1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
damaskus [11]
3 years ago
8

As a newly hired professional engineer your an executive with the firm strongly suggests that you attend a monthly dinner party

that is held by an important client and major contractor. These parties are extravagant, paid for by the client and contractor, and attended by the executives from your company, the client, and the contractor. You are already working on one of the projects for this client. You want to please management and feel it would be a good way to learn more about the client and contractor in a casual setting, but have some reservations. Which of the following is allowed by the NCEES Model Rules of Professional Conduct?
A. Accept the invitation, but come up with creative excuses every month to not attend.
B. Attend the parties every 3 months, but not every month.
C. Decline the invitation and explain to your manager that to do otherwise is inappropriate for a registered professional engineer.
D. Attend the monthly parties to demonstrate to management that he understands the importance of pelasing the client
Engineering
1 answer:
Stolb23 [73]3 years ago
4 0

Answer:

C. Decline the invitation and explain to your manager that to do otherwise is inappropriate for a registered professional engineer.

Explanation:NCEES has 3 major rules with some sub sections which helps to uphold the professional conducts of it's members.

The rule that supports this professional conduct is rule II. LICENSEE’S OBLIGATION TO EMPLOYER AND CLIENTS subsection(d)

Subsection d states that a licensee shall not reveal any information about a client,contractor or his employer to a another party except it is required by Law.

You might be interested in
You are working as an electrical technician. One day, out in the field, you need an inductor but cannot find one. Looking in you
telo118 [61]

Answer:

a) the inductance of the coil is 6 mH

b) the emf generated in the coil is 18 mV  

Explanation:

Given the data in the question;

N = 570 turns

diameter of tube d = 8.10 cm = 0.081 m

length of the wire-wrapped portion l =  35.0 cm = 0.35 m

a) the inductance of the coil (in mH)

inductance of solenoid

L = N²μA / l

A = πd²/4  

so

L = N²μ(πd²/4) / l

L = N²μ(πd²) / 4l

we know that μ = 4π × 10⁻⁷ TmA⁻¹

we substitute

L = [(570)² × 4π × 10⁻⁷× ( π × (0.081)² )] / 4(0.35)

L =  0.00841549 / 1.4

L = 6 × 10⁻³ H    

L = 6 × 10⁻³ × 1000 mH

L = 6 mH

Therefore, the inductance of the coil is 6 mH

b)

Emf ( ∈ ) = L di/dt

given that; di/dt = 3.00 A/sec

{∴ di = 3 - 0 = 3 and dt = 1 sec}

Emf ( ∈ ) = L di/dt

we substitute

⇒ 6 × 10⁻³ ( 3/1 )

= 18 × 10⁻³ V

= 18 × 10⁻³ × 1000

= 18 mV  

Therefore, the emf generated in the coil is 18 mV  

7 0
2 years ago
A standard penetration test has been conducted on a coarse sand at a depth of 16 ft below the ground surface. The blow counts ob
scoray [572]

Solution :

Given :

The number of blows is given as :

0 - 6 inch = 4 blows

6 - 12 inch = 6 blows

12 - 18 inch = 6 blows

The vertical effective stress $=1500 \ lb/ft^2$

                                              $= 71.82 \ kN/m^2$

                                             $ \sim 72 \ kN/m^2 $

Now,

$N_1=N_0 \left(\frac{350}{\bar{\sigma}+70} \right)$

$N_1 = $ corrected N - value of overburden

$\bar{\sigma}=$ effective stress at level of test

0 - 6 inch, $N_1=4 \left(\frac{350}{72+70} \right)$

                      = 9.86

6 - 12 inch, $N_1=6 \left(\frac{350}{72+70} \right) $

                        = 14.8

12 - 18 inch, $N_1=6 \left(\frac{350}{72+70} \right) $

                         = 14.8

$N_{avg}=\frac{9.86+14.8+14.8}{3}$

       = 13.14

       = 13

8 0
2 years ago
) A shaft encoder is to be used with a 50 mm radius tracking wheel to monitor linear displacement. If the encoder produces 256 p
andrey2020 [161]

Answer:

number of pulses produced =  162 pulses

Explanation:

give data

radius = 50 mm

encoder produces = 256 pulses per revolution

linear displacement = 200 mm

solution

first we consider here roll shaft encoder on the flat surface without any slipping

we get here now circumference that is

circumference = 2 π r .........1

circumference = 2 × π × 50

circumference = 314.16 mm

so now we get number of pulses produced

number of pulses produced = \frac{linear\ displacement}{circumference} × No of pulses per revolution .................2

number of pulses produced = \frac{200}{314.16} × 256

number of pulses produced =  162 pulses

5 0
3 years ago
Were is carbon monoxide located in a car
Semenov [28]

Answer:

all exhaust gases from all gasoline engines

Explanation:

if u look at the back of ur car when its on u can feel the heat from the exhaust and whT ur feeling is the heat coming from the carbon monoxide gases

6 0
3 years ago
Read 2 more answers
The exhaust steam from a power station turbine is condensed in a condenser operating at 0.0738 bar(abs). The surface of the heat
lozanna [386]

Answer:

Percentage change 5.75 %.

Explanation:Given ;

Given

 Pressure of condenser =0.0738 bar

Surface temperature=20°C

Now from steam table

Properties of steam at 0.0738 bar  

Saturation temperature corresponding to saturation pressure =40°C      

 h_f= 167.5\frac{KJ}{Kg},h_g= 2573.5\frac{KJ}{Kg}

So Δh=2573.5-167.5=2406 KJ/kg

Enthalpy of condensation=2406 KJ/kg

So total heat=Sensible heat of liquid+Enthalpy of condensation

Total\ heat\ =C_p\Delta T+\Delta h

Total heat =4.2(40-20)+2406

Total heat=2,544 KJ/kg

Now film coefficient before inclusion of sensible heat

  h_1=\dfrac{\Delta h}{\Delta T}

  h_1=\dfrac{2406}{20}

h_1=120.3\frac{KJ}{kg-m^2K}

Now film coefficient after inclusion of sensible heat

 h_2=\dfrac{total\ heat}{\Delta T}

 h_2=\dfrac{2,544}{20}

h_2=127.2\frac{KJ}{kg-m^2K}

So\ Percentage\ change=\dfrac{h_2-h_1}{h_1}\times 100

             =\dfrac{127.2-120.3}{120.3}\times 100

                   =5.75 %

So Percentage change 5.75 %.

3 0
3 years ago
Other questions:
  • According to the zeroth law of thermodynamics, which of the following cannot occur?
    7·1 answer
  • The temperature of a flowing gas is to be measured with a thermocouple junction and wire stretched between two legs of a sting,
    8·1 answer
  • What are the seven problem solving steps?
    12·1 answer
  • Three single-phase, 10 kVA, 2400/280 V, 60-Hz transformers are connected to form a three-phase, 2400/480 V transformer The equiv
    15·1 answer
  • What is the key to being a good engineer?
    15·2 answers
  • For an AC machine, what percentage of power is at the negative terminal?
    14·1 answer
  • List the parts of a manual transmission <br><br> List the parts of a typical clutch assembly?
    14·1 answer
  • Consider a single crystal of some hypothetical metal that has the FCC crystal structure and is oriented such that a tensile stre
    7·1 answer
  • What are the philological elements of interior design most like?
    15·1 answer
  • From the list of problems below, check all that are known to be NP-complete. You do not need to justify your answer. (Set cover)
    11·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!