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Stolb23 [73]
2 years ago
5

Jasmine is a mechanical engineer at ACME Engineering. ACME currently buys most of their industrial parts from Old Guard Industri

al, but Jasmine has been unhappy about the quality and price of the parts from Old Guard. Shaun is a sales representative for an up-and-coming parts supplier called Paradigm Parts. Shaun and Jasmine, who had worked together previously, go to the same church, and after the Sunday service, Shaun offers to buy Jasmine lunch to discuss the ways they can work together.
If Jasmine were to claim that she should refuse to have lunch with Shaun because it violates ACME’s code of ethics, this claim would be a(n):_________.
(A) application claim
(B) conceptual claim
(C) factual claim
(D) moral claim
Engineering
1 answer:
madam [21]2 years ago
8 0

Answer:

(D) moral claim

Explanation:

Ethics is an aspect of philosophy that dwells on the study of morality. Morality on the other hand refers to how right or wrong a given situation or thing is.

When Jasmine who is a Mechanical Engineer in ACME Engineering refuses to have lunch with Shaun, a Sales Representative in another company that her company does not do business with, she has made a moral claim because this claim or decision is based on the rightfulness or wrongness of her actions on the basis of her organizations' principles. This is an ethical consideration and therefore a moral claim.

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A three-point bending test was performed on an aluminum oxide specimen having a circular cross section of radius 3.5 mm (0.14 in
RideAnS [48]

To resolve this problem we have,

R=3.5mm\\F_f1=950N\\L_1=50mm\\b=12mm\\L_2=40mm

F_{f2} is unknown.

With these dates we can calculate the Flexural strenght of the specimen,

\sigma{fs}=\frac{F_{f1}L}{\pi R^3}\\\sigma{fs}=\frac{(950)(50*10^{-3})}{\pi 3.5*10^{-3}}\\\sigma{fs}=352.65Mpa

After that, we can calculate the flexural strenght for the square cross section using the previously value.

\sigma{fs}=\frac{F_{f2}L}{\pi R^3}\\(352.65*10^6)=\frac{3Ff(40*10^{-3})}{2(12*10^{-10})}\\F_{f2}=\frac{352.65*10^6}{34722.22}\\F_{f2}=10156.32N\\F_{f2}=10.2kN

6 0
2 years ago
A fatigue test was conducted in which the mean stress was 46.2 MPa and the stress amplitude was 219 MPa.
sleet_krkn [62]

Answer:

a)σ₁ = 265.2 MPa

b)σ₂ = -172.8 MPa

c)Stress\ ratio =-0.65

d)Range = 438 MPa

Explanation:

Given that

Mean stress ,σm= 46.2 MPa

Stress amplitude ,σa= 219 MPa

Lets take

Maximum stress level = σ₁

Minimum stress level =σ₂

The mean stress given as

\sigma_m=\dfrac{\sigma_1+\sigma_2}{2}

2\sigma_m={\sigma_1+\sigma_2}

2 x 46.2 =  σ₁ +  σ₂

 σ₁ +  σ₂ = 92.4 MPa    --------1

The amplitude stress given as

\sigma_a=\dfrac{\sigma_1-\sigma_2}{2}

2\sigma_a={\sigma_1-\sigma_2}

2 x 219 =  σ₁ -  σ₂

 σ₁ -  σ₂ = 438 MPa    --------2

By adding the above equation

2  σ₁ = 530.4

σ₁ = 265.2 MPa

-σ₂ = 438 -265.2 MPa

σ₂ = -172.8 MPa

Stress ratio

Stress\ ratio =\dfrac{\sigma_{min}}{\sigma_{max}}

Stress\ ratio =\dfrac{-172.8}{265.2}

Stress\ ratio =-0.65

Range = 265.2 MPa - ( -172.8 MPa)

Range = 438 MPa

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Answer:

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The detailed calculation is as shown in the attached file.

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B is correct! in this senecio
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