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OverLord2011 [107]
2 years ago
14

Technician a says that personal protective equipment (ppe) does not include clothing. technician b says that the ppe used should

be based on the task you are performing. who is correct?
Engineering
1 answer:
frez [133]2 years ago
4 0

Based on the information provided, the technician who is correct is: B. Technician B.

<h3>What is PPE?</h3>

PPE is an abbreviation for personal protective equipment and it can be defined as a terminology that is used to denote any piece of equipment which offer protection to different parts of the body while working in a potentially hazardous environment.

Additionally, some examples of personal protective equipment (PPE) that are used to protect the different parts of the body are:

  • Respirators
  • Face mask
  • Face shield
  • Gloves
  • Boots
  • Helmet

Based on the information provided, the technician who is correct is Technician B only because PPE should be used based on the task being performed.

Read more on PPE here: brainly.com/question/19131588

#SPJ1

Complete Question:

Technician a says that personal protective equipment (ppe) does not include clothing. technician b says that the ppe used should be based on the task you are performing. who is correct?

Technician A.

Technician B.

Neither Technician A nor Technician B

Both Technician A and Technician B

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hram777 [196]

Answer:

N = 38546.82 rpm

Explanation:

D_{1} = 150 mm

A_{1}= \frac{\pi }{4}\times 150^{2}

              = 17671.45 mm^{2}

D_{2} = 250 mm

A_{2}= \frac{\pi }{4}\times 250^{2}

              = 49087.78 mm^{2}

The centrifugal force acting on the flywheel is fiven by

F = M ( R_{2} - R_{1} ) x w^{2} ------------(1)

Here F = ( -UTS x A_{1} + UCS x A_{2} )

Since density, \rho = \frac{M}{V}

                        \rho = \frac{M}{A\times t}

                        M = \rho \times A\times tM = 7100 \times \frac{\pi }{4}\left ( D_{2}^{2}-D_{1}^{2} \right )\times t

                        M = 7100 \times \frac{\pi }{4}\left ( 250^{2}-150^{2} \right )\times 37

                        M = 8252963901

∴ R_{2} - R_{1} = 50 mm

∴ F = 763\times \frac{\pi }{4}\times 250^{2}-217\times \frac{\pi }{4}\times 150^{2}

  F = 33618968.38 N --------(2)

Now comparing (1) and (2)

33618968.38 = 8252963901\times 50\times \omega ^{2}

∴ ω = 4036.61

We know

\omega = \frac{2\pi N}{60}

4036.61 = \frac{2\pi N}{60}

∴ N = 38546.82 rpm

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Answer:

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Explanation:

Given:

S = 150 MVA

Vline = 24 kV = 24000 V

X_{s} =1.23(\frac{V_{line}^{2}  }{s} )=1.23\frac{24000^{2} }{1500} =4723.2 ohms

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the line induced voltage is

|E_{line} |=\sqrt{3} *|E|=\sqrt{3} *34.98=60.59kV

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