after coolant moves through the water jackets, it finds its way to the _______, where it returns to the radiator. group of answer choices radiator

Answer:

The changes in media and modern technology have had a significant impact on how candidates campaign. One of the most significant changes is the increasing use of social media by candidates to reach voters. This has made it easier for candidates to directly engage with voters and share their message without having to rely on traditional forms of media, such as television and print ads.

Additionally, the rise of online news sources and the ability for individuals to access information instantly on their phones has made it important for candidates to be able to clearly and concisely explain their positions in written articles. This has led to a greater emphasis on messaging and communication strategies in campaigns.

Overall, the changes in media and technology have made it easier for candidates to connect with voters and share their message, but have also raised the bar for how effectively they must do so in order to be successful.

Explanation:

i] The density of the pellet is 0.977 g/cm^{3}. ii] The macropore volume in cm^{3}/g is 0.205 cm^{3}/g. iii] The macropore void fraction in the pellet is 25.1%.iv] The micropore void fraction in the pellet is 49.0%. v] The solid fraction of the pellet is 25.9%. vi] The density of the particles is 1.222 g/cm^{3}.

i] To determine the density of the pellet, we can use the formula:

Density = Mass / Volume

Given that the mass of the pellet is 3.15 g and the volume is 3.22cm^{3}, we can calculate the density as follows:

Density = 3.15 g / 3.22 cm^{3}≈ 0.977 \(g/cm^{3\)

ii] The macropore volume in cm3/g can be calculated by dividing the macropore volume of the pellet (0.645 cm3) by the mass of the pellet (3.15 g):

Macropore volume = 0.645 cm^{3} / 3.15 g ≈ 0.205 \(cm^{3} /g\)

iii] The macropore void fraction in the pellet can be calculated using the formula:

Macropore void fraction = Macropore volume / Total volume of the pellet

Total volume of the pellet = Volume - Macropore volume = 3.22 cm^{3}- 0.645 cm^{3} = 2.575 cm^{3}

Macropore void fraction = 0.645 cm^{3} / 2.575 \(cm^{3}\)≈ 0.251 or 25.1%

iv] The micropore void fraction in the pellet can be calculated using the given micropore volume of the particles (0.40 cm^{3} /g) and the mass of the pellet (3.15 g):

Micropore volume in the pellet = Micropore volume/g x Mass

Micropore volume in the pellet = 0.40 \(cm^{3} /g\) x 3.15 g = 1.26 cm3

Micropore void fraction = Micropore volume in the pellet / Total volume of the pellet

Micropore void fraction = 1.26 \(cm^{3}\) / 2.575 \(cm^{3}\) ≈ 0.490 or 49.0%

v] The solid fraction of the pellet can be calculated by subtracting the sum of macropore and micropore void fractions from 1:

Solid fraction = 1 - (Macropore void fraction + Micropore void fraction)

Solid fraction = 1 - (0.251 + 0.490) ≈ 0.259 or 25.9%

vi] The density of the particles can be determined using the mass of the pellet (3.15 g) and the total volume of the particles:

Total volume of the particles = Volume - Macropore volume = 3.22 \(cm^{3}\)- 0.645 \(cm^{3}\) = 2.575\(cm^{3}\)

Density of the particles = Mass / Total volume of the particles

Density of the particles = 3.15 g / 2.575\(cm^{3}\) ≈ 1.222 \(g/cm^{3}\)

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The true power is obtained as 70.7 VA.

We define the power a the rate of doing work, we know that the power in a circuit is the product of the current and the voltage. In this case, we want to find the true power thus we have to involve the use of the phase shift in degrees.

Thus;

True power = PcosΦ

P =  100 VA

Φ = 45 degrees

True power =  100 VA * cos 45 degrees

True power = 70.7 VA

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Answer:

i think 1 is c and im so sorry but i dont know whats the rest

Explanation:

To perform the tasks mentioned in the lab, follow the steps below:

On CorpDC:

Log in to the CorpDC server.

Open the Active Directory Users and Computers management console.

Navigate to the Engineering OU.

Create a new group with the following specifications:

Group name: Lead-Engineers

Group members: Emma Tomco and Treg Reese

On CorpServer:

Log in to the CorpServer.

