All categories

2 years ago

What should be your strongest tool be for gulding your ethical decisions making process

Engineering

1 answer:

valkas [14]2 years ago

7 0

Answer:

Recognize that there is a moral dilemma.

Determine the actor. ...

Gather the relevant facts. ...

Test for right versus wrong issues. ...

Test for right versus right paradigms. ...

Apply the resolution principles. ...

Investigate the trilemma options. ...

Make the decision.

You might be interested in

With a brief description, What are the 14 principles of management by fayol.

Rudik [331]

Answer:

is all about focus and unity.

Explanation:

All employees deliver the same activities that can be linked to the same objectives.

7 0

3 years ago

In an experiment, the local heat transfer over a flat plate were correlated in the form of local Nusselt number as expressed by

Stells [14]

Answer:

$\dfrac{\bar{h}}{h}=\dfrac{5}{4}$

Explanation:

Given that

$Nu_x=0.035Re_x^{0.8} Pr^{1/3}$

We know that

Rex=ρvx/μ

$Nu_x=0.035Re_x^{0.8} Pr^{1/3}$

$Nu_x=0.035\times\left(\dfrac{\rho vx}{\mu}\right)^{0.8}Pr^{1/3}$

All other quantities are constant only x is a variable in the above equation .so lets take all other quantities as a constant C

$Nu_x=C.x^{0.8}=C.x^{4/5}$

We also know that

Nux=hx/K

$C.x^{4/5}=\dfrac{hx}{k}$

m is the constant

$h=mx^{-1/5}$

This is local heat transfer coefficient

The average value of h given as

$\bar{h}=\dfrac{\int_{0}^{L}hdx}{L}$

$\bar{h}=\dfrac{5m}{4}\times\dfrac{L^{4/5}}{L}$

$\bar{h}=\dfrac{5m}{4}L^{-1/5}$ ---------1

The value of local heat transfer coefficient at x=L

$h=mx^{-1/5}$

$h=mL^{-1/5}$ -----------2

From 1 and 2 we can say that

$\dfrac{\bar{h}}{h}=\dfrac{5}{4}$

8 0

3 years ago

A crude fermenter is set up in a shed in the backyard of a suburban house. Under anaerobic conditions with ammonia as the nitrog

Aleks04 [339]

Answer:

using calculations Heat losses will be 4512 J

5 0

3 years ago

A cylindrical specimen of a hypothetical metal alloy is stressed in compression. If its original and final diameters are 30.00 a

IrinaVladis [17]

Answer:

The original length of the specimen $l_{o} = 104.7 mm$

Explanation:

Original diameter $d_{o}$ = 30 mm

Final diameter $d_{1}$ = 30.04 mm

Change in diameter Δd = 0.04 mm

Final length $l_{1}$ = 105.20 mm

Elastic modulus E = 65.5 G pa = 65.5 × $10^{3}$ M pa

Shear modulus G = 25.4 G pa = 25.4 × $10^{3}$ M pa

We know that the relation between the shear modulus & elastic modulus is given by

$G = \frac{E}{2(1 + \mu)}$

$25.5 = \frac{65.5}{2 (1 + \mu)}$

$\mu = 0.28$

This is the value of possion's ratio.

We know that the possion's ratio is given by

$\mu = \frac{\frac{0.04}{30} }{\frac{change \ in \ length}{l_{o} } }$

${\frac{change \ in \ length}{l_{o} } = \frac{\frac{0.04}{30} }{0.28}$

${\frac{change \ in \ length}{l_{o} } =$ 0.00476

$\frac{l_{1} - l_{o} }{l_{o} } = 0.00476$

$\frac{l_{1} }{l_{o} } = 1.00476$

Final length $l_{o}$ = 105.2 m

Original length

$l_{o} = \frac{105.2}{1.00476}$

$l_{o} = 104.7 mm$

This is the original length of the specimen.

5 0

3 years ago

For the following circuit diagram, if A=010 , B= 101.

Fantom [35]

Answer:

cgghhhh chick jjkkkkkki

4 0

2 years ago