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34kurt
2 years ago
12

Which of the following types of protective equipment protects workers who are passing by from stray sparks or metal while anothe

r worker is welding?
A. Visors
B. Goggles
C. Protective Clothing
D. Dark Walls
E. Welding Screens
Engineering
1 answer:
lawyer [7]2 years ago
4 0

A protective equipment which protects workers who are passing by from stray sparks or metal while another worker is welding is: E. Welding Screens.

A wielder refers to an individual who is saddled with responsibility of joining two or more metals together by wielding.

During the process of wielding, sparks and minute metallic objects are produced, which are usually hazardous to both the wielder and other workers within the vicinity.

Hence, the following protective equipment are meant to be worn or used directly by a wielder (worker) who is wielding:

  • Visors.
  • Goggles.
  • Protective clothing.
  • Dark walls.

However, a protective equipment which protects other workers who are passing by from stray sparks or metallic objects while wielder (worker) is welding is referred to as welding screens.

Find more information: brainly.com/question/15442363

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3 years ago
What are the main microsoft ware packages widely used today​
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2 years ago
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In a home, air infiltrates from the outside through cracks around doors and windows. Consider a residence where the total length
masya89 [10]

Answer:

Time period  = 41654.08 s

Explanation:

Given data:

Internal volume is 210 m^3

Rate of air infiltration  9.4 \times 10^{-5} kg/s

length of cracks 62 m

air density = 1.186 kg/m^3

Total rate of air infiltration = 9.4\times 10^{-5} \times 62 = 582.8\times 10{-5} kg/s

total volume of air  infiltration= \frac{582.8\times 10{-5}}{1.156} = 5.04\times 10^{-3} m^3/s

Time period = \frac{210}{5.04\times 10^{-3}} = 41654.08 s

3 0
3 years ago
After a strong storm, a worker does not realize that a power transmission line has fallen on his car and is electrocuted while o
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3 0
2 years ago
What is the metal removal rate when a 2 in-diameter hole 3.5 in deep is drilled in 1020 steel at cutting speed of 120 fpm with a
Studentka2010 [4]

Answer:

a) the metal removal rate is 14.4 in³/min

b) the cutting time is 0.98 min

Explanation:

Given the data from the question

first we find the rpm for the spindle of the drilling tool, using the equation

Ns = 12V/πD

V is the cutting speed(120 fpm) and D is the diameter of the hole( 2 in)

so we substitute

Ns = 12 × 120 / π2

Ns = 1440 / 6.2831

Ns = 229.18 rmp

Now we find the metal removal rate using the equation

MRR = (πD²/4) Fr × Ns

Fr is the feed rate( 0.02 ipr ),

so we substitute

MRR = ((π × 2²)/4) × 0.02 × 229.18

MRR = 14.3998 ≈ 14.4 in³/min

Therefore the metal removal rate is 14.4 in³/min

Next we find the allowance for approach of the tip of the drill

A = D/2

A = 2/2

= 1 in

now find the time required to drill the hole

Tm = (L + A) / (Fr × Ns)

Lis the the depth of the hole( 3.5 in)

so we substitute our values

Tm = (3.5 + 1) / (0.02 × 229.18  )

Tm = 4.5 / 4.5836

Tm = 0.98 min

Therefore the cutting time is 0.98 min

8 0
2 years ago
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