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3 years ago

Which of the following factors does not promote safety in the shop?

1 answer:

6 0

“Thinking about pleasant things to pass the time” would not promote safety in the shop because it would be taking the focus away from important tasks, which in turn decreases safety.

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What’s another name for a service overcurrent device?

alexandr402 [8]

Overcurrent protective devices, or OCPDs

4 0

3 years ago

Evan notices a small fire in his workplace. Since the fire is small and the atmosphere is not smoky he decides to fight the fire

Norma-Jean [14]

Answer:

not calling the firemean

Explanation:

7 0

3 years ago

How are project deliverables determined?

Greeley [361]

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.

7 0

3 years ago

Read 2 more answers

3. It's

jeyben [28]

Answer:

right answer is option no d

6 0

3 years ago

A cylindrical tank is required to contain a gage pressure 560 kPa . The tank is to be made of A516 grade 60 steel with a maximum

adoni [48]

Answer:

5.6 mm

Explanation:

Given that:

A cylindrical tank is required to contain a:

Gage Pressure P = 560 kPa

Allowable normal stress $\sigma$ = 150 MPa = 150000 Kpa.

The inner diameter of the tank = 3 m

In a closed cylinder there exist both the circumferential stress and the longitudinal stress.

Circumferential stress $\sigma = \dfrac{pd}{2t}$

Making thickness t the subject; we have

$t = \dfrac{pd}{2* \sigma}$

$t = \dfrac{560000*3}{2*150000000}$

t = 0.0056 m

t = 5.6 mm

For longitudinal stress.

$\sigma = \dfrac{pd}{4t}$

$t= \dfrac{pd}{4*\sigma }$

$t = \dfrac{560000*3}{4*150000000}$

t = 0.0028 mm

t = 2.8 mm

From the above circumferential stress and longitudinal stress; the stress with the higher value will be considered ; which is circumferential stress and it's minimum value with the maximum thickness = 5.6 mm

8 0

3 years ago