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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.

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The complete stress distribution obtained by superposing the stresses produced by an axial force and a bending moment is correct
Zanzabum

The complete stress distribution obtained by superposing the stresses produced by an axial force and a bending moment is correctly represented by F/A - (My)/(Iz).

<h3>What is the distribution of pressure at some stage in bending?</h3>

Compressive and tensile forces expand withinside the path of the beam axis beneath neath bending loads. These forces set off stresses at the beam. The most compressive pressure is observed on the uppermost fringe of the beam whilst the most tensile pressure is positioned on the decrease fringe of the beam.

The bending pressure is computed for the rail through the equation Sb = Mc/I, wherein Sb is the bending pressure in kilos in keeping with rectangular inch, M is the most bending second in pound-inches, I is the instant of inertia of the rail in (inches)4, and c is the space in inches from the bottom of rail to its impartial axis.

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7 0
2 years ago
What happens to resistance in the strain gage and voltage drop from a connected Wheatstone bridge if you were to pull the strain
snow_tiger [21]
The resistance and voltage drop will still increase but at a smaller rate than the intended axis such as the long axis.
3 0
3 years ago
What is the weight of a steel plate in the shape of a circle with a diameter of 10'? The steel weighs 14 Ib per Ft2
S_A_V [24]
Width * Length * Thickness * Density = Weight.
48″ * 96″ * . 1875″ * 0.284 lb/in3 = 245 lb.
3 0
3 years ago
In the first-order process,a blue dye reacts to form a purple dye. The amount of blue dye at the end of 1 hr is 480 g and the en
Tasya [4]

Answer:

The answer is 960 kg

Explanation:

Solution

Given that:

Assume the initial dye concentration as A₀

We write the expression for the dye concentration for one hour as follows:

ln (C₁) = ln (A₀) -kt

Here

C₁ = is the concentration at 1 hour

t =time

Now

Substitute 480 g for C₁ and 1 hour for t

ln (480) = ln (A₀) -k(1) ------- (1)

6.173786 =  ln (A₀) -k

Now

We write the expression for the dye concentration for three hours as follows:

ln (C₃) = ln (A₀) -k

Here

C₃ = is the concentration at 3 hour

t =time

Thus

Substitute 480 g for C₃ and 3 hour for t

ln (120) = ln (A₀) -k(3) ------- (2)

4.787492 = ln (A₀) -3k

Solve for the equation 1 and 2

k =0.693

Now

Calculate the amount of blue present initially using the expression:

Substitute 0.693 for k in equation (2)

4.787492 = ln (A₀) -3 (0.693)

ln (A₀) =6.866492

A₀ =e^6.866492

= 960 kg

Therefore, the amount of the blue dye present from the beginning is  960 kg

6 0
4 years ago
The rectangular frame is composed of four perimeter two-force members and two cables AC and BD which are incapable of supporting
Rama09 [41]

Answer:

Your question is lacking some information attached is the missing part and the solution

A) AB = AD = BD = 0, BC = LC

    AC = \frac{5L}{3}T, CD = \frac{4L}{3} C

B) AB = AD = BC = BD = 0

   AC = \frac{5L}{3} T, CD = \frac{4L}{3} C

Explanation:

A) Forces in all members due to the load L in position A

assuming that BD goes slack from an inspection of Joint B

AB = 0 and BC = LC from Joint D, AD = 0 and CD = 4L/3 C

B) steps to arrive to the answer is attached below

AB = AD = BC = BD = 0

AC = \frac{5L}{3} T,  CD = \frac{4L}{3}C

7 0
4 years ago
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