Open the Hyper-V Manager.

Locate the CorpDC2 virtual machine.

Move the storage for the CorpDC2 virtual machine to the H:\HYPERV2 location.

On CorpCluster1:

Log in to the CorpCluster1 server.

Open the Server Manager.

Add the File Server role to the CorpCluster1 server.

Specify the appropriate file server type for virtual machines with extended file open periods.

Name the clustered role as CorpApps.

On CorpCluster1 and CorpCluster2:

Log in to both CorpCluster1 and CorpCluster2 servers.

Open the Failover Cluster Manager.

Configure the cluster quorum settings for CorpCluster using the following specifications:

Add a file share witness.

Use the file share path: \CorpServer\Witness.

Note: Ensure that CorpCluster1 and CorpCluster2 servers are physically located in the Networking Closet of Building A.

By following these steps, you will complete the specified tasks in the lab.

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Create a group named "Lead-Engineers" in the Engineering OU of the BizCorp domain on CorpDC, with members Emma Tomco and Treg Reese. Move CorpDC2's storage to H:\HYPERV2 on CorpServer. Add the File Server role to CorpCluster1, naming it "CorpApps." Configure CorpCluster's quorum settings on CorpCluster1 with a file share witness at I\CorpServer\Witness.corpCluster1.

To perform the tasks outlined in the lab, follow these steps:

1. On CorpDC, create a group within the Engineering OU of the BizCorp domain. Name the group "Lead-Engineers" and add the members Emma Tomco and Treg Reese.

2. On CorpServer, move the storage for the CorpDC2 virtual machine to the H:\HYPERV2 location.

3. On CorpCluster1, add the File Server role to the CorpCluster. Both CorpCluster1 and CorpCluster2 servers are part of the CorpCluster cluster. Name the clustered role "CorpApps."

4. On CorpCluster1, configure the cluster quorum settings for CorpCluster using the following specifications:

  - Add a file share witness.

  - Set the file share path as I\CorpServer\Witness.corpCluster1.

Note: CorpCluster1 and CorpCluster2 are located in the Networking Closet of Building A.

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Answer:

1964

Explanation:

It was discovered in 1964 when a pair of Geiger counters were carried on board a sub-orbital rocket launched from New Mexico.

Answer:

Explanation:

Temperature of atmospheric air To = 25°C = 298 K

Free  stream velocity of air Vo = 25 m/s

Length and width of plate = 1m

Temperature of plate Tp = 125°C = 398 K

We know for air, Prandtl number Pr = 1

And for air, thermal conductivity K = 24.1×10?³ W/mK

Here, charectorestic dimension D = 1m

Given value of Reynolds number Re = 105

For laminar boundary layer flow over flat plate

= 3.402

Therefore, hx = 0.08199 W/m²K

So, heat transfer rate q = hx×A×(Tp – To)

                                          = 0.08199×1×(398 – 298)

Answer:

25 psig

Explanation:

The net oil pressure = 85 - 60 = 25 psig

Answer:

i play it

Explanation:

The idea of creating computers with human-level intelligence has been a topic of discussion for a long time.

While it's an exciting prospect, it's also important to consider the areas where we should push to accelerate the building of such computers.
One area where we should focus on accelerating the building of such computers is the medical field. With the help of these computers, doctors can diagnose diseases more accurately and efficiently, and even predict future health issues. Additionally, these computers can analyze medical data faster, which could lead to the development of new drugs and treatments.
Another area where we can push for the development of human-level intelligent computers is the field of engineering. These computers can simulate complex structures and designs, leading to the creation of better and more efficient machines.
However, there are also certain areas where we should avoid building such computers. For example, creating autonomous weapons or robots with human-level intelligence can have disastrous consequences. Such weapons or robots could make decisions that could harm humans, which is not something we should take lightly.
In conclusion, while the development of computers with human-level intelligence is an exciting prospect, it's important to focus on the areas where they can be used to improve human lives. At the same time, we must be cautious about the potential risks associated with their development in certain areas.

