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Sliva [168]
3 years ago
15

A customer executive from Southeast Asia will visit your HQ facility and meet with your

Engineering
1 answer:
Rufina [12.5K]3 years ago
7 0

<u>Answer</u>:

Say no!

Explanation:

<u>Remember</u>, this scenario is an ethical dilemma. Been ethical implies doing what's professionally right, and as such saying no to that request is the right need to do. In the workplace a company's policies should be obeyed always by individual employees even though some may resent it.

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A(n) ____ is an object setting used to control the visible display of objects.
KatRina [158]
Remote?? maybe I’m not really sure
3 0
2 years ago
Timken rates its bearings for 3000 hours at 500 rev/min. Determine the catalog rating for a ball bearing running for 10000 hours
svet-max [94.6K]

Answer:

C₁₀ = 6.3 KN

Explanation:

The catalog rating of a bearing can be found by using the following formula:

C₁₀ = F [Ln/L₀n₀]^1/3

where,

C₁₀ = Catalog Rating = ?

F = Design Load = 2.75 KN

L = Design Life = 1800 rev/min

n = No. of Hours Desired = 10000 h

L₀ = Rating Life = 500 rev/min

n₀ = No. of Hours Rated = 3000 h

Therefore,

C₁₀ = [2.75 KN][(1800 rev/min)(10000 h)/(500 rev/min)(3000 h)]^1/3

C₁₀ = (2.75 KN)(2.289)

<u>C₁₀ = 6.3 KN</u>

3 0
3 years ago
During his military campaign in what is now Germany, Julius Caesar lead his army of 40,000 soldiers to the western bank of the R
CaHeK987 [17]

Answer:

identifying a problem

Explanation:

its right

5 0
3 years ago
An inventor claims to have devised a cyclical power engine that operates with a fuel whose temperature is 750 °C and radiates wa
Phantasy [73]

Answer:

Yes

Explanation:

Given Data

Temprature of source=750°c=1023k

Temprature of sink =0°c=273k

Work produced=3.3KW

Heat Rejected=4.4KW

Efficiency of heat engine(η)=\frac{Work produced}{Heat supplied}

and

Heat Supplied {\left (Q_s\right)}=Work Produced(W)+Heat rejected\left ( Q_r \right )

{Q_s}=3.3+4.4=7.7KW

η=\frac{3.3}{7.7}

η=42.85%

Also the maximum efficiency of a heat engine operating between two different Tempratures i.e. Source & Sink

η=1-\frac{T_ {sink}}{T_ {source}}

η=1-\frac{273}{1023}

η=73.31%

Therefore our Engine Efficiency is less than the maximum efficiency hence the given claim is valid.

5 0
3 years ago
At steady state, a reversible refrigeration cycle discharges energy at the rate QH to a hot reservoir at temperature TH, while r
ludmilkaskok [199]

Answer:

a) COP_{R} = 25.014, b) T_{H} = 327.78\,K\,(54.63\,^{\textdegree}C)

Explanation:

a) The coefficient of performance of a reversible refrigeration cycle is:

COP_{R} = \frac{T_{L}}{T_{H}-T_{L}}

Temperatures must be written on absolute scales (Kelvin for SI units, Rankine for Imperial units)

COP_{R} = \frac{275.15\,K}{286.15\,K-275.15\,K}

COP_{R} = 25.014

b) The respective coefficient of performance is determined:

COP_{R} = \frac{Q_{L}}{Q_{H}-Q_{L}}

COP_{R} = \frac{8.75\,kW}{10.5\,kW-8.75\,kW}

COP_{R} = 5

But:

COP_{R} = \frac{T_{L}}{T_{H}-T_{L}}

The temperature at hot reservoir is found with some algebraic help:

COP_{R} \cdot (T_{H}-T_{L})=T_{L}

T_{H}-T_{L} = \frac{T_{L}}{COP_{R}}

T_{H} = T_{L}\cdot \left(1+\frac{1}{COP_{R}}  \right)

T_{H} = 273.15\,K \cdot \left(1+\frac{1}{5}  \right)

T_{H} = 327.78\,K\,(54.63\,^{\textdegree}C)

8 0
3 years ago
Read 2 more answers
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