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Novosadov [1.4K]
2 years ago
6

Question 2: (a) In your own words, clearly distinguish and differentiate between Ethics in Engineering and Ethics in Computing (

9 Marks) (b) Draw a table of comparison to illustrate your answer in (a) and provide in each case five practical examples in the fields outlined above.
Engineering
1 answer:
zlopas [31]2 years ago
7 0

Engineering ethics is not without abstraction, but in contrast with computing, it is animated by a robust and active movement concerned with the seamless identification of ethics with practice.

<h3 /><h3>What is engineering?</h3>

This is a branch of science and technology concerned with the design, building, and use of engines, machines, and structures that uses scientific principles.

Comparing ethics in engineering and ethics in computing:

  • Engineering ethics are a set of rules and guidelines. While computing ethics deals with procedures, values and practices.
  • In engineering ethics, engineers must adhere to these rules as a moral obligation to their profession While in computing ethics, the ethics govern the process of consuming computer technology.
  • Following these ethics for the two professions will NOT cause damage, but disobeying them causes damage.

Some practical examples in the computing field:

  • Avoid using the computer to harm other people such as creating a bomb or destroying other people's work.
  • Users also should not use a computer for stealing activities like breaking into a bank or company.
  • Make sure a copy of the software had been paid for by the users before it is used.

Some practical examples in the engineering field:

  • Integrity for oneself.
  • Respect for one another.
  • Pursuit of excellence and accountability.

Hence, Engineering ethics is the field of system of moral principles that apply to the practice of engineering and following them is important to the profession.

Read more about <em>engineering</em> here:

brainly.com/question/17169621

#SPJ1

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A fuel cell vehicle draws 50 kW of power at 70 mph and is 40% efficient at rated power. You are asked to size the fuel cell syst
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Answer:

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  b) fuel tank weight and volume: 321 kg, 643 L

Explanation:

<u>Fuel Cell</u>

Delivery of 50 kW from a source with a power density of 1.5 kW/L requires a source that has a volume of ...

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The weight of the power source is 1 kW/kg, so will be ...

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<u>Fuel Tank</u>

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_____

<em>Comment on fuel tank volume</em>

It appears the fuel tank would need to be equivalent in size to a sphere about 1.1 m in diameter.

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Do heavier cars really use more gasoline? Suppose a car is chosen at random. Let x be the weight of the car (in hundreds of poun
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Answer:

Answer is explained in the explanation section below.

Explanation:

Solution:

Note: This question is incomplete and lacks necessary data to solve. But I have found the similar question on the internet. So, I will be using the data from that question to solve this question for the sack of concept and understanding.

Data Given:

x = 27 , 44 , 32 , 47, 23 , 40, 34, 52

y = 30, 19,  24,  13 , 29,  19,  21,  14

It is given that,

∑x = 299

∑y = 167

∑x^{2} = 11887

∑y^{2} = 3773

We are asked to verify the above values manually in this question.

So,

1. ∑x = 299

Let's verify it:

∑x = 27 + 44 + 32 + 47 + 23 + 40 + 34 + 52

∑x = 299

Yes, it is equal to the given value. Hence, verified.

2. ∑y = 167

Let's verify it:

∑y = 30 + 19 +  24 + 13 + 29 + 19 +  21 +  14

∑y = 169

No, it is not equal to the given value.

3. ∑x^{2} = 11887

Let's verify it:

For this to find,  first we need to square all the value of x individually and then add them together to verify.

∑x^{2} = 27^{2} + 44^{2} + 32^{2} + 47^{2} + 23^{2} + 40^{2} + 34^{2} + 52^{2}

∑x^{2} = 11,887

Yes, it is equal to the given value. Hence, verified.

4. ∑y^{2} = 3773

Let's verify it:

Again, for this we need to find the squares of all the y values and then add them together to verify it.

∑y^{2} = 30^{2} + 19^{2} +  24^{2} + 13^{2} + 29^{2} + 19^{2} +  21^{2} +  14^{2}

∑y^{2}  = 3,845

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Answer:

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PV^n = constant.

That is, P1V1^n = P2V2^n

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Work = (8000 - 6080)/ -0.3

Work = -1920/0.3

Work = -6400 J

Work = -6.4 kJ

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