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Brilliant_brown [7]
3 years ago
12

Pls list up to five key lessons and knowledge areas that you have acquired in this course about operations management. How do yo

u believe they help you in your future professional career?
Try to describe your response in brief detail.
Engineering
1 answer:
Aneli [31]3 years ago
3 0

Explanation:

Production planning: Planning is ideal so that there are the right resources, at the right time and in the right quantity that can meet the production needs of a period.

Strategies: The strategic development of production is the area that will assist in organizational competitiveness and in meeting consumer demand and needs.

Product and service design: Development of new products and services and their improvement, innovations and greater benefits

Production systems: Study of physical arrangements so that production takes place effectively according to the ideal layout for each type of product or service.

Production capacity planning: Analysis of the short, medium and long term related to production, and identification if necessary to obtain more resources, increase in staff, machinery, etc., to meet present and future demands.

<em>Each area of ​​knowledge acquired will assist in the development of a professional career, as technical knowledge is essential in decision-making, provision, problem solving, the development of new ideas and innovation.</em>

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A customer comes in with a vehicle that needs tire rotation and wheel alignment. Your bay has a two post frame engaging lift is
just olya [345]

Answer:

D

Explanation:

5 0
3 years ago
Oil with a density of 850 kg/m3 and kinematic viscosity of 0.00062 m2/s is being discharged by an 8-mm-diameter, 42-m-long horiz
sladkih [1.3K]

Answer:

The flow rate of oil through the pipe is 1.513E-7 m³/s.

Explanation:

Given

Density, ρ = 850 kg/m³

Kinematic viscosity, v = 0.00062 m²/s

Diameter, d = 8-mm = 0.008m

Length of horizontal pipe, L = 42-m

Height, h = 4-m.

We'll solve the flow rate of oil through the pipe by using Hagen-Poiseuille equation.

This is given as

∆P = (128μLQ)/πD⁴

Where ∆P = Rate of change of pressure

μ = Dynamic Viscosity

Q = Flow rate of oil through the pipe.

First, we need to determine the dynamic viscosity and the rate of change in pressure

Dynamic Viscosity, μ = Density (ρ) * Kinematic viscosity (v)

μ = 850 kg/m³ * 0.00062 m²/s

μ = 0.527kg/ms

Then, we calculate the rate of change of pressure.

Assuming that the velocity through the pipe is so small;

∆P = Pressure at the bottom of the tank

∆P = Density (ρ) * Acceleration of gravity (g) * Height (h)

Taking g = 9.8m/s²

∆P = 850kg/m³ x 9.8m/s² x 4m

∆P = 33320N/m²

Recall that Hagen-Poiseuille equation.

∆P = (128μLQ)/πD⁴ --- Make Q the subject of formula

Q = (πD⁴P)/(128μL)

By substituton;

Q = (π * 0.008⁴ * 33320)/(128 * 0.527 * 42)

Q = 0.00000015133693643099

Q = 1.513E-7 m³/s.

Hence, the flow rate of oil through the pipe is 1.513E-7 m³/s.

8 0
3 years ago
Economics is the study of how individuals and societies make choices under the condition of
k0ka [10]

Answer:

ig scarcity

Explanation:

Hope it Helps!!!

3 0
2 years ago
Design a half-wave recti er which provides a peak voltage of 15 V, and anaverage voltage of 3.8 V when driven by a 120 V (rms) a
nirvana33 [79]

Answer:

You need a 120V to 24V commercial transformer  (transformer 1:5), a 100 ohms resistance, a 1.5 K ohms resistance and a diode with a minimum forward current of 20 mA (could be 1N4148)

Step by step design:

  1. Because you have a 120V AC voltage supply you need an efficient way to reduce that voltage as much as possible before passing to the rectifier, for that I recommend a standard 120V to 24V transformer.  120 Vrms = 85 V and 24 Vrms = 17V = Vin
  2. Because 17V is not 15V you still need a voltage divider to step down that voltage, for that we use R1 = 100Ω and R2 = 1.3KΩ. You need to remember that more than 1 V is going to be in the diode, so for our calculation we need to consider it. Vf = (V*R2)/(R1+R2), V = Vin - 1 = 17-1 = 16V and Vf = 15, Choosing a fix resistance R1 = 100Ω and solving the equation we find R2 = 1.5KΩ
  3. Finally to select the diode you need to calculate two times the maximum current and that would be the forward current (If) of your diode. Imax = Vf/R2 = 10mA and If = 2*Imax = 20mA

Our circuit meet the average voltage (Va) specification:

Va = (15)/(pi) = 4.77V considering the diode voltage or 3.77V without considering it

6 0
4 years ago
Sea water with a density of 1025 kg/m3 flows steadily through a pump at 0.21 m3 /s. The pump inlet is 0.25 m in diameter. At the
myrzilka [38]

Answer:

\dot W_{pump} = 16264.922\,W\,(16.265\,kW)

Explanation:

The pump is modelled after applying Principle of Energy Conservation, whose form is:

\frac{P_{1}}{\rho\cdot g}+ \frac{v_{1}^{2}}{2\cdot g} +z_{1} + h_{pump}=\frac{P_{2}}{\rho\cdot g}+ \frac{v_{2}^{2}}{2\cdot g} +z_{2}

The head associated with the pump is cleared:

h_{pump} = \frac{P_{2}-P_{1}}{\rho\cdot g}+\frac{v_{2}^{2}-v_{1}^{2}}{2\cdot g}+(z_{2}-z_{1})

Inlet and outlet velocities are found:

v_{1} = \frac{0.21\,\frac{m^{3}}{s} }{\frac{\pi}{4}\cdot (0.25\,m)^{2} }

v_{1} \approx 4.278\,\frac{m}{s}

v_{2} = \frac{0.21\,\frac{m^{3}}{s} }{\frac{\pi}{4}\cdot (0.152\,m)^{2} }

v_{2} \approx 11.573\,\frac{m}{s}

Now, the head associated with the pump is finally computed:

h_{pump} = \frac{175\,kPa-81.326\,kPa}{(1025\,\frac{kg}{m^{3}} )\cdot (9.807\,\frac{m}{s^{2}} )} +\frac{(11.573\,\frac{m}{s} )^{2}-(4.278\,\frac{m}{s} )^{2}}{2\cdot (9.807\,\frac{m}{s^{2}} )} + 1.8\,m

h_{pump} = 7.705\,m

The power that pump adds to the fluid is:

\dot W_{pump} = \dot V \cdot \rho \cdot g \cdot h_{pump}

\dot W_{pump} = (0.21\,m^{3})\cdot (1025\,\frac{kg}{m^{3}})\cdot (9.807\,\frac{m}{s^{2}})\cdot(7.705\,m)

\dot W_{pump} = 16264.922\,W\,(16.265\,kW)

4 0
3 years ago
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