Answer:
Not sure
Explanation:
Looking now
Create an array of 10 size and assign 10 random numbers. Now find the sum of the array using for and while loop.
Answer:
10
Explanation:
A single phase, 50-KVA, 2400-240-volt, 60 Hz distribution transformer has the following parameters: •
Resistance of the 2400-volt winding Ri=0.75 ohm
Resistance of the 240-volt winding R=0.0075 ohm
Leakage reactance of the 2400-volt winding X:=1 ohm
Leakage reactance of the 240-volt winding X2=0.01 ohm
Core loss resistance on the 2400 side Rc=733.5 ohms
Magnetizing reactance on the 2400 side X=4890 ohms
a. Draw the equivalent circuit referred to the high voltage side and referred to the low- voltage side. Label the impedances numerically.
b. The transformer is used as a step-down transformer at the load end of a feeder having impedance of (0.5+j2.0) ohms. Determine the voltage V. at the ending end of the feeder when the transformer delivers rated load at rated secondary voltage and 0.8 laggin power factor. Neglect the "exciting current" l. of the transformer (this is the current into the parallel Rc/jXm combination), which is another way of saying that you should use what we called in class the "approximate circuit #2).
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
Compute the number of kilograms of hydrogen that pass per hour through a 5-mm-thick sheet of palladium having an area of 0.20 m^2 at 500°C. Assume a diffusion coefficient of 1.0 x 10^8 m^2 /s, that the concentrations at the high- and low-pressure sides of the plate are 2.4 and 0.6 kg of hydrogen per cubic meter of palladium, and that steady-state conditions have been attained.
Answer:
The answer is "[tex]\bold{ 259.2 \times 10^{11} }[/tex]".
Explanation:
The amount of kilograms, which travel in a thick sheet of hydrogen:
[tex]M= -DAt \frac{\Delta C}{ \Delta x} \\\\[/tex]
[tex]D =1.0 \times 10^{8} \ \ \ \frac{m^2}{s} \\\\ A = 0.20 \ m^2\\\\t = 1\ \ h = 3600 \ \ sec \\\\[/tex]
calculating the value of [tex]\Delta C:[/tex]
[tex]\Delta C =C_A -C_B[/tex]
[tex]= 2.4 - 0.6 \\\\ = 1.8 \ \ \frac{kg}{m^3}[/tex]
calculating the value of [tex]\Delta X:[/tex]
[tex]\Delta x = x_{A} -x_{B}[/tex]
[tex]= 0 - (5\ mm) \\\\ = - 5 \ \ mm\\\\= - 5 \times 10^{-3} \ m[/tex]
[tex]M = -(1.0 \times 10^{8} \times 0.20 \times 3600 \times (\frac{1.8}{-5 \times 10^{-3}})) \\\\[/tex]
[tex]= -(1.0 \times 10^{8} \times 720 \times (\frac{1.8}{-5 \times 10^{-3}})) \\\\= -(1.0 \times 10^{8} \times \frac{ 1296}{-5 \times 10^{-3}})) \\\\= (1.0 \times 10^{8} \times 259.2 \times 10^3)) \\\\= 259.2 \times 10^{11} \\\\[/tex]
What makes a particular sector of the residential construction market dependent on globalization?
its need for innovation to spur growth
its embrace of renewable and sustainable technology
its commitment to multiculturalism
its requirement of labor or materials from far-flung sources
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.
Mechanic... Mechanical Engineer... What's the difference?
Instructions: Answer the question below with at least TWO complete sentences.
Answer:
Mechanic: a person who repairs and maintains machinery
Mechanical engineers: design power-producing machines
Explanation:
Design a U-tube manometer that can measure gage pressures up to 69 kPa of air. You will want to choose a manometer fluid with good static sensitivity but will not result in an unreasonably tall manometer. Further, the manometer fluid should be mostly immiscible with the air. The two design parameters you should consider are manometer fluid (impacts manometer fluid density) as well as the manometer height.
Required:
Compute the static sensitivity, K, in mmHg/Pa
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]
technician A says that in any circuit, electrical current takes the path of least resistance. technician B says that while this is true in a series circuit, it's not entirely true in a parallel circuit. who is correct?
Answer:
technician A is correct
Explanation:
Technician B has circuit topologies confused. In a series circuit, there is only one path for electrical current to take. In a parallel circuit, the current will divide between paths in proportion to the inverse of their resistance. The least resistance path will have the most current.
Technician A is mostly correct.
James is planning on registering for a course in electrical engineering. Which of the following sub-disciplines could he opt for?
Instrumentation engineering
Prosthetics
Plant design
Signal processing
Answer:
I hope this helps
Explanation:
Instrmentation engineering
Answer:
Instrumentation engineering
Explanation:
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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:
Check the 2nd, 3rd and 4th statements.
Explanation:
The router provides two basic functions traffic management between these networks through the transmission of data packets to the designated IP address, and the use of the same Internet connection by many devices.
A networking device is a router that transmits data packets between computer nets. Traffic control functions are executed on the internet via routers. Data sent via the internet are in data packets, for example, a web page or email.A router is a device connecting two or more packets or subnetworks. It is used to sets the components from the networking of your computer.Therefore, the final answer is "the second choice".
Learn more:
brainly.com/question/2162586
Compare automation and autonomous
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:
Martha has been running a small business for two years. She now seeks additional investment to finance her business. She has found potential investors who want to check how much Martha's business owns and owes others. Which financial statement should Martha provide to her investors? A. cash flow statement B. balance sheet C. trial balance D. journal
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
PLZ HELP A person hired by a pharmaceutical company to streamline the company’s drug production process would most likely be an electrical engineer? True or False
Answer:
False
Explanation:
False, though they could do it.
It's most likely and Industrial Engineer, it could also be a Chemical Engineer.
Large wind turbines with blade span diameters of over 100 m are available for electric power generation. Consider a wind turbine with a blade span diameter of 100 m installed at a site subjected to steady winds at 8 m/s. Taking the overall efficiency of the wind turbine to be 32 percent and the air density to be 1.25 kg/m3 , determine the electric power generated by this wind turbine. Also, assuming steady winds of 8 m/s during a 24-hour period, determine the amount of electric energy and the revenue generated per day for a unit price of $0.09/kWh for electricity
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.