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3 years ago

Which option identifies why Ethan’s skills are valuable to his team in the following scenario?

Engineering

1 answer:

larisa [96]3 years ago

4 0

Answer:

Explanation:

The options are:

- In an isometric drawing, multiple angles and axes can be shown in one sketch.

- There is no room for detail in an isometric drawing, so the detail is shown in the orthographic projection.

- Only one sketch will be needed since all other previous designs will no longer be necessary.

- Computer programs will not be necessary to create the exact dimensions of the design.

Orthographic projections are in either the First or Third Angles but the angles are fixed and do not provide perspective view. Isometric drawings are perspective views from different angles.

So Ethan's skill is valuable because "In an isometric drawing, multiple angles and axes can be shown in one sketch."

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Which of the following are true about algorithms? (You may select more than one)

r-ruslan [8.4K]

Have a orderf. Fr

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4 years ago

Read 2 more answers

Solve the problem with conditions if a wall has inner and outer surface temperatures of 16 and 6 C, respectively. The interior a

Angelina_Jolie [31]

Answer:

Heat flux is 20 W/m^2

Explanation:

Heat flux (Q) is computed as

$Q = h \, \Delta T$

where h is heat transfer coefficient and ΔT is the difference between body's temperature

From the interior air to the inner wall

$Q = 5 \frac{W}{m^2 K} \, 4 K$

$Q = 20 \frac{W}{m^2}$

From the the outer wall to the exterior air

$Q = 20 \frac{W}{m^2 K} \, 1 K$

$Q = 20 \frac{W}{m^2}$

The wall is under steady-state condition because heat flux is constant

7 0

3 years ago

Steam enters a turbine at 8000 kPa, 440oC. At the exit, the pressure and quality are 150 kPa and 0.19, respectively.

levacccp [35]

Answer:

$\dot W_{out} = 3863.98\,kW$

Explanation:

The turbine at steady-state is modelled after the First Law of Thermodynamics:

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

The specific enthalpies at inlet and outlet are, respectively:

Inlet (Superheated Steam)

$h_{in} = 3353.1\,\frac{kJ}{kg}$

Outlet (Liquid-Vapor Mixture)

$h_{out} = 890.1\,\frac{kJ}{kg}$

The power produced by the turbine is:

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

$\dot W_{out} = -2.93\,kW + (1.57\,\frac{kg}{s} )\cdot (3353.1\,\frac{kJ}{kg} - 890.1\,\frac{kJ}{kg} )$

$\dot W_{out} = 3863.98\,kW$

8 0

3 years ago

Describe the importance of ferrite and austenite stabilizing elements in steels

podryga [215]

Answer:

The importance of ferrite and austenite stabilizing elements in steels .

Explanation:

Alloying -

The process which improves the properties of the steel by changing the chemical composition of the steel via adding some elements .

The properties can be improved by - Stabilizing Austenite and Stabilizing Ferrite .

Stabilizing austenite -

The process by which temperature is increased , in which Austenite exists .

Elements with the same crystal structure as of the austenite ( FCC ) raises its A4 value i.e. the temperature of the formation of austenite from its liquid phase and reduces the value of A3 .

Hence, the elements are -

Cobalt , Nickel , Manganese , Copper.

The examples of the Austenitic steels are -

Hadfield Steel ( 13% Mn , 1.2% Cr , 1% C ) and Austenitic Stainless steel.

Stabilizing ferrite –

The process by which temperature is decreased , in which austenite exists .

Elements with the same crystal structure as of the ferrite (BCC - Cubic body centered ) lowers its A4 value i.e. the temperature of the formation of austenite from its liquid phase and increases the value of A3 .These elements have lower solubility of carbon in austenite, that lead to increase in the amount of carbides in the steel.

Hence, the elements are -

Aluminium , Silicon , Tungsten , Chromium , Molybdenum , Vanadium

The examples of the Ferritic steels are -

F-Cr alloys , transformer sheets steel ( 3% Si ).

3 0

4 years ago

. A normal-weight concrete has an average compressive strength of 20 MPa. What is the estimated flexure strength

bulgar [2K]

Answer:

2.77mpa

Explanation:

compressive strength = 20 MPa. We are to find the estimated flexure strength

We calculate the estimated flexural strength R as

R = 0.62√fc

Where fc is the compressive strength and it is in Mpa

When we substitute 20 for gc

Flexure strength is

0.62x√20

= 0.62x4.472

= 2.77Mpa

The estimated flexure strength is therefore 2.77Mpa

4 0

3 years ago