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Bogdan [553]
4 years ago
6

IAs a relatively new hire within a large multinational firm, you are extremely pleased with the many challenging assignments tha

t have come your way. Now another new hire with whom you have become friends is seeking your input on an important decision that she must make within the next week. She has been told that the firm has decided to cut costs in her department by outsourcing a large portion of the department's work to an offshore resource firm that has an excellent reputation. Your friend would remain with the firm to oversee the transition of work to the outsourcing firm including training the new workers, establishing performance benchmarks and measures, tracking results, and providing ongoing feedback to management both at your firm and the outsourcing firm. She is very concerned that there will be great tension as some 25 workers are replaced and feels that some of their anger will be direct at her. She is also concerned that once the start-up issues are resolved and things are running smoothly, she will be let go. What advice would you offer your friend? Ethics in Computer
Engineering
1 answer:
Usimov [2.4K]4 years ago
4 0

Answer:

I would strongly not to lay all her eggs in  one basket and considering the state it is now. seeking for other opportunities while still holding on to the current job is necessary. i  advise that  research on company that has   good standing, integrity and uprightness in terms of service delivery, years of existence and achievement and most importantly how they treat their worker and tender an application if need be.  usually, the way a company handles and treat their workers is quite important for service delivery and optimum production so it is necessary that it should be one of the top priorities of any firm, company or any establishment, because  the worker do put their best, knowledge and their all at times to the company's growth and development so it is right that they deserve it and without them, companies are just dead.

Explanation:

You might be interested in
A large well-mixed tank of unknown volume, open to the atmosphere initially, contains pure water. The initial height of the solu
trasher [3.6K]

Answer:

The exact time when the sample was taken is = 0.4167337 hr

Explanation:

The diagram of a sketch of the tank is shown on the first uploaded image

Let A denote the  first inlet

Let B denote the second inlet

Let C denote the single outflow from the tank

From the question we are given that the diameter of A is = 1 cm = 0.01 m

                              Area of  A is  = \frac{\pi}{4}(0.01)^{2} m^{2}

                                                    = 7.85 *10^{-5}m^{2}

Velocity of liquid through A = 0.2 m/s

  The rate at which the liquid would flow through the first inlet in terms of volume  = \frac{Volume of Inlet }{time} = Velocity * Area i.e is m^{2} * \frac{m}{s}   = \frac{m^{3}}{s}

             = 0.2 *7.85*10^{-5} \frac{m^{3}}{s}

  The rate at which the liquid would flow through the first inlet in terms of mass of the liquid = mass of liquid × the rate of flow in terms of volume

                              =  1039.8 * 0.2 * 7.85 *10^{-5} Kg/s

                              = 0.016324 \frac{Kg}{s}

From the question the diameter of B = 2 cm = 0.02 m

                                           Area of B = \frac{\pi}{4} * (0.02)^{2} m^{2} = 3.14 * 10^{-4}m^{2}

                                     Velocity of liquid through B = 0.01 m/s

The rate at which the liquid would flow through the first inlet in terms of volume  = \frac{Volume of Inlet }{time} = Velocity * Area i.e is m^{2} * \frac{m}{s}   = \frac{m^{3}}{s}

             = 3.14*10^{-4} *0.01 \frac{m^{3}}{s}

The rate at which the liquid would flow through the second inlet in terms of mass of the liquid = mass of liquid × the rate of flow in terms of volume

                              = 1053 * 3.14*10^{-6} \frac{Kg}{s}

                              = 0.00330642 \frac{Kg}{s}

From the question The flow rate in term of volume of the outflow at the time of measurement is given as  = 0.5 L/s

And also from the question the mass of  potassium chloride  at the time of measurement is given as 13 g/L

So The rate at which the liquid would flow through the outflow in terms of mass of the liquid = mass of liquid × the rate of flow in terms of volume

                              = 13\frac{g}{L} * 0.5 \frac{L}{s}

                              =  \frac{6.5}{1000}\frac{Kg}{s}       Note (1 Kg = 1000 g)

                              = 0.0065 kg/s

Considering potassium chloride

         Let denote the  rate at which liquid flows in terms of mass as   as \frac{dm}{dt} i.e change in mass with respect to time hence

           Input(in terms of mass flow ) - output(in terms of mass flow ) = Accumulation in the Tank(in terms of mass flow )

         

      (0.016324 + 0.00330642) - 0.0065 = \frac{dm}{dt}

          \int\limits {\frac{dm}{dt} } \, dx  =\int\limits {0.01313122} \, dx

      => 0.01313122 t = (m - m_{o})

  From the question  (m - m_{o})  is given as = 19.7 Kg

Hence the time when the sample was taken is given as

               0.01313122 t = 19.7 Kg

      =>  t = 1500.2414 sec

            t = .4167337 hours (1 hour = 3600 seconds)

5 0
4 years ago
Poached eggs are cooked in bath of boiling water at 100°C. Over time they reach thermal equilibrium with the bath. They are then
barxatty [35]

Answer:

Check the explanation

Explanation:

Kindly check the attached images below to see the step by step explanation to the question above and also note that I derived the whole expressions in variable firm that are required to solve this question and giving reasons for each step so that you can employ these steps in solving new problems.