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Answer:

False

Explanation:

In this problem, we are given the weight of soil in its natural state and after being dried, as well as the specific gravity of the soil. From this information, we can determine various properties such as degree of saturation, void ratio, porosity, and water content.

Additionally, we can calculate the volume of water required to saturate the soil in its natural state.

a) In the phase diagram, we have:

Wt (total weight) = 113 lbs

Ww (weight of water) = 113 - 96 = 17 lbs

Ws (weight of solids) = 96 lbs

Vt (total volume) = Vv (volume of voids) + Vs (volume of solids)

Vv is the volume of water, and Vs is the volume of solids.

We need additional information to determine the values of Vv and Vs.

b) To determine the degree of saturation (S), we use the equation:

S = (Ww / Ws) * 100

Substituting the given values, we can calculate the degree of saturation.

The void ratio (e) is the ratio of the volume of voids to the volume of solids. It can be calculated as:

e = Vv / Vs

The porosity (n) is the ratio of the volume of voids to the total volume. It can be calculated as:

n = Vv / Vt

The water content (ω) is the ratio of the weight of water to the weight of solids. It can be calculated as:

ω = (Ww / Ws) * 100

Using the given information, we can calculate the values of S, e, n, and ω.

c) The volume of water required to saturate 1 ft3 of soil in its natural state can be determined by multiplying the volume of voids (Vv) by the specific gravity (Gs) of the soil. Since the specific gravity is given as 2.70, we can calculate the volume of water required.

By applying the relevant equations and using the given information, we can determine the desired properties and volume of water required for the soil in its natural state.

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The skin surface temperature is 307.2 K and the  rate of heat loss is 146 W.

The skin surface temperature may be obtained by performing an energy balance at the skin surface to give the formula;

k(T_i - T_s)/L = h(T_s - T_∞) + h_r(T⁴_s - T⁴_surr)

Now, T_surround = T_s and so making T_s the subject gives us;

T_s = [(k*T_i)/L + (h + h_r)T_∞]/((k/l)

Using the equation h_r = εσ(T_s + T_surr)(T²_s + T²_surr) we can find h_r with a guessed value of T_s = 305 K and T∞ = 297 K, to yield h_r = 5.9 W/m².K.

Then, substituting numerical values into the T_s equation, we have;

T_s = [(0.3 * 308)/(3 * 10⁻³) + (2 + 5.9)297]/[((0.3 * 308)/(3 * 10⁻³)) + (2 + 5.9)]

T_s = 307.2 K

Thus the skin temperature is 307.2 K.

The rate of heat loss can be found by evaluating the conduction through the skin/fat layer:

q_s = kA(T_i - T_s)/L

q_s = 0.3*1.8(308 - 307.2)/(3 * 10⁻³)

q_s = 146 W

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Answer:

7.7 kN

Explanation:

The capacity of a material having a crack to withstand fracture is referred to as fracture toughness.

It can be expressed by using the formula:

\(K = \sigma Y \sqrt{\pi a}\)

where;

fracture toughness K = 137 MPa\(m^{1/2}\)

geometry factor Y = 1

applied stress \(\sigma\) = ???

crack length a = 2mm = 0.002

∴

\(137 =\sigma \times 1 \sqrt{ \pi \times 0.002 }\)

\(137 =\sigma \times 0.07926\)

\(\dfrac{137}{0.07926} =\sigma\)

\(\sigma = 1728.489 MPa\)

Now, the tensile impact obtained is:

\(\sigma = \dfrac{P}{A}\)

P = A × σ

P = 1728.289 × 4.5

P = 7777.30 N

P = 7.7 kN

The performance of a first-order system is characterized by the time constant, which determines the speed of the system's response to changes in input. The location of the poles in the s-plane affects the transient and steady-state responses, as well as the frequency and damping characteristics of the system.

For second-order systems, there are five performance specifications: natural frequency, damping ratio, rise time, settling time, and overshoot or undershoot. These specifications describe the system's frequency and damping characteristics, as well as its transient response to changes in input.

