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3 years ago

Which option identifies what engineers will do to help in the following scenario?

Engineering

2 answers:

Tasya [4]3 years ago

4 0

Answer:

Explanation:

MariettaO [177]3 years ago

3 0

Answer:

Explanation:

"Engineers and scientists immediately worked diligently to design a system that would cap off the pipeline and stop the flow of the petroleum."

Trust me I know this is accurate because I have a reference.

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- massmaster34

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A "scale" is constructed with a 4-ft-long cord and the 10-lb block D. The cord is fixed to a pin at A and passes over two small

Anarel [89]

Answer:

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Explanation:

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3 years ago

Bridge Reflection:

Nataliya [291]

Answer:

IT feels proud to be done making a bridge

Explanation:

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3 years ago

Derive the expression ε=ln(1+e), where ε is the true strain and e is the engineering strain. Note that this expression is not va

Rudiy27

The formula for true strain after derivation from basic terms is; ε_t = In(1 + ε_e)

<h3>How to derive the expression for True Strain?</h3>

Formula for Engineering Stress is;

σ_e = Load/Area

Formula for true stress is;

σ_t = Force/Instantaneous Area

Formula for Engineering Strain is;

ε_e = ΔL/L₀

Formula for true strain is;

dε_t = dL/L

Total true strain is gotten from;

ε_t = ∫(dL/L) between boundaries of L_f and L_o

When we integrate between those boundaries, we have;

ε_t = In[(L₀ + ΔL)/L₀

⇒ ε_t = In[(1+ ΔL/L₀)

⇒ ε_t = In(1 + ε_e)

Read more about True Strain at; brainly.com/question/20717759

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6 0

2 years ago

An angle is observed repeatedly using the same equipment and procedures producing the data below:35 ∘ 40'00",35 ∘ 40'10",35 ∘ 40

Helen [10]

Answer: (a) +/- 7.5° (b) +/- 3.75°

Explanation:

See attachment

6 0

3 years ago

Water enters the pump of a steam power plant as saturated liquid at 20 kPa at a rate of 45 kg/s and exits at 6 MPa. Neglecting t

Natasha2012 [34]

Answer:

$\dot W_{in} = 273.69\,kW$

Explanation:

The pump is modelled after the First Law of Thermodynamics. A reversible process means that fluid does not report any positive change in entropy:

$\dot W_{in} + \dot m \cdot (h_{in}-h_{out}) = 0$

The properties of the fluid at entrance and exit are, respectively:

Inlet (Saturated Liquid)

$P = 20\,kPa$

$T = 60.06\,^{\textdegree}C$

$h = 251.42\,\frac{kJ}{kg}$

$s = 0.8320\,\frac{kJ}{kg\cdot K}$

Outlet (Subcooled Liquid)

$P = 6000\,kPa$

$T = 60.06\,^{\textdegree}C$

$h = 257.502\,\frac{kJ}{kg}$

$s = 0.8320\,\frac{kJ}{kg\cdot K}$

The power input to the pump is computed hereafter:

$\dot W_{in} = \left(45\,\frac{kg}{s} \right)\cdot \left(257.502\,\frac{kJ}{kg} -251.42\,\frac{kJ}{kg} \right)$

$\dot W_{in} = 273.69\,kW$

8 0

4 years ago