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3 years ago

What is electrical energy used for?

Engineering

2 answers:

Stolb23 [73]3 years ago

8 0

Answer:

Electrical energy is used throughout the world to power devices, appliances and methods of transportation utilized in daily life. To make things operate, electrical energy must be emitted from energy sources such as power plants, to enable an object to consume the power it needs to function.

Explanation:

FromTheMoon [43]3 years ago

6 0

Answer:

to power devices appliances and some methods of transportation

Explanation:

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The constant A in Equation 17.2 is 12π4 R/5(θD) 3 where R is the gas constant and θD is the Debye temperature (K). Estimate θD f

kirill115 [55]

Answer:

The Debye temperature for aluminum is 375.2361 K

Explanation:

Molecular weight of aluminum=26.98 g/mol

T=15 K

The mathematical equation for the specific heat and the absolute temperature is:

$C_{v} =AT^{3}$

Substituting in the expression of the question:

$C_{v} =(\frac{12\pi ^{4}R }{5\theta _{D}^{3} } )T^{3}$

$\theta _{D} =(\frac{12\pi ^{4}RT^{3} }{5C_{v} } ) ^{1/3}$

Here

$C_{v} =4.6\frac{J}{kg-K} *\frac{1kg}{1000g} *\frac{26.98g}{1mol} =0.1241J/mol-K$

Replacing:

$\theta _{D} =(\frac{12\pi ^{4}*8.31*15^{3} }{5*0.1241} )^{1/3} =375.2361K$

3 0

3 years ago

How are isometric drawings and orthographic drawings similar?

sashaice [31]

Answer:

The horizontal lines of the orthographic drawing

are NOT horizontal in the isometric drawing but are projected at 30-degree and 60. ... degree angles, and the length of the lines remain the same in the isometric as they were in the orthographic.

Hope it's helpful to you

pls mark me as brain list

5 0

4 years ago

10 POINTS!!

marusya05 [52]

Answer:

disable yahoo from activating.

Explanation:

either force quit it or add chrome to your user bar at the bottom of the screen (if ure on a computer) if yahoo is on ur bar make sure to force quit it by right clicking and clicking "force quit" and it should stop

5 0

3 years ago

Consider a very long rectangular fin attached to a flat surface such that the temperature at the end of the fin is essentially t

kodGreya [7K]

Answer:

$\frac{T-20}{130-20}= e^{-14.28*0.05}$

And if we solve for T we got:

$T= 20 + 110e^{-14.28*0.05} = 73.86 C$

The answer for this case would be $T = 73.86 C$ at 5cm from the base of the fin.

Explanation:

Data given

For this case we have the following data given:

$h = 20 \frac{W}{m^2 K}$ represent the heat transfer coefficient.

$p$ represent the perimeter for this case and would be given by:

$p = 2*0.05m +2*0.001m= 0.102m$

$k = 200 \frac{W}{m C}$ represent the thermal conductivity

$w = 5cm =0.05 m$ represent the width

$h = 1mm =0.001m$ represent the thickness

$A= wh= 0.05m *0.001m = 0.00005 m^2$

Solution to the problem

For this case we assume that we have steady conditions, the temperature of the fins varies just in one direction, the heat transfer coefficient not changes with the time and the thermal properties of the fin not change.

We can determine the temperature if the fin at x=5 cm=0.05 m from the base with the following formula:

$\frac{T-T_{\infty}}{T_b -T_{\infty}} = e^{-mx}$

Where m is a coefficient given by:

$m = \sqrt{\frac{hp}{kA}}=\sqrt{\frac{20 W/m^2 C 0.102 m}{200 W/ mC 0.00005 m^2}}= 14.28 m^{-1}$

The value of x for this case represent the distance $x =5 cm =0.05m$

$T_b =130 C$ represent the base temperature

$T_{\infty}= 20$ represent the temperature of the sorroundings or the ambient.

If we replace we have this:

$\frac{T-20}{130-20}= e^{-14.28*0.05}$

And if we solve for T we got:

$T= 20 + 110e^{-14.28*0.05} = 73.86 C$

The answer for this case would be $T = 73.86 C$ at 5cm from the base of the fin.

3 0

3 years ago

Explain difference(s) between strength, hardness and toughness.

harkovskaia [24]

Explanation:

Step1

Strength is the maximum stress induces in the material under applied load condition. More the strength more will be ability to sustain load. Strength can be measured from tensile test. It has same unit as stress. Generally ductile material has more strength as compare to brittle material.

Step2

Hardness is the resistance to scratch on the material under given load condition. More the hardness more will be the resistance towards scratching of material. Hardness can be measured by Rockwell or Birnell hardness test. This property of metal is opposite to the strength.

Step3

Toughness is the ability to absorb energy under given loading condition up to its fracture point. It is a type of strain energy that is stored in the metal. Generally ductile metal has more toughness as compare to brittle.

4 0

4 years ago