Explanation:

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Answer:

by hìuuf5ëcz

Explanation:

6tgiïuggd

Answer:

The net work done is 272.38 kJ

Explanation:

The parameters given are;

Mass of water = 5 kg

p₁ = 20 bar

T₁ = 360°C

v₁ = 0.141147 m³/kg  

Process 1 to 2 = Constant pressure process

p₂ = 20 bar

Process 2 to 3 = Constant volume process

p₃ = 5 bar

Process 3 to 4 = Constant pressure process

Process 4 to 1 = Polytropic process pv = Constant

For Stage 1 to 2, we have;

p₂ = 20 bar

From the steam tables for superheated steam, we have;

T₂ = 212.385°C

v₂ = 0.0995805 m³/kg

Work done = p₂×(v₂ - v₁) = 2×10⁶ × (0.0995805 - 0.141147 ) = -83133 J/kg

For the 5 kg, we have;

\(W_{1-2}\) = -83133 J/kg × 5 = -415,665 J

Stage 2 to 3: Constant volume cooling

v₂ = v₃ = 0.0995805 m³/kg

p₃ = 5 bar

T₃ = 151.836°C

(0.0995805 - 0.00109256)/(0.374804 - 0.00109256) = 0.2635 liquid vapor mixture

Work done, \(W_{2-3}\) = 0

Stage 3 to 4: Constant pressure heating

p₃ = p₄ = 5 bar

v₄/T₄ = v₃/T₃

v₄ =  0.374804 m³/kg

T₄ = v₄×T₃/v₃ = 0.374804*(273.15 + 151.836)/0.0995805 = 1599.6 K = 1326.4 °C

Work done = p₄×(v₄ - v₃) = 5×10⁵ × (0.374804  - 0.0995805 ) = 137611.75 J/kg

For the 5 kg, we have;

\(W_{3-4}\) = 137,611.75  J/kg × 5 = 688,058.75 J

Stage 4 to 1: Polytropic process    

\(\dfrac{p_{4}}{p_{1}} = \left (\dfrac{V_{1}}{V_{4}} \right )^{n} = \left (\dfrac{T_{4}}{T_{1}} \right )^{\dfrac{n}{n-1}}\)

Which gives;

\(\dfrac{5}{20} = \left (\dfrac{0.141147 }{0.374804} \right )^{n}\)

n = log(5/20) ÷log(0.141147/0.374804) = 1.42

Work done, \(W_{pdv}\), is given as follows;

\(W_{pdv} = \dfrac{p_4 \times v_4 -p_4 \times v_4 }{n-1}\)

Which gives;

\(W_{pdv} = \dfrac{5\times 0.374804 -20\times 0.141147 }{1.42-1} = -2.259 \, J\)

For the 5 kg, we have;

\(W_{4-1}\) = -2.259 J/kg × 5 = -11.2967 J

The net work done, \(W_{Net}\), is therefore;

\(W_{Net}\) = \(W_{1-2}\)  + \(W_{3-4}\) + \(W_{4-1}\)

-415,665  + 688,058.75 -11.2967 = 272,382.45 J = 272.38 kJ.

A system contains two components A and B that are connected in series. The two components are required to function together in order for the system to work. Let’s assume that A and B work independently from each other. The probability that component A will function is 0.8, while the probability that component B will function is 0.9. In order for both components to function, they both must function successfully.

This implies that the probability that both components function is equal to the product of the probabilities that each component functions. Here's how it works:P(A and B) = P(A) × P(B) = 0.8 × 0.9 = 0.72.Now that we know the probability that both components will work is 0.72, let's double-check. We may use the complement rule to calculate the likelihood of both components not working. Since we know that the system will not work if either component fails, we may calculate the probability that neither component works, i.e., P(A’ and B’). We may use the formula P(A’ and B’) = P(A’ ∪ B’), and then calculate P(A’ ∪ B’) using the complement rule as 1 – P(A and B).P(A' ∪ B') = 1 - P(A and B) = 1 - 0.72 = 0.28.Therefore, the probability that the system will not work is 0.28, and the probability that it will work is 1 – 0.28 = 0.72.

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If the input frequency to a 6-bit counter is 10 MHz The frequency at the output terminals is 5 MHz.

