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Ivenika [448]
4 years ago
11

What should a technician do before entering a confined space? Question 1 options: A) Post another worker outside the confined sp

ace. B) Take a radio or cell phone for emergency calls. C) Wear clothes that are resistant to spontaneous arcing. D) Leave a note at the entry point.
Engineering
1 answer:
Lady bird [3.3K]4 years ago
3 0

Answer:

C) Wear clothes that are resistant to spontaneous arcing.

Explanation:

A confined area or space is one that has strict restriction against movement of people due to the availability of hazardous substance or material within the confined space. If there is need for anyone to enter the space, it is very important that the person make use of personal protective equipment (PPE).

Personal protective equipment are materials that are capable of inhibiting injury or hazard when used appropriately. Examples are: helmet, respirators, gloves, goggles, ear muffs etc.

Before a technician should enter a confined space, he/ she should use a PPE by wearing clothes that are resistant to spontaneous arcing.

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5. A non-cold-worked brass specimen of average grain size 0.01 mm has a yield strength of 150 MPa. Estimate the yield strength o
Papessa [141]

Answer:

97.17 MPa

Explanation:

Given:-

- The nominal strength of the grain, σ0  = 25 MPa

- The average grain size of the brass specimen, d* = 0.01 m

- The yield strength of the non-cold worked specimen, σy = 150 MPa

- Conditions of cold-working: T = 500°C , t = 1000 s

Find:-

Estimate the yield strength of this alloy after cold - working process

Solution:-

- The nominal strength of the grain is a function of yield strength of the material, grain yield factor ( Ky ) and the grain size.

- the following relation is used to determine the grain strength:

                             σ0  = σy  - ( Ky / √( d ) )

- We will use the above relation to determine the grain yield factor ( Ky ) for the alloy as follows. Note: here we will use the average value of grain size:

                            Ky = ( σy  - σy )*√( d* )

                            Ky = ( 150 - 25 ) * √0.01

                            Ky = 12.5 MPa - √mm

- Now we will use the cold working conditions of T = 500 C and time of the process is t = 1000 s. We will look up the elongated size of the grain after the cold-working process in lieu with its yield factor ( Ky ). Use figure 7.25.

- The cold-worked grain size with the given conditions can be read off from the figure 7.25. The new size comes out to be d = 0.03 mm.

- We will again use the nominal grain strength relation expressed initially. And compute for the new yield strength of the cold-worked alloy.

                            σ0  = σy  - ( Ky / √( d ) )

                            σy = σ0 + ( Ky / √( d ) )

                            σy = 25MPa + ( 12.5 / √( 0.03 mm ) )

                            σy = 97.17 MPa

- We see that the yield strength of the alloy decreases after cold-working process. This happens because the cold working process leaves with inter-granular strain ( dislocation of planes ) in the material structure which results from the increase in grain size.

6 0
3 years ago
The driveshaft of an automobile is being designed to transmit 238 hp at 3710 rpm. Determine the minimum diameter d required for
AURORKA [14]

Explanation:

Below is an attachment containing the solution.

4 0
4 years ago
The “Sun-Star” Company has purchased new office furniture for their offices at a retail price of $100,000. An additional $12,000 h
Zinaida [17]

Based on the cost of the furniture, the following are true:

  • a. $10,400.
  • b. $93,600
  • c. $20,800

<h3>Depreciation in second year</h3>

Depreciation per year = (Cost of furniture - Salvage value) / Useful life

Cost will include both the purchase price and the charge for insurance and shipping.

= (100,000 + 12,000 - 8,000) / 10

= $10,400

<h3>BV at end of first year</h3>

= Cost - Depreciation

= 104,000 - 10,400

= $93,600

<h3>BV after 8 years </h3>

= Cost - (Depreciation x 8 years )

= 104,000 - (10,400 x 8)

= $20,800

In conclusion, depreciation is $10,400 per year.

Find out more about SL depreciation at brainly.com/question/13734742.

4 0
3 years ago
To be safe, the engineers making the ride want to be sure the normal force does not exceed 1.8 times each persons weight - and t
Yuri [45]

Answer:

μ = 0.55

Explanation:

Given that

Normal weight = 1.8 x weight of person

N= 1.8 mg

We know that friction force Fr

Fr= μ N

μ=Coefficient of friction

N=Normal force

To find  μ We have to equate friction and gravity force

Fr= Wt

μ N = m g

μ  x 1.8 m g = m g

μ = 0.55

So the coefficient of friction will be 0.55.

5 0
3 years ago
If welding is being done in the vertical position, the torch should have a travel angle of?
siniylev [52]

Answer:

Between 35°– 45°

Explanation:

In the vertical position, Point the flame in the direction of travel. Keep the flame tip at the correct height above the base metal. An angle of 35°–45° should be maintained between the torch tip and the base metal. This angle may be varied up or down to heat or cool the weld pool if it is too narrow or too wide

4 0
2 years ago
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