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denpristay [2]
3 years ago
7

Strong performance in your engineering discipline ordinarily is one necessary condition for becoming a successful engineering ma

nager. What other conditions are there?
Engineering
1 answer:
zhenek [66]3 years ago
4 0

Answer:

1.  Manpower Management

2. Productive Meetings

3.Establish administrative policies, procedures, and standards.

4. High command over assets management.

5. Specific field related knowledge of procurement.

6. Compliance and EHS  knowledge.

Explanation:

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The first one is d or the 4th answer choice and the second one is false. Hope this helps!
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You can assume there is no pressure drop between the exit of the compressor and the entrance of the turbine. All the power from
Eddi Din [679]

Answer:

s6rt5x11j4fgu

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6 0
3 years ago
A semiconductor is a solid substance that has a conductivity between that of an insulator and that of most metals. (True , False
tiny-mole [99]

The answer is : True

4 0
3 years ago
A ductile hot-rolled steel bar has a minimum yield strength in tension and compression of Syt = 60 kpsi and Syc = 75 kpsi. Using
kow [346]

Answer:

2.135

Explanation:

Lets make use of these variables

Ox 16.5 kpsi, and Oy --14,5 kpsi

To determine the factor of safety for the states of plane stress. We have to first understand the concept of Coulomb-Mohr theory.

Mohr–Coulomb theory is a mathematical model describing the response of brittle materials such as concrete, or rubble piles, to shear stress as well as normal stress.

Please refer to attachment for the step by step solution.

4 0
3 years ago
An asbestos pad is square in cross section, measuring 5 cm on a side at its small end increasing linearly to 10 cm on a side at
polet [3.4K]

Answer:

q = 1.73 W

Explanation:

given data

small end  = 5 cm

large end = 10 cm

high = 15 cm

small end is held = 600 K

large end at = 300 K

thermal conductivity of asbestos  = 0.173 W/mK

solution

first we will get here side of cross section that is express as

S = S1 + \frac{S2-S1}{L} x     ...............1

here x is distance from small end and S1 is side of square at small end

and S2 is side of square of large end and L is length

put here value and we get

S = 5 + \frac{10-5}{15} x

S = \frac{0.15 + x}{3}    m

and  

now we get here Area of section at distance x is

area A = S²    ...............2

area A = (\frac{0.15 + x}{3})^2    m²

and

now we take here small length dx and temperature difference is dt

so as per fourier law

heat conduction is express as

heat conduction q = \frac{-k\times A\  dt}{dx}      ...............3

put here value and we get

heat conduction q = -k\times (\frac{0.15 + x}{3})^2 \   \frac{dt}{dx}  

it will be express as

q \times \frac{dx}{(\frac{0.15 + x}{3})^2} = -k (dt)  

now we intergrate it with limit 0 to 0.15 and take temp 600 to 300 K

q \int\limits^{0.15}_0 {\frac{dx}{(\frac{0.15 + x}{3})^2 } = -0.173 \int\limits^{300}_{600} {dt}          

solve it and we get

q (30)  = (0.173) × (600 - 300)

q = 1.73 W

5 0
3 years ago
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