6. Question
Which statements are true about routers? Check all that apply.
A router can only send data to another computer that's on the same network.
A router is a set of components that makes up computer networking.
A router connects devices together and helps direct network traffic.
A router utilizes network protocols to help determine where to send data packets.
7.
Question

Answer:

Automation generally means “a process performed without human assistance”, while autonomy implies “satisfactory performance under significant uncertainties in the environment and the ability to compensate for system failures without external intervention [emphasis mine].”

Explanation:

Answer:

False

Explanation:

False, though they could do it.

It's most likely and Industrial Engineer, it could also be a Chemical Engineer.

Answer:

I hope this helps

Explanation:

Instrmentation engineering

Answer:

Instrumentation engineering

Explanation:

Answer:

The wind turbine generates [tex]19297.222[/tex] kilowatt-hours of electricity daily.

The wind turbine makes a daily revenue of 1736.75 US dollars.

Explanation:

First, we have to determine the stored energy of wind ([tex]E_{wind}[/tex]), measured in Joules, by means of definition of Kinetic Energy:

[tex]E_{wind} = \frac{1}{2}\cdot \dot m_{wind}\cdot \Delta t \cdot v_{wind}^{2}[/tex] (Eq. 1)

Where:

[tex]\dot m_{wind}[/tex] - Mass flow of wind, measured in kilograms per second.

[tex]\Delta t[/tex] - Time in which wind acts in a day, measured in seconds.

[tex]v_{wind}[/tex] - Steady wind speed, measured in meters per second.

By assuming constant mass flow and volume flows and using definitions of mass and volume flows, we expand the expression above:

[tex]E_{wind} = \frac{1}{2}\cdot \rho_{air}\cdot \dot V_{air} \cdot \Delta t \cdot v_{wind}^{2}[/tex] (Eq. 1b)

Where:

[tex]\rho_{air}[/tex] - Density of air, measured in kilograms per cubic meter.

[tex]\dot V_{air}[/tex] - Volume flow of air through wind turbine, measured in cubic meters per second.

[tex]E_{wind} = \frac{1}{2}\cdot \rho_{air}\cdot A_{c}\cdot \Delta t\cdot v_{wind}^{3}[/tex] (Eq. 2)

Where [tex]A_{c}[/tex] is the area of the wind flow crossing the turbine, measured in square meters. This area is determined by the following equation:

[tex]A_{c} = \frac{\pi}{4}\cdot D^{2}[/tex] (Eq. 3)

Where [tex]D[/tex] is the diameter of the wind turbine blade, measured in meters.

If we know that [tex]\rho_{air} = 1.25\,\frac{kg}{m^{3}}[/tex], [tex]D = 100\,m[/tex], [tex]\Delta t = 86400\,s[/tex] and [tex]v_{wind} = 8\,\frac{m}{s}[/tex], the stored energy of the wind in a day is:

[tex]A_{c} = \frac{\pi}{4}\cdot (100\,m)^{2}[/tex]

[tex]A_{c} \approx 7853.982\,m^{2}[/tex]

[tex]E_{wind} = \frac{1}{2}\cdot \left(1.25\,\frac{kg}{m^{3}} \right) \cdot (7853.982\,m^{2})\cdot (86400\,s)\cdot \left(8\,\frac{m}{s} \right)^{3}[/tex]

[tex]E_{wind} = 2.171\times 10^{11}\,J[/tex]

Now, we proceed to determine the quantity of energy from wind being used by the wind turbine in a day ([tex]E_{turbine}[/tex]), measured in joules, with the help of the definition of efficiency:

[tex]E_{turbine} = \eta\cdot E_{wind}[/tex] (Eq. 4)

Where [tex]\eta[/tex] is the overall efficiency of the wind turbine, dimensionless.

If we get that [tex]E_{wind} = 2.171\times 10^{11}\,J[/tex] and [tex]\eta = 0.32[/tex], then the energy is:

[tex]E_{turbine} = 0.32\cdot (2.171\times 10^{11}\,J)[/tex]

[tex]E_{turbine} = 6.947\times 10^{10}\,J[/tex]

The wind turbine generates [tex]6.947\times 10^{10}[/tex] joules of electricity daily.