The output frequency depends on which output terminal is being considered. Given that the counter is 6-bit counter, the number of output terminals is 6. The output frequency can be calculated by considering the input frequency and the modulus of the counter. Divide input frequency by the modulus of the counter to get the output frequency. So, for the 6-bit counter, the modulus is 2⁶ = 64.

Therefore, the output frequency at the following output terminals would be:

Q0 = 10 MHz / 64 = 156.25 kHzQ1 = 10 MHz / 32 = 312.5 kHzQ2 = 10 MHz / 16 = 625 kHzQ3 = 10 MHz / 8 = 1.25 MHzQ4 = 10 MHz / 4 = 2.5 MHzQ5 = 10 MHz / 2 = 5 MHz

The output frequency depends on which output terminal is being considered. Given that the counter is 6-bit counter, the number of output terminals is 6. The output frequency can be calculated by considering the input frequency and the modulus of the counter.

Divide input frequency by the modulus of the counter to get the output frequency. So, for the 6-bit counter, the modulus is 2⁶ = 64.

Therefore, the output frequency at the following output terminals would be:

Q0 = 10 MHz / 64 = 156.25 kHzQ1 = 10 MHz / 32 = 312.5 kHzQ2 = 10 MHz / 16 = 625 kHzQ3 = 10 MHz / 8 = 1.25 MHzQ4 = 10 MHz / 4 = 2.5 MHzQ5 = 10 MHz / 2 = 5 MHz

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Answer:

Weight Percentage of Ash = 3.4

Explanation:

Given - Coal containing 21% ash is completely combusted, and the ash is 100% removed in a water contact scrubber. If 10,000 kg of coal is burned per hour with a scrubber flow rate of 1.0 m3/min.

To find - the weight percentage of the ash in the water/ash stream leaving the scrubber is most nearly ?

Solution -

Given that,

Coal Burned Rate = 10,000 kg/hr

                              = \(\frac{10,000}{60 min} * 1 hr *\frac{kg}{hr}\)

                              = 166.6666 kg/min

⇒Coal Burned Rate = 166.6666 kg/min

Now,

Given that,

Ash content in coal = 21 %

⇒Ash in (coal that burned) = 166.6666 × \(\frac{21}{100}\) kg/min

                                             = 34.9999 ≈ 35 kg/min

⇒Ash in (coal that burned) = 35 kg/min

Now,

We know,

Density of water = 1000 kg/m³

Now,

Water flow Rate = \(1\frac{m^{3} }{min} * density\)

                           = 1000 kg/min

⇒Water flow Rate = 1000 kg/min

Now,

Total Mass flow Rate of (Water + Ash stream) = ( 1000 + 35) kg/min

                                                                           = 1035 kg/min

⇒Total Mass flow Rate of (Water + Ash stream) = 1035 kg/min

So,

Weight Percentage of Ash = (Weight of Ash ÷ Total weight of Stream) × 100

                                            = (35 ÷ 1035) × 100

                                            = 3.38 ≈ 3.4

∴ we get

Weight Percentage of Ash = 3.4

Answer:

Sole Proprietorship, General Partnership , Limited Partnership, corporation

Explanation:

Business in something that an individual or a group of people do for a living and produce products and services that benefits the society and the people. There are several entities that can be common in business. Some common form of entities are :

Sole Proprietorship : One to one

-- here there is only one owner in the business and he maintains and manages the entire business functions under his control.

Limited Partnership : many to one

-- here two or more than two partners establish business and runs it but only one or more is liable to the amount of the investments.

General Partnership : many to many

-- It is a business partnership where all the partners shares the profits, the assets, legal liabilities and financial liabilities, etc.

Corporation : many to one

It is a business entity where a group of individual or a group of companies run a single business which is generally authorized by the state.

Some of the typical types of the entities that one may think to be common in term of the business entities are about the relationships that are held within many to many and one to many.

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Answer:

E = 132.69 sin(ωt -11.56)

i(t) = 6.64 sin (ωt +48.44) A

Explanation:

given data

e1 = 80 sin ωt volts                       80 < 0

e2 = 60 sin (ωt + π/2) volts          60 < 90

e3 = 100 sin (ωt – π/3) volts        100 < -60

solution

resultant will be  = e2 + e2 + e3

E =  80 < 0 + 60 < 90 + 100 < -60

\(\bar E\) = 80 + j60 + 50 - j50\(\sqrt{3}\)

\(\bar E\) = 130 + (-j26.60)

\(\bar E\) = 132.69  that is less than -11.56

so

E = 132.69 sin(ωt -11.56)

and

as we have given the impedance

z = (10-j17.3)Ω

z = 19.982 < -60

and

i(t) = \(\frac{132.69}{19.982}\) sin(ωt -11.56 + 60)  

i(t) = 6.64 sin (ωt +48.44) A

Answer:

HF I am writing this email with your company is it a try and make them easier for us and we will need a new job and then email

Formation damage control is crucial in petroleum engineering to maintain well productivity. It involves mitigating factors that hinder fluid flow, such as solids deposition, clay swelling, and fluid incompatibility.

Formation damage control is essential in petroleum engineering to prevent or minimize the impairment of well productivity caused by various mechanisms. It involves identifying and addressing factors that can hinder fluid flow within the reservoir formation. Formation damage can occur due to factors such as solids deposition, clay swelling, emulsion formation, fluid incompatibility, and organic/inorganic scale precipitation. These issues can restrict the flow of hydrocarbons from the reservoir to the wellbore, reducing production rates and overall recovery. Basic principles of formation damage control include implementing effective drilling and completion practices, utilizing suitable drilling fluids and additives, optimizing well stimulation techniques, and conducting thorough reservoir characterization and analysis. Proper formation damage control strategies aim to maximize reservoir permeability, minimize the impact of drilling and production operations on the formation, and enhance overall well productivity and hydrocarbon recovery.

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Answer:

  0 ft

Explanation:

The equation of the plane can be found from the cross product AC×BC. That vector is ...

  N = (2, 11, -6) × (-3, -6, 4) = (8, 10, 21)

Then the equation of the plane is ...

  8x +10y +21z = 14 . . . . . 14 = N·A

Point P satisfies this equation, so is on the plane. The distance is 0 feet.

  8(9) +10(11) -8(21) = 72 +110 -168 = 14

Answer:

b. satellite surveying

c. aerial surveying

d. optical tooling

e. marketing surveying

f. control surveying

h. statistical surveying

i. telephone surveying

k. alignment surveying

l. mine surveying

m. solar surveying

Explanation:

A Survey is an act of examination of the features of a subject or material under consideration. Land surveying refers to the examination of the natural and man-made features of a piece of land using scientific and mathematical methods.

Land surveying finds application in construction where a survey is made on all the structures found in a constructed property. Topographic surveying deals with examining the natural and man-made feature of a piece of land. As-built survey as the name implies examines the features and location of a building during or recently after construction. These three are examples of land surveys.

Some type of surveying other than land surveying includes satellite, aerial, optical tooling, marketing, control surveying, statistical, telephone, alignment, mine, and solar surveying

A Survey means an examination of a features, subject or material under consideration.

Land surveying refers to the examination of the natural and man-made features of a piece of land using scientific and mathematical methods.

However, some other type of surveying other than land surveying includes satellite, aerial, optical tooling, marketing, control surveying, statistical, telephone, alignment, mine, and solar surveying.

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The highest-paid Biomedical Engineer is biomedical engineers in the United States.

Biomedicаl engineering is а discipline thаt аdvаnces knowledge in engineering, biology аnd medicine, аnd improves humаn heаlth through cross-disciplinаry аctivities thаt integrаte the engineering sciences with biomedicаl sciences аnd clinicаl prаctice.

Аccording to PаyScаle, biomedicаl engineers in the United Stаtes with less thаn five yeаrs of experience mаke аn аverаge of $62,000 per yeаr. However, U.S. bаsed biomedicаl engineers with five to 10 yeаrs of experience mаy mаke neаrly 31 percent more with аverаge eаrnings between $80,000 to $81,000 per yeаr.

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Answer: Also known as an Ethernet adapter, a network interface card (NIC) is a chip or circuit board that generally comes preinstalled on a computer or device.

Explanation: lmk if im right