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3 years ago

Explain why advances in technology can actually create more problems for engineers.

Engineering

1 answer:

andrew11 [14]3 years ago

3 0

Explanation:

Engineering is science in practical terms. It is the application of scientific findings in problem solving and creating a better world.

How does technological advancements create more problems for engineers?

Loss of job to automation: the world is driving at automating work processes through the use of specially designed and crafted machinery. Work is now properly being done using machines with little to no human input in the whole process. This is a huge let off for engineers. Engineers have to compete with machines which are their own inventions for jobs now.
Fast paced work environment: machines can handle work more efficiently and faster than the people making them. There is an increasing race between engineers and their own inventions today for better product delivery. Unless a machine is faulty, they are more productive and efficient than man. This can cause engineers to want to catch up with their own inventions leading to a work life of stress.
Environmental problems they cannot solve: most inventions use components from the environment. They release effluents that are very difficult to be properly disposed or stored. This is a huge problem for engineers and can lead to ethical calls from the government and the populace. In short, they can create problems they are expected to solve but cannot solve.
Social problems: engineers can be portrayed as terrible beings for their own inventions. This leads to psychological problems on a good and creative invention. For example, rare earth metals in DR Congo are instrumental in making solar panels, but mining of these metals have forced several thousands of people into hard and intense labor on mines; there is a call on technological firms to stop exploiting people this way for their own gains.
Misuse of technology: any good technology can be put into the wrong use. A nuclear reaction can be packaged into a bomb and also, it can be the center of electricity generation on a commercial scale. How can engineers solve this kind of problem? Technological inventions are subjective in their usage.

Learn more:

New technology brainly.com/question/5768621

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Some connecting rods have ____ to help lubricate the cylinder wall or piston pin.

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Answer:

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4 years ago

Read 2 more answers

A four-lane divided multilane highway (two lanes in each direction) in rolling terrain has five access points per mile and 11-ft

grandymaker [24]

Answer:

peak-hour volume = 1890 veh/h

Explanation:

<u>Determine the peak-hour Volume </u>

Applying the equation below

Vp = v / ( PHF * N * Fg * Fdp ) -------------- ( 1 )

where :

Vp = 1250

v ( peak - hour volume ) = ?

PHF ( peak hour factor ) = 0.84

N = 2 lanes per direction

Fg ( grade adjustment for rolling terrain ) = 0.99 ≈ 1

Fdp = 0.90

<u>Back to equation 1 </u>

v = Vp ( PHF * N * Fg * Fdp )

= 1250 ( 0.84 * 2 * 1 * 0.90 )

= 1890 veh/h

5 0

3 years ago

A thin aluminum sheet is placed between two very large parallel plates that are maintained at uniform temperatures T1 = 900 K, T

Maru [420]

The net radiation heat transfer between the two plates per unit surface area of the plates with shield and without shied are respectively; 2282.76 W/m² and 9766.75 W/m²

<h3>How to find the net radiation heat transfer?</h3>

We are given;

Temperature 1; T₁

Temperature 2; T₂

Temperature 3; T₃

Emissivity 1; ε₁ = 0.3

Emissivity 2; ε₂ = 0.7

Emissivity 3; ε₃ = 0.2

The net rate of radiation heat transfer with a thin aluminum shield per unit area of the plates with shield is;

Q'₁₂ = σ(T₁⁴ - T₂⁴)]/[((1/ε₁) + (1/ε₂) - 1) + ((1/ε₃,₁) + (1/ε₃,₂) - 1)]

Q'₁₂ = 5.67 * 10⁻⁸(900⁴ - 300⁴)/[((1/0.3) + (1/0.7) - 1) + ((1/0.15) + (1/0.15) - 1)]

Q'₁₂,shield = 2282.76 W/m²

The net rate of radiation heat transfer with a thin aluminum shield per unit area of the plates with no shield is;

Q'₁₂,no shield = σ(T₁⁴ - T₂⁴)]/((1/ε₁) + (1/ε₂) - 1))

Q'₁₂,no shield = 5.67 * 10⁻⁸(900⁴ - 300⁴)/[(1/0.3) + (1/0.7) - 1)]

Q'₁₂,no shield = 9766.75 W/m²

Then the ratio of radiation heat transfer for the two cases becomes;

Q'₁₂,shield/Q'₁₂,no shield = 2282.76/9766.75 = 0.2337 or 4/17

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2 years ago

Air is compressed by a 40-kW compressor from P1 to P2. The air temperature is maintained constant at 25Â°C during this process a

AlexFokin [52]

Answer:

the rate of entropy change of the air is -0.1342 kW/K

the assumptions made in solving this problem

- Air is an ideal gas.

- the process is isothermal ( internally reversible process ). the change in internal energy is 0.

- It is a steady flow process

- Potential and Kinetic energy changes are negligible.

Explanation:

Given the data in the question;

From the first law of thermodynamics;

dQ = dU + dW ------ let this be equation 1

where dQ is the heat transfer, dU is internal energy and dW is the work done.

from the question, the process is isothermal ( internally reversible process )

Thus, the change in internal energy is 0

dU = 0

given that; Air is compressed by a 40-kW compressor from P1 to P2

since it is compressed, dW = -40 kW

we substitute into equation 1

dQ = 0 + ( -40 kW )

dQ = -40 kW

Now, change in entropy of air is;

ΔS $_{air$ = dQ / T

given that T = 25 °C = ( 25 + 273.15 ) K = 298.15 K

so we substitute

ΔS $_{air$ = -40 kW / 298.15 K

ΔS $_{air$ = -0.13416 ≈ -0.1342 kW/K

Therefore, the rate of entropy change of the air is -0.1342 kW/K

the assumptions made in solving this problem

- Air is an ideal gas.

- the process is isothermal ( internally reversible process ). the change in internal energy is 0.

- It is a steady flow process

- Potential and Kinetic energy changes are negligible.

7 0

3 years ago

Three identical fatigue specimens (denoted A, B, and C) are fabricated from a nonferrous alloy. Each is subjected to one of the

Law Incorporation [45]

Answer:

B A and C

Explanation:

Given:

Specimen σ $_{max}$ σ $_{min}$

A +450 -150

B +300 -300

C +500 -200

Solution:

Compute the mean stress

σ $_{m}$ = (σ $_{max}$ + σ $_{min}$ )/2

σ $_{mA}$ = (450 + (-150)) / 2

= (450 - 150) / 2

= 300/2

σ $_{mA}$ = 150 MPa

σ $_{mB}$ = (300 + (-300))/2

= (300 - 300) / 2

= 0/2

σ $_{mB}$ = 0 MPa

σ $_{mC}$ = (500 + (-200))/2

= (500 - 200) / 2

= 300/2

σ $_{mC}$ = 150 MPa

Compute stress amplitude:

σ $_{a}$ = (σ $_{max}$ - σ $_{min}$ )/2

σ $_{aA}$ = (450 - (-150)) / 2

= (450 + 150) / 2

= 600/2

σ $_{aA}$ = 300 MPa

σ $_{aB}$ = (300- (-300)) / 2

= (300 + 300) / 2

= 600/2

σ $_{aB}$ = 300 MPa

σ $_{aC}$ = (500 - (-200))/2

= (500 + 200) / 2

= 700 / 2

σ $_{aC}$ = 350 MPa

From the above results it is concluded that the longest fatigue lifetime is of specimen B because it has the minimum mean stress.

Next, the specimen A has the fatigue lifetime which is shorter than B but longer than specimen C.

In the last comes specimen C which has the shortest fatigue lifetime because it has the higher mean stress and highest stress amplitude.

7 0

4 years ago