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2 years ago

Which of the following describes what a manufacturing engineer does

2 answers:

7 0

Manufacturing Engineers focus on the design and operation of integrated systems for the production of high-quality, economically competitive products.

Anit [1.1K]2 years ago

6 0

Answer:

a manufacturing engineer researches and develops tools to update existing processes in order to make the machine make products faster

Explanation:

hope this helps! :))

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1. Which of the following will cause a spark knock?

zlopas [31]

Answer:

I couldn't find options for your question online, but I can give you an explanation so you can choose the correct option.

Explanation:

A spark knock is a form of unpredictable behavior that occurs in combustion, that is, in the chemical reaction that occurs between oxygen and an oxidizable material. Such combustion is usually manifested by incandescence or flame.

The spark knock is a detonation that occurs when there is a lot of pressure in the fuel.

<u>Some situations in which this can happen are: </u>

Engine overloaded.
Maximum pressure in the cylinders.
Engine overheated.
Overheated air.
Long and excessive engine ignition timing.
Spark plug at high temperatures.

5 0

3 years ago

- Viscoelastic stress relaxation

My name is Ann [436]

Explanation:

The correct answers to the fill in the blanks would be;

1. Viscoelastic stress relaxation refers to scenarios for which the stress applied to a polymer must decay over time in order to maintain a constant strain. Otherwise, over time, the polymer chains will slip and slide past one another in response to a constant applied load and the strain will increase (in magnitude).

2. Viscoelastic creep refers to scenarios for which a polymer will permanently flow over time in response a constant applied stress.

The polymer whose properties have been mentioned above is commonly known as Kevlar.

It is mostly used in high-strength fabrics and its properties are because of several hydrogen bonds between polymer molecules.

5 0

3 years ago

The link acts as part of the elevator control for a small airplane. If the attached aluminum tube has an inner diameter of 25 mm

aksik [14]

Answer:

Tmax=14.5MPa

Tmin=10.3MPa

Explanation:

T = 600 * 0.15 = 90N.m

$T_max =\frac{T_c}{j} = \frac{x}{y} = \frac{90 \times 0.0175}{\frac{\pi}{2} \times (0.0175^4-0.0125^4)}$

=14.5MPa

$T_{min} =\frac{T_c}{j} = \frac{x}{y} = \frac{90 \times 0.0125}{\frac{\pi}{2} \times (0.0175^4-0.0125^4)}$

=10.3MPa

7 0

3 years ago

Water flows steadily through the pipe as shown below, such that the pressure at section (1) and at section (2) are 300 kPa and 1

steposvetlana [31]

Answer:

The velocity at section is approximately 42.2 m/s

Explanation:

For the water flowing through the pipe, we have;

The pressure at section (1), P₁ = 300 kPa

The pressure at section (2), P₂ = 100 kPa

The diameter at section (1), D₁ = 0.1 m

The height of section (1) above section (2), D₂ = 50 m

The velocity at section (1), v₁ = 20 m/s

Let 'v₂' represent the velocity at section (2)

According to Bernoulli's equation, we have;

$z_1 + \dfrac{P_1}{\rho \cdot g} + \dfrac{v^2_1}{2 \cdot g} = z_2 + \dfrac{P_2}{\rho \cdot g} + \dfrac{v^2_2}{2 \cdot g}$

Where;

ρ = The density of water = 997 kg/m³

g = The acceleration due to gravity = 9.8 m/s²

z₁ = 50 m

z₂ = The reference = 0 m

By plugging in the values, we have;

$50 \, m + \dfrac{300 \ kPa}{997 \, kg/m^3 \times 9.8 \, m/s^2} + \dfrac{(20 \, m/s)^2}{2 \times 9.8 \, m/s^2} = \dfrac{100 \ kPa}{997 \, kg/m^3 \times 9.8 \, m/s^2} + \dfrac{v_2^2}{2 \times 9.8 \, m/s^2}$ 50 m + 30.704358 m + 20.4081633 m = 10.234786 m + $\dfrac{v_2^2}{2 \times 9.8 \, m/s^2}$

50 m + 30.704358 m + 20.4081633 m - 10.234786 m = $\dfrac{v_2^2}{2 \times 9.8 \, m/s^2}$

90.8777353 m = $\dfrac{v_2^2}{2 \times 9.8 \, m/s^2}$

v₂² = 2 × 9.8 m/s² × 90.8777353 m

v₂² = 1,781.20361 m²/s²

v₂ = √(1,781.20361 m²/s²) ≈ 42.204308 m/s

The velocity at section (2), v₂ ≈ 42.2 m/s

3 0

2 years ago

A cylindrical specimen of steel has an original diameter of 12.8 mm. It is tested in tension its engineering fracture strength i

Mama L [17]

Answer:

a) The ductility = -30.12%

the negative sign means reduction

Therefore, there is 30.12% reduction

b) the true stress at fracture is 658.26 Mpa

Explanation:

Given that;

Original diameter $d_{o}$ = 12.8 mm

Final diameter $d_{f}$ = 10.7

Engineering stress $\alpha _{E}$ = 460 Mpa

a) determine The ductility in terms of percent reduction in area;

Ai = π/4( $d_{o}$ )² ; Ag = π/4( $d_{f}$ )²

% = π/4 [ ( ( $d_{f}$ )² - ( $d_{o}$ )²) / ( π/4 ( $d_{o}$ )²) ]

= ( ( $d_{f}$ )² - ( $d_{o}$ )²) / ( $d_{o}$ )² × 100

we substitute

= [( (10.7)² - (12.8)²) / (12.8)² ] × 100

= [(114.49 - 163.84) / 163.84 ] × 100

= - 0.3012 × 100

= -30.12%

the negative sign means reduction

Therefore, there is 30.12% reduction

b) The true stress at fracture;

True stress $\alpha _{T}$ = $\alpha _{E}$ ( 1 + $E_{E}$ )

$E_{E}$ is engineering strain

$E_{E}$ = dL / Lo

= (do² - df²) / df² = (12.8² - 10.7²) / 10.7² = (163.84 - 114.49) / 114.49

= 49.35 / 114.49

$E_{E}$ = 0.431

so we substitute the value of $E_{E}$ into our initial equation;

True stress $\alpha _{T}$ = 460 ( 1 + 0.431)

True stress $\alpha _{T}$ = 460 (1.431)

True stress $\alpha _{T}$ = 658.26 Mpa

Therefore, the true stress at fracture is 658.26 Mpa

6 0

2 years ago