8 0
3 years ago
Base course aggregate has a target density of 121.8 lb/ft3 in place It will be laid down and compacted in a rectangular street r
irakobra [83]

Answer:

total weight of the aggregate = 3594878.28 lbs

Explanation:

given data

density = 121.8 lb/ft³

area =  1000 ft x 60 ft x 6 in = 1000 ft x 60 ft x 0.5 ft

moisture = 3.5 %

compaction = 95%

solution

we get here first volume of the space that is filled with the aggregate that is

volume = 1000 ft x 60 ft x 0.5 ft  = 30,000 cu ft

now we get fill space with aggregate that compact to 95% of dry density.

so we fill space with aggregate of density that is = 95% of 121.8

= 115.71 lb/ cu ft

so now dry weight of aggregate is

dry weight of aggregate = 30,000  × 115.71 = 3471300 lb

when we assume that moisture percentage is by weight

then weight of the moisture in aggregate will be

weight of the moisture in aggregate = 3.56 % of 3471300 lb  

weight of the moisture in aggregate = 123578.28 lbs

and

we get total weight of the aggregate to fill space that is

total weight of the aggregate = 3471300 lb +123578.28 lb

total weight of the aggregate = 3594878.28 lbs

4 0
3 years ago
A power washer is being used to clean the siding of a house. Water enters at 20 C, 1 atm, with a velocity of 0.2 m/s .A jet of w
BabaBlast [244]

A power washer is being used to clean the siding of a house. Water at the rate of 0.1 kg/s enters at 20°c and 1 atm, with the velocity 0.2m/s. The jet of water exits at 23°c, 1 atm with a velocity 20m/s at an elevation of 5m. At steady state, the magnitude of the heat transfer rate from power unit to the surroundings is 10% of the power input. Determine the power input to the motor in kW.

Answer:

Net power of 1.2 KW is being extracted

Explanation:

We are given;

Mass flow rate; m' = 0.1kg/s

Inlet temperature; T1 = 20°C = 293K

Inlet pressure; P1 = 1 atm = 10^(5) pa

Inlet velocity; v1 = 0.2 m/s

Exit Pressure; P2 = 1 atm = 10^(5) pa

Exit Temperature; T2 = 1 atm = 296K

Exit velocity; V2 = 20m/s

Change in elevation; h = Z2 - Z1 = 5m

We are told that the magnitude of the heat transfer rate from the power unit to the surroundings is 10% of the power input.

Thus;

Q = -0.1W

From Bernoulli equation;

Q - W = ∆Potential energy + ∆Kinetic energy + ∆Pressure energy

Where;

∆Potential energy = mg(z2 - z1)

∆Kinetic energy = ½m(v2² - v1²)

∆Pressure energy = mc_p(T2 - T1)

Thus;

-0.1W - W = [m'g(z2 - z1)] + [½m'(v2² - v1²)] + [m'c_p(T2 - T1)]

Where C_p is specific heat capacity of water = 4200 J/Kg.k

Plugging in the relevant values, we have;

-1.1W = (0.1 × 9.81 × 5) + (½ × 0.1(20² - 0.2²)) + (0.1 × 4200 × (296 - 293))

-1.1W = 4.905 + 19.998 + 1260

-1.1W = 1284.903

W = -1284.903/1.1

W ≈ -1168 J/s ≈ -1.2 KW

The negative sign means that work is extracted from the system.

7 0
3 years ago
During an unsteady-flow process, the amount of mass and energy entering a control volume must be equal to the amount of mass and
SSSSS [86.1K]

Answer:

False

Explanation: An unstable flow is a flow process where the properties of the flow is unstable, in this case the flow input and flow output are not equal.

Unstable flow is also known as transient flow process because the main properties of interest which includes the time of flow and the volume of the flow through the pipe is not the same so this type of low gives more concern to the person carrying out the study.

3 0
4 years ago
Read 2 more answers
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