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Bad White [126]
3 years ago
15

With reference to the NSPE Code of Ethics, which one of the following statements is true regarding the ethical obligations of th

e engineers involved in the VW Emissions Cheating Scandal.
a. The VW engineers involved conducted themselves honorably, responsibly, ethically, and lawfully so as to enhance the honor, reputation, and usefulness of the profession.
b. As faithful agents and trustees of Volkswagen the engineers involved could not ethically report the emissions cheating violations to public authorities.
c. The VW engineers involved were ethically obligated to hold paramount the health, welfare and safety of the public even if their supervisors directed them to implement software and hardware that enabled cheating on the emissions testing software.
d. The VW engineers involved were not ethically obligated to report to their supervisors and to upper management the emissions cheating violations being implemented in the control system hardware and software.
Engineering
1 answer:
34kurt3 years ago
4 0

Answer: c. The VW engineers involved were ethically obligated to hold paramount the health, welfare and safety of the public even if their supervisors directed them to implement software and hardware that enabled cheating on the emissions testing software.

Explanation: The National Society of professional Engineers, NSPE define the code of ethics which must guide engineers in their duty. These codes act as principles of personal conduct, towards the public and their employers.  

One of the areas covered by these codes is overriding importance of the safety and health of the public to any other factor. In addition, engineers are to avoid deception and maintain the reputation of their profession. These cannot be sacrificed for the financial gain of their employers or explained away by saying they are following the direction of their employers. While they have certain responsibilities to their employers, the health welfare and safety of the public is more important.  

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For methyl chloride at 100°C the second and third virial coefficients are: B = −242.5 cm 3 ·mol −1 C = 25,200 cm 6 ·mol −2 Calcu
bogdanovich [222]

Answer:

a)W=12.62 kJ/mol

b)W=12.59 kJ/mol

Explanation:

At T = 100 °C the second and third virial coefficients are

B = -242.5 cm^3 mol^-1

C = 25200 cm^6  mo1^-2

Now according isothermal work of one mole methyl gas is

W=-\int\limits^a_b {P} \, dV

a=v_2\\

b=v_1

from virial equation  

\frac{PV}{RT}=z=1+\frac{B}{V}+\frac{C}{V^2}\\   \\P=RT(1+\frac{B}{V} +\frac{C}{V^2})\frac{1}{V}\\

And  

W=-\int\limits^a_b {RT(1+\frac{B}{V} +\frac{C}{V^2}\frac{1}{V}  } \, dV

a=v_2\\

b=v_1

Now calculate V1 and V2 at given condition

\frac{P1V1}{RT} = 1+\frac{B}{v_1} +\frac{C}{v_1^2}

Substitute given values P_1\\ = 1 x 10^5 , T = 373.15 and given values of coefficients we get  

10^5(v_1)/8.314*373.15=1-242.5/v_1+25200/v_1^2

Solve for V1 by iterative or alternative cubic equation solver we get

v_1=30780 cm^3/mol

Similarly solve for state 2 at P2 = 50 bar we get  

v_1=241.33 cm^3/mol

Now  

W=-\int\limits^a_b {RT(1+\frac{B}{V} +\frac{C}{V^2}\frac{1}{V}  } \, dV

a=241.33

b=30780

After performing integration we get work done on the system is  

W=12.62 kJ/mol

(b) for Z = 1 + B' P +C' P^2 = PV/RT by performing differential we get  

         dV=RT(-1/p^2+0+C')dP

Hence work done on the system is  

W=-\int\limits^a_b {P(RT(-1/p^2+0+C')} \, dP

a=v_2\\

b=v_1

by substituting given limit and P = 1 bar , P2 = 50 bar and T = 373 K we get work  

W=12.59 kJ/mol

The work by differ between a and b because the conversion of constant of virial coefficients are valid only for infinite series  

8 0
3 years ago
Consider that a system has two entities, Students, Instructors and Course. The Student has the following properties: student nam
tekilochka [14]

Answer:

There's no answer ?

Explanation:

5 0
3 years ago
A masonry facade consisting of 3,800 square feet is to be constructed for a building. The total cost per worker hour is estimate
lana66690 [7]

Answer:

Days: 6.9444 days

Production rate: 547.2035 ft²/s

Explanation:

the solution is attached in the Word file

Download docx
6 0
3 years ago
If you are interested only in the temperature range of 20° to 40°C and the ADC has a 0 to 3V input range, design a signal condit
mario62 [17]

Explanation:

Temperature range → 0 to 80'c

respective voltage output → 0.2v to 0.5v

required temperature range 20'c to 40'c

Where T = 20'c respective voltage

\begin{aligned}v_{20} &=0.2+\frac{0.5-0.8}{80} \times 20 \\&=0.2+\frac{0.3}{80} \times 20 \\V_{20} &=0.275 v\end{aligned}

\begin{aligned}\text { when } T=40^{\circ} C & \text { . } \\v_{40} &=0.2+\frac{0.5-0.2}{80} \times 40 \\&=0.35 V\end{aligned}

Therefore, Sensor output changes from 0.275v to 0.35volts for the ADC the required i/p should cover the dynamic range of ADC (ie - 0v to 3v)

so we have to design a circuit which transfers input voltage 0.275volts - 0.35v to 0 - 3v

Therefore, the formula for the circuit will be

\begin{array}{l}v_{0}=\left(v_{i n}-0.275\right) G \\\sigma=\ldots \frac{3-0}{0.35-0.275}=3 / 0.075=40 \\v_{0}=\left(v_{i n}-0.275\right) 40\end{array}

The simplest circuit will be a op-amp

NOTE: Refer the figure attached

Vs is sensor output

Vr is the reference volt, Vr = 0.275v

\begin{aligned}v_{0}=& v_{s}-v_{v}\left(1+\frac{R_{2}}{R_{1}}\right) \\\Rightarrow & \frac{1+\frac{R_{2}}{R_{1}}}{2}=40 \\& \frac{R_{2}}{R_{1}}=39 \quad \Rightarrow\end{aligned}

choose R2, R1 such that it will maintain required  ratio

The output Vo can be connected to voltage buffer if you required better isolation.

3 0
3 years ago
Determine the general lighting load for a two-story office building that measures 125 feet by 150 feet.
Irina-Kira [14]

The general lighting load for a two-story office building that measures 125 feet by 150 feet is 112, 500 sq ft.

<h3>What is lighting load?</h3>

Lighting loads are the energy used to power electric lights and they make up nearly a third of US commercial building energy use.

Lighting load = n(LW)

where;

  • L is length of the building
  • W is width of the building
  • n is number of story building

For one story building, = 3

For two story building, n = 6

Lighting load = 6 x 125 x 150 = 112, 500 sq ft.

Learn more about lighting load here: brainly.com/question/14070748

#SPJ12

7 0
2 years ago
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