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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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Mazyrski [523]

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They are combination structures made up of solid, and shell structures.

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4 0
3 years ago
If the maximum allowable shear stress is 70 MPa, find the shaft diameter needed to transmit 40 kW when the shaft speed is 250 rp
victus00 [196]

Answer:

The diameter is 50mm

Explanation:

The answer is in two stages. At first the torque (or twisting moment) acting on the shaft and needed to transmit the power needs to be calculated. Then the diameter of the shaft can be obtained using another equation that involves the torque obtained above.

T=(P×60)/(2×pi×N)

T is the Torque

P is the the power to be transmitted by the shaft; 40kW or 40×10³W

pi=3.142

N is the speed of the shaft; 250rpm

T=(40×10³×60)/(2×3.142×250)

T=1527.689Nm

Diameter of a shaft can be obtained from the formula

T=(pi × SS ×d³)/16

Where

SS is the allowable shear stress; 70MPa or 70×10⁶Pa

d is the diameter of the shaft

Making d the subject of the formula

d= cubroot[(T×16)/(pi×SS)]

d=cubroot[(1527.689×16)/(3.142×70×10⁶)]

d=0.04808m or 48.1mm approx 50mm

7 0
3 years ago
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