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kobusy [5.1K]
1 year ago
7

A common boundary-crossing problem for engineers is when their home country' values come into sharp contrast with the host count

rys" values O True o fare 10 pts Question 10 Which of the following are common ethical problems that engineers might encounter when working abroad, particularly among cultures that are considered vulnerable populations? Select that oly Uncomenda
Engineering
1 answer:
nikdorinn [45]1 year ago
8 0

A dual relationship that is acceptable, maybe inevitable, or impossible to avoid is often classified as a boundary crossing.

<h3>What is Boundary Crossing?</h3>
  • A dual relationship that is acceptable, maybe inevitable, or impossible to avoid is often classified as a boundary crossing.
  • For instance, if you reside in a small town or rural area, it is likely that you will cross boundaries and interact with clients who are not in a professional setting. It's just a reality of life; it's not necessarily unethical, and managing those boundary problems well is necessary.
  • Another illustration is when your spouse or partner collaborates with your customer. A boundary has been crossed. You might attend a holiday party that your spouse's or partner's company sponsored since you were invited, only to find out afterwards that your client is there because they both work for the same company or have done business together.

To Learn more About  boundary crossing refer to:

brainly.com/question/8344761

#SPJ4

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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

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Answer:

The source code files for this question have been attached to this response.

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Plan to refuel when your fuel gauge reads __________.
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Answer:

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