A kilowatt-hours equals 3.6 million joules. We calculate the equivalent amount of energy generated by wind turbine in kilowatt-hours:

[tex]E_{turbine} = 6.947\times 10^{10}\,J\times\frac{1\,kWh}{3.6\times 10^{6}\,J}[/tex]

[tex]E_{turbine} = 19297.222\,kWh[/tex]

The wind turbine generates [tex]19297.222[/tex] kilowatt-hours of electricity daily.

Lastly, the revenue generated per day can be found by employing the following:

[tex]C_{rev} = c\cdot E_{turbine}[/tex] (Eq. 5)

Where:

[tex]c[/tex] - Unit price, measured in US dollars per kilowatt-hour.

[tex]C_{rev}[/tex] - Revenue generated by the wind turbine in a day, measured in US dollars.

If we know that [tex]c = 0.09\,\frac{USD}{kWh}[/tex] and [tex]E_{turbine} = 19297.222\,kWh[/tex], then the revenue is:

[tex]C_{rev} = \left(0.09\,\frac{USD}{kWh} \right)\cdot (19297.222\,kWh)[/tex]

[tex]C_{rev} = 1736.75\,USD[/tex]

The wind turbine makes a daily revenue of 1736.75 US dollars.

Answer:

  its requirement of labor or materials from far-flung sources

Explanation:

"Globalization" generally means having or creating sources of supply in foreign countries, usually those with lower labor or material costs. If some industry sector is dependent upon that, then it's probably because it depends on labor or materials from low-cost sources.

Answer:

B) voltage at the sending end of the feeder = 2483.66 v

Explanation:

attached below is the the equivalent circuits and the remaining solution  for option A

B) voltage = 2400 v

   I  = [tex]\frac{50*10^3}{2400}[/tex]  =  20.83 A

calculate voltage at sending end ( Vs )

Vs = 2400 +  20.83 ∠ -cos^-1 (0.8) ( 0.75*2 + 0.5 + j 2 + j2 )

hence Vs = 2483.66 ∠ 0.961

therefore voltage at the sending end = 2483.66 v

Answer:

Not sure

Explanation:

Looking now

Answer:

The correct option is B) Balance Sheet

Explanation:

A Balance Sheet offers a description of a company's obligations, assets, and investments as well as net income over a given span of time such as a period of 6 months or 12 months, for instance.

Also known as the Statement of Financial Position, it contains sufficient information for investors and business owners to determine the company's financial performance in that period as well as to compare the performance of that company with industry norms or competition.

Cheers

Answer:

The answer "K = 0.0075"

Explanation:

If we try to measure up to 69 kPa of air, find mercury or fluid for gauge.  

While mercury was its largest liquid with a density of 13600 kg / m3 at normal room temperature.  

Let's all measure for 69 kPa that height of the  mercury liquid column.

[tex]\to P = 69 \ kPa[/tex]

       [tex]= 69000 Pa \\\\[/tex]

[tex]\to \rho = 13600 \ \ \frac{kg}{m^3} \\\\\\to g = 9.81 \ \ \frac{m}{s^2} \\\\[/tex]

Formula:

[tex]\to P=\rho \ gh[/tex]

[tex]\to 69000 = 13600\times9.81 \times h\\\\\to h= \frac{69000}{13600\times9.81} \\\\\to h= \frac{69000}{133416} \\\\\to h= 0.517179349 \\\\ \to h= 517 \ mm \\\\[/tex]

The right choice for pressure measurements up to 69 kPa is mercury.  

Atmospheric Mercury up to 69 kPa Air 517 mm  

The relationship of Hg to Pa is = 134.22 Pa 1 mm Hg  

Static sensitivity to Pa of mm hg = change of mercury height to Pa:

[tex]= \frac{\Delta Hg }{ \Delta P }\\\\= \frac{1 }{ 133.3 }\\\\= 0.0075[/tex]

Answer:

Mechanic: a person who repairs and maintains machinery

Mechanical engineers: design power-producing machines

Explanation:

Answer:

10

Explanation: