 mm

Now we find the elongation;

Elongation = l_i - l_0

we substitute

Elongation = 501.998 mm - 480 mm

Elongation = 21.998 mm

Therefore, the elongation of the metal alloy is 21.998 mm

6 0
3 years ago
Which of the following approaches to lean engineering consists of teams of people from different departments sharing
AURORKA [14]

Answer:

PDSA, where you work as a group

Explanation:

8 0
3 years ago
A square steel bar has a length of 7.2 ft and a 2.5 in by 2.5 in cross section and is subjected to axial tension. The final leng
Nataly_w [17]

Answer:

A) ν = 0.292

B) ν = 0.381

Explanation:

Poisson's ratio = - (Strain in the direction of the load)/(strain in the direction at right angle to the load)

In axial tension, the direction of the load is in the length's direction and the direction at right angle to the load is the side length

Strain = change in length/original length = (Δy)/y or (Δx)/x or (ΔL/L)

A) Strain in the direction of the load = (2.49946 - 2.5)/2.5 = - 0.000216

Strain in the direction at right angle to the load = (7.20532 - 7.2)/7.2 = 0.0007389

Poisson's ratio = - (-0.000216)/(0.0007389) = 0.292

B) Strain in the direction of the load = (2.09929 - 2.1)/2.1 = - 0.0003381

Strain in the direction at right angle to the load = (5.30470 - 5.3)/5.3 = 0.0008868

Poisson's ratio = - (-0.0003381)/(0.0008868) = 0.381

7 0
3 years ago
This is just so I can flip the picture
Pepsi [2]

Answer:

ok

Explanation:

thx for points

7 0
4 years ago
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