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Brilliant_brown [7]
3 years ago
8

_________ is an example of an electrical device. A) PlugB) CordC) CableD) All of the above

Engineering
2 answers:
MA_775_DIABLO [31]3 years ago
6 0

Answer:

<h3>D ) All of the above</h3>

have a good day ~

9966 [12]3 years ago
3 0

Answer:

D ) All of the above

<h3>brainliest please.....</h3>

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The entrance of the space module into atmosphere is one of the very critical phases of any space mission. Due to the formation o
Hatshy [7]

Answer:

Check the explanation

Explanation:

The transferred energy calculations can be done using the equation:

This is when:

• power is calculated in watts (W)

• energy is calculated in joules (J)

• time is calculated in seconds (s)

it can also be calculated by utilizing the same equation but with different units:

• when energy is measured in kilowatt-hours (kWh)

• when power is measured in kilowatts (kW)

• and when time is measured in hours (h)

kindly check the comprehensive step by step calculation to the question in the attached images below

5 0
4 years ago
3 MAOP stands for which of the following?
raketka [301]

Answer:

MAOP Master of Arts in Organizational Psychology (various universities)

MAOP Maximum Allowable Operating Pressure

MAOP Mid-Atlantic Association of Oracle Professionals

MAOP Mid Atlantic Oncology Program

MAOP Maryland Association of Osteopathic Physicians

MAOP Master Air Operations Planner

MAOP Meharry Association of Office Personnel

MAOP Managers' Annual Operating Plan

MAOP Military Assistance Officer Program (US DoD)

Explanation:

3 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
Atmospheric air at 25 °C and 8 m/s flows over both surfaces of an isothermal (179C) flat plate that is 2.75m long. Determine the
vekshin1

Answer:

Re=100,000⇒Q=275.25 \frac{W}{m^2}

Re=500,000⇒Q=1,757.77\frac{W}{m^2}

Re=1,000,000⇒Q=3060.36 \frac{W}{m^2}

Explanation:

Given:

For air      T_∞=25°C  ,V=8 m/s

  For surface T_s=179°C

     L=2.75 m    ,b=3 m

We know that for flat plate

Re⇒Laminar flow

Re>30\times10^5⇒Turbulent flow

<u> Take Re=100,000:</u>

 So this is case of laminar flow

  Nu=0.664Re^{\frac{1}{2}}Pr^{\frac{1}{3}}

From standard air property table at 25°C

  Pr= is 0.71  ,K=26.24\times 10^{-3}

So    Nu=0.664\times 100,000^{\frac{1}{2}}\times 0.71^{\frac{1}{3}}

Nu=187.32   (\dfrac{hL}{K_{air}})

187.32=\dfrac{h\times2.75}{26.24\times 10^{-3}}

     ⇒h=1.78\frac{W}{m^2-K}

heat transfer rate =h(T_∞-T_s)

                           =275.25 \frac{W}{m^2}

<u> Take Re=500,000:</u>

So this is case of turbulent flow

  Nu=0.037Re^{\frac{4}{5}}Pr^{\frac{1}{3}}

Nu=0.037\times 500,000^{\frac{4}{5}}\times 0.71^{\frac{1}{3}}

Nu=1196.18  ⇒h=11.14 \frac{W}{m^2-K}

heat transfer rate =h(T_∞-T_s)

                             =11.14(179-25)

                           = 1,757.77\frac{W}{m^2}

<u> Take Re=1,000,000:</u>

So this is case of turbulent flow

  Nu=0.037Re^{\frac{4}{5}}Pr^{\frac{1}{3}}

Nu=0.037\times 1,000,000^{\frac{4}{5}}\times 0.71^{\frac{1}{3}}

Nu=2082.6  ⇒h=19.87 \frac{W}{m^2-K}

heat transfer rate =h(T_∞-T_s)

                             =19.87(179-25)

                           = 3060.36 \frac{W}{m^2}

7 0
3 years ago
Which of the following is an example of a hardwood? A maple B spruce C pine D fir
bearhunter [10]

Answer:

A. Maple

Explanation:

Maple is a hardwood.

Hope that helps!

7 0
2 years ago
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