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LuckyWell [14K]
3 years ago
7

Ldentiy three industries that often need the skills of mechanical engineers. Briefly explain the skills that mechanical engineer

s bring to these
industries.
Engineering
1 answer:
const2013 [10]3 years ago
4 0

Answer:

3 industries that often need the skills of mechanical engineers are:

  • Automotive industry
  • Construction industry
  • Aerospace industry

The key skills mechanical engineers bring to these industries are effective technical skills, the ability to work under pressure, problem-solving skills, creativity and teamwork.

Explanation:

Automotive industry: The skills mechanical engineers bring to automotive industry include designing new cars for development, conducting laboratory testing for performance safety, and troubleshooting design or manufacturing issues with recalled vehicles. Automotive engineers have:

  • good mathematical skills, for instance in calculating the stresses power trains and other parts have to withstand;  
  • understanding and application of principles of physics and chemistry to properly design engines, electrical systems and other car components;  
  • good computer skills, because 21st century engineers rely on computer-assisted design software;
  • knowledge of ergonomics, which is applied in the process of designing a car so that the driver and passengers have a comfortable and functional environment, is another skill mechanical; engineers need.

Construction industry: Mechanical engineers are responsible for designing, building, establishing, and maintaining all kinds of mechanical machinery, tools, and components in the construction industry.

Aerospace industry: Mechanical engineers in aerospace industry produce specifications for design, development, manufacture and installing of new or modified mechanical components or systems. They design more fuel-efficient aircraft that cut emissions and build the fleets of satellites that power modern GPS technology.

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Oliver is designing a new children's slide to increase speed at which a child can descend.His first design involved steel becaus
goldenfox [79]

Answer:

Steel can rusts easily in the presence of moisture and oxygen, and tarnishes as the rust progresses.

Explanation:

Steel is an alloy of carbon and iron. it is a very useful alloy that is found in almost any engineered piece. The problem with steel is that it is very susceptible to rust, and the cost of maintaining it in order to prevent rust is very high. Steel rusts in the presence of moisture and oxygen (rust is an oxidation-reduction process). Using it as a water slide exposes it constantly to moisture, and to prevent it from rusting in this case will involve a lot of maintenance cost, which is why steel is not advisable to be used in this case.

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4 years ago
what i the maximum flow rate of glycerine at 20C in a 10cm diameter pipe that can be assumed to remain laminar
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Answer: tube flow

Explanation:

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3 years ago
A small submarine has a triangular stabilizing fin on its stern. The fin is 1 ft tall and 2 ft long. The water temperature where
Arturiano [62]

Answer:

\mathbf{F_D \approx 1.071 \ lbf}

Explanation:

Given that:

The height of a  triangular stabilizing fin on its stern is 1 ft tall

and it length is 2 ft long.

Temperature = 60 °F

The objective is to determine the drag on the fin when the submarine is traveling at a speed of 2.5 ft/s.

From these information given; we can have a diagrammatic representation describing how the  triangular stabilizing fin looks like as we resolve them into horizontal and vertical component.

The diagram can be found in the attached file below.

If we recall ,we know that;

Kinematic viscosity v = 1.2075 \times 10^{-5} \ ft^2/s

the density of water ρ = 62.36 lb /ft³

Re_{max} = \dfrac{Ux}{v}

Re_{max} = \dfrac{2.5 \ ft/s \times 2  \  ft }{1.2075 \times 10 ^{-5} \ ft^2/s}

Re_{max} = 414078.6749

Re_{max} = 4.14 \times 10^5 which is less than < 5.0 × 10⁵

Now; For laminar flow;  the drag on  the fin when the submarine is traveling at 2.5 ft/s can be determined by using the expression:

dF_D = (\dfrac{0.664 \times \rho  \times U^2 (2-x) dy}{\sqrt{Re_x}})^2

where;

(2-x) dy = strip area

Re_x = \dfrac{2.5(2-x)}{1.2075 \times 10 ^{-5}}

Therefore;

dF_D = (\dfrac{0.664 \times 62.36  \times 2.5^2 (2-x) dy}{\sqrt{ \dfrac{2.5(2-x)}{1.2075 \times 10 ^{-5}}}})

dF_D = 1.136 \times(2-x)^{1/2} \ dy

Let note that y = 0.5x from what we have in the diagram,

so , x = y/0.5

By applying the rule of integration on both sides, we have:

\int\limits \  dF_D =  \int\limits^1_0 \  1.136 \times(2-\dfrac{y}{0.5})^{1/2} \ dy

\int\limits \  dF_D =  \int\limits^1_0 \  1.136 \times(2-2y)^{1/2} \ dy

Let U = (2-2y)

-2dy = du

dy = -du/2

F_D =  \int\limits^0_2 \  1.136 \times(U)^{1/2} \ \dfrac{du}{-2}

F_D = - \dfrac{1.136}{2} \int\limits^0_2 \ U^{1/2} \ du

F_D = -0.568 [ \dfrac{\frac{1}{2}U^{ \frac{1}{2}+1 }  }{\frac{1}{2}+1}]^0__2

F_D = -0.568 [ \dfrac{2}{3}U^{\frac{3}{2} }   ] ^0__2

F_D = -0.568 [0 -  \dfrac{2}{3}(2)^{\frac{3}{2} }   ]

F_D = -0.568 [- \dfrac{2}{3} (2.828427125)}   ]

F_D = 1.071031071 \ lbf

\mathbf{F_D \approx 1.071 \ lbf}

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Read 2 more answers
Determine the depreciation expense for 2018 and 2019 using the following​ methods: (a)​ Straight-line (SL),​ (b) Units of produc
prohojiy [21]

Answer:

Check Explanation.

Explanation:

(1). The straight-line method: the general clue with this method is that in the two years, depreciation is the same. The formula for Calculating depreciation is given below;

straight-line method = (cost - Residual value)/ useful life in years.

From the question we know that the cost of acquisition is $30,000,000, the residual value of the asset is $4,000,000 and useful life is 7 years. Therefore;

straight-line method = ($30,000,000 - $4,000,000)/ 7.

= $3, 714,285.71 Per year.

That is $3, 714,285.71 for 2018 and 2019.

(2).Units of production​ (UOP) = (cost - Residual value)/ useful life in units.

= ($30,000,000 - $4,000,000)/ 4,375, 000.

Units of production​ (UOP) = $6 per mile.

Hence, the depreciation in 2018 = Depreciation per unit × 2018 year usage.

= 6 × 1,100,000 mile.

= $6,600,000.

depreciation in 2019 = Depreciation per unit × 2019 year usage.

= 6 × 1,200,000.

= $7,200,000.

Double-declining-balance​ (DDB)= (cost - accumulated depreciation) × 2 × 1/(useful life years).

Double-declining-balance​ (DDB) = (30,000,000 - 0)× 2 × (1/7).

= $8,571,428.57 depreciation in 2018.

= $8,571,428.6 depreciation in 2018

Double-declining-balance​ (DDB) = (30,000,000 - 8,571,428.57) × 2 × 1/7.

= $6,122,449.00 depreciation in 2019.

====================================================================

Total depreciation for straight-line method(2018 and 2019) = $7,428,571.42.

Total depreciation for Units of production​ (UOP)(2018 and 2019) = $13,800,000.

Total depreciation for Double-declining-balance (DDB)= $ 14,693,877.6.

5 0
3 years ago
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