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Igoryamba
3 years ago
8

FAST MECHANICAL LLC how many main parts does a motor have ​

Engineering
1 answer:
SSSSS [86.1K]3 years ago
4 0

Answer:

three main parts

Explanation:

Electric motor designs can vary quite a lot, though in general they have three main parts: a rotor, a stator and a commutator. These three parts use the attractive and repulsive forces of electromagnetism, causing the motor to spin continually as long as it receives a steady flow of electric current.

I HOPE THIS HELPED!!

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The equilibrium fraction of lattice sites that are vacant in electrum (a silver gold alloy) at 300K C is 9.93 x 10-8. There are
IrinaK [193]

Answer:

the density of the electrum is 14.30 g/cm³

Explanation:

Given that:

The equilibrium fraction of lattice sites that are vacant in electrum  = 9.93*10^{-8}

Number of vacant atoms = 5.854 * 10^{15} \ vacancies/cm^3

the atomic mass of the electrum = 146.08 g/mol

Avogadro's number = 6.022*10^{23

The Number of vacant atoms = Fraction of lattice sites × Total number of sites(N)

5.854*10^{15} = 9.93*10^{-8}  × Total number of sites(N)

Total number of sites (N) = \dfrac{5.854*10^{15}}{9.93*10^{-8}}

Total number of sites (N) = 5.895*10^{22}

From the expression of the total number of sites; we can determine the density of the electrum;

N = \dfrac{N_A \rho _{electrum}}{A_{electrum}}

where ;

N_A = Avogadro's Number

\rho_{electrum} = density of the electrum

A_{electrum} = Atomic mass

5.895*10^{22} = \dfrac{6.022*10^{23}* \rho _{electrum}}{146.08}

5.895*10^{22} *146.08}= 6.022*10^{23}* \rho _{electrum}

8.611416*10^{24}= 6.022*10^{23}* \rho _{electrum}

\rho _{electrum}=\dfrac{8.611416*10^{24}}{6.022*10^{23}}}

\mathbf{  \rho _{electrum}=14.30 \ g/cm^3}

Thus; the density of the electrum is 14.30 g/cm³

7 0
3 years ago
Manufacturers frequently make choices about their suppliers of raw materials based on their impact on society and the environmen
love history [14]
Manufacturers seek attention
4 0
3 years ago
Two loads connected in parallel draw a total of 2.4 kW at 0.8 pf lagging from a 120-V rms, 60-Hz line. One load absorbs 1.5 kW a
stealth61 [152]

Answer: a) 0.948 b) 117.5µf

Explanation:

Given the load, a total of 2.4kw and 0.8pf

V= 120V, 60 Hz

P= 2.4 kw, cos θ= 80

P= S sin θ - (p/cos θ) sin θ

= P tan θ(cos^-1 (0.8)

=2.4 tan(36.87)= 1.8KVAR

S= 2.4 + j1. 8KVA

1 load absorbs 1.5 kW at 0.707 pf lagging

P= 1.5 kW, cos θ= 0.707 and θ=45 degree

Q= Ptan θ= tan 45°

Q=P=1.5kw

S1= 1.5 +1.5j KVA

S1 + S2= S

2.4+j1.8= 1.5+1.5j + S2

S2= 0.9 + 0.3j KVA

S2= 0.949= 18.43 °

Pf= cos(18.43°) = 0.948

b.) pf to 0.9, a capacitor is needed.

Pf = 0.9

Cos θ= 0.9

θ= 25.84 °

(WC) V^2= P (tan θ1 - tan θ2)

C= 2400 ( tan (36. 87°) - tan (25.84°)) /2 πf × 120^2

f=60, π=22/7

C= 117.5µf

7 0
3 years ago
Motorcycles are extremely hard to see if they are _______. powered by quiet motors approaching from the side driving on the shou
Assoli18 [71]

Answer: uve earned 5 b point for helping us

Explanation:well know I'm not helping u

7 0
3 years ago
What are the basic types of heat exchangers?
liraira [26]

Answer:

The two large divisions of heat exchangers are direct contact between fluids and indirect contact between fluids.

Explanation:

A heat exchanger is one of the most used equipment at the level of thermal installations, both at the building, tertiary and industrial levels. A heat exchanger is a device designed to transfer heat between two fluids. These two fluids (liquids, gases) can be in contact or separated by a solid barrier. Its use is basic in all types of air conditioning or refrigeration, air conditioning, energy transfer or chemical processes. Heat transmission occurs through convection and conduction.

Classifying heat exchange systems can be carried out using many different criteria. When classifying different types of heat exchangers, different criteria can be taken into account. Taking into account the degree of contact between the fluids, they are grouped into two different types:

Direct Contact Heat Exchanger:

In direct contact exchangers, heat transfer occurs through a physical mixture of the fluids involved in the process. An example of this type of exchangers are the cooling towers. In this case, direct contact occurs between a stream of hot water (fluid to be cooled) using dry and colder air.

Indirect Contact Heat Exchanger :

In a direct type exchanger there is no direct contact between the fluids and they never mix. The fluids are separated by a solid barrier and may also not coincide at the same time.

Indirect contact heat exchangers can be of various types, being the most used, according to their constructive typology:

  • Concentric tubes or double tube .
  • Shell and tubes .
  • Of plates .
  • Compact heat exchangers .
  • Regenerators .

The concentric tube equipments are the simplest that exist since they are composed of two concentric tubes of different diameter so that one of the fluids circulates inside the smaller one and the other does it through the annular space between both tubes.

Shell and tube exchangers are widely used at an industrial level and use a housing with a multitude of tubes inside.

The equipment of plates are formed by a succession of sheets of metal, armed in a frame and separated by joints, which are fixed with a steel shell. The fluid circulates between these sheets.

5 0
4 years ago
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