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2 years ago

In most states, an engineer is not expected to be licensed as a professional engineer (pe) in order to work. true true false

Engineering

1 answer:

tigry1 [53]2 years ago

7 0

Answer:

True

Explanation:

Although Engineers need their Profession Engineering license (PE) in order to sign off on design drawings, they are not required to have their PE in order to work. The majority of engineering careers do not require a PE.

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g If the rails are originally laid in contact, what is the stress in them on a summer day when their temperature is 31.0

Anettt [7]

A) The amount of space must be left between adjacent rails if they are just to touch on a summer day when their temperature is 31.0 °C is; 0.6048 cm

B) The stress in the rails on a summer day when their temperature is 31.0 °C is; 86.4 × 10⁶ Pa

<h3>Linear Thermal Expansion</h3>

We are given;

Length; L = 14 m

Initial Temperature; T_i = −5 °C

Final Temperature; T_f = 31 °C

The formula for Linear Thermal Expansion is;

ΔL = L_i * α * ΔT

where;

L_i is initial length

α is thermal expansion

ΔL is change in length

ΔT is change in temperature

Now, the thermal expansion of steel from online tables is α = 1.2 × 10⁻⁵ C⁻¹

Thus;

ΔL = 14 * 1.2 × 10⁻⁵ * (31 - (-5))

ΔL = 6.048 × 10⁻³ m = 0.6048 cm

The formula to get the stress is;

σ = Y * α * ΔT

where;

Y is young's modulus of steel = 20 × 10¹⁰ Pa

α is thermal expansion

ΔT is change in temperature

Thus;

σ = 20 × 10¹⁰ × 1.2 × 10⁻⁵ × (31 - (-5))

σ = 86.4 × 10⁶ Pa

The complete question is;

Steel train rails are laid in 14.0-m long segments placed end to end. The rails are laid on a winter day when their temperature is −5 °C.

(a) How much space must be left between adjacent rails if they are just to touch on a summer day when their temperature is 31.0 °C?

(b) If the rails are originally laid in contact, what is the stress in them on a summer day when their temperature is 31.0 °C?

Read more about Linear Thermal Expansion at; brainly.com/question/6985348

4 0

2 years ago

Who was part of dempwolf his firm when he first started

Zinaida [17]

Explanation:

Dempwolf created by John Augustus, Among the most prominent innovative solutions in Southern California Pennsylvania was established by Dempwolf with brother Reinhardt or uncle's son Frederick entered the company of J.A. Dozens of structures in 10 states were engineered by Dempwolf.

5 0

3 years ago

A model shows a city located between a warm ocean and next to coastal mountains. Which statement best describes average weather

DanielleElmas [232]

Answer:

The city will experience high levels of rainfall.

8 0

3 years ago

Read 2 more answers

What type of slab and beam used in construction of space neddle

Sav [38]

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5 0

3 years ago

An ideal Rankine cycle operates with turbine inlet steam at 90 bar and 500°C, and a condenser at 40 °C. Calculate the efficiency

lilavasa [31]

Answer:

40.8%

Explanation:

A rankine cycle is a generation cycle using water as a working fluid, when heat enters the boiler the water undergoes a series of changes in state and energy until generating power through the turbine.

This cycle is composed of four main components, the boiler, the pump, the turbine and the condenser as shown in the attached image

To solve any problem regarding the rankine cycle, enthalpies in all states must be calculated using the thermodynamic tables and taking into account the following.

• The pressure of state 1 and 4 are equal

• The pressure of state 2 and 3 are equal

• State 1 is superheated steam

• State 2 is in saturation state

• State 3 is saturated liquid at the lowest pressure

• State 4 is equal to state 3 because the work of the pump is negligible.

Once all enthalpies are found, the following equations are used using the first law of thermodynamics

Wout = m (h1-h2)

Qin = m (h1-h4)

Win = m (h4-h3)

Qout = m (h2-h1)

The efficiency is calculated as the power obtained on the heat that enters

Efficiency = Wout / Qin

Efficiency = (h1-h2) / (h1-h4)

first we calculate the enthalpies in all states

h1=3386kJ/Kg

h2=2073kJ/Kg

h2=h3=167.5kJ/Kg

we use the efficiency ecuation

$Efficiency =\frac{(h1-h2) }{(h1-h4)} =\frac{3386-2073}{3386-167.5} =0.408=40.8%$

8 0

3 years ago