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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

Define prisms. How to draw true shape of a prism?

aksik [14]

Answer:

In geometry, a prism is a polyhedron comprising an n-sided polygonal base, a second base which is a translated copy of the first, and n other faces joining corresponding sides of the two bases. All cross-sections parallel to the bases are translations of the bases.

4 0

3 years ago

The less surface area of gasoline exposed to the air, the faster a given amount will burn. True of False

romanna [79]

Answer:

true

Explanation:

it depends on how fast the car goes because the gas burns faster or slower depending on the speed you go or drive. if your drive 50 MPH, the gas will burn slowly. if your drive 45 MPH, the gas will burn faster if you go slow too much. if you press the glass button and hold it for 5 minutes, the more you hold the gas for too long, the more the gas will burn inside the gas engine. go look at your car or your parents car and see how it goes and that will help.

4 0

3 years ago

Oil with a density of 850 kg/m3 and kinematic viscosity of 0.00062 m2 /s is being discharged by a 5 mm diameter, 40 m long horiz

ra1l [238]

Answer:

Flow rate is $1.82\times 10^{-8} m^{3}/s$

Explanation:

Given information

Density of oil, $\rho_{oil}= 850 Kg/m^{3}$

kinematic viscosity, $v= 0.00062 m^{2} /s$

Diameter of pipe, D= 5 mm= 0.005 m

Length of pipe, L=40 m

Height of liquid, h= 3 m

Volume flow rate for horizontal pipe will be given by

$\bar v=\frac {\triangle P\pi D^{4}}{128\mu L}$ where $\mu$ is dynamic viscosity and $\triangle P$ is pressure drop

At the bottom of the tank, pressure is given by

$P_{bottom}=\rho_{oil} gh=850 Kg/m^{3}\times 9.81 m/s^{2}\times 3 m= 25015.5 N/m^{2}$

Since at the top pressure is zero, therefore change in pressure a difference between the pressure at the bottom and the top. It implies that change in pressure is still $25015.5 N/m^{2}$

Dynamic viscosity, $\mu=\rho_{oil}v= 850 Kg/m^{3}\times 0.00062 m^{2}/s=0.527 Kg/m.s$

Now the volume flow rate will be

$\bar v=\frac {25015.5 N/m^{2}\times \pi \times 0.005^{4}}{128\times 0.527 Kg/m.s \times 40}=1.82037\times 10^{-8} m^{3}/s\approx 1.82\times 10^{-8} m^{3}/s$

Proof of flow being laminar

The velocity of flow is given by

$V_{flow}=\frac {\bar v}{A}=\frac {1.82\times 10^{-8} m^{3}/s}{0.25\times \pi\times 0.005^{2}}=0.000927104 m/s$

Reynolds number, $Re=\frac {\rho_{oil} v_{flow} D}{\mu}=\frac {850 Kg/m^{3}\times 0.000927104 m/s\times 0.005}{0.527 kg/m.s}=0.007476648$

Since the Reynolds number is less than 2300, the flow is laminar and the assumption is correct.

5 0

3 years ago

An automotive fuel cell consumes fuel at a rate of 28m3/h and delivers 80kW of power to the wheels. If the hydrogen fuel has a h

EastWind [94]

Answer:

The efficiency of this fuel cell is 80.69 percent.

Explanation:

From Physics we define the efficiency of the automotive fuel cell ( $\eta$ ), dimensionless, as:

$\eta = \frac{\dot W_{out}}{\dot W_{in}}$ (Eq. 1)

Where:

$\dot W_{in}$ - Maximum power possible from hydrogen flow, measured in kilowatts.

$\dot W_{out}$ - Output power of the automotive fuel cell, measured in kilowatts.

The maximum power possible from hydrogen flow is:

$\dot W_{in} = \dot V\cdot \rho \cdot L_{c}$ (Eq. 2)

Where:

$\dot V$ - Volume flow rate, measured in cubic meters per second.

$\rho$ - Density of hydrogen, measured in kilograms per cubic meter.

$L_{c}$ - Heating value of hydrogen, measured in kilojoules per kilogram.

If we know that $\dot V = \frac{28}{3600}\,\frac{m^{3}}{s}$ , $\rho = 0.0899\,\frac{kg}{m^{3}}$ , $L_{c} = 141790\,\frac{kJ}{kg}$ and $\dot W_{out} = 80\,kW$ , then the efficiency of this fuel cell is: