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TiliK225 [7]
4 years ago
9

According to the article "Edward R. Murrow: Inventing Broadcast Journalism," how did Murrow perceive the threat of Adolf Hitler?

Engineering
1 answer:
aliya0001 [1]4 years ago
8 0

Answer:

d. To Murrow, Hitler was a threat to all of civilization

Explanation:

For Murrow, Hitler's rise was a serious problem and a major threat to the entire civilization. For this reason, he believed that covering news about Hitler's advance and the battles of Nazi Germany was essential, even if the American population did not see Hitler as a threat and the European population, believed that these reprotations were only a way to denigrate the image of the continent to the world.

For Murrow reporting on Hitler's actions was as important as reporting on natural disasters.

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3 MAOP stands for which of the following?
raketka [301]

Answer:

MAOP Master of Arts in Organizational Psychology (various universities)

MAOP Maximum Allowable Operating Pressure

MAOP Mid-Atlantic Association of Oracle Professionals

MAOP Mid Atlantic Oncology Program

MAOP Maryland Association of Osteopathic Physicians

MAOP Master Air Operations Planner

MAOP Meharry Association of Office Personnel

MAOP Managers' Annual Operating Plan

MAOP Military Assistance Officer Program (US DoD)

Explanation:

3 0
3 years ago
can anyone help me with this please.i have the current and pf for branch 1 and 2 but cant figure out the total current, pf and a
anyanavicka [17]

Answer:

  • branch 1: i = 25.440∠-32.005°; pf = 0.848 lagging
  • branch 2: i = 21.466∠63.435°; pf = 0.447 leading
  • total: i = 31.693∠10.392° leading; pf = 0.984 leading

Explanation:

To calculate the currents in the parallel branches, we need to know the impedance of each branch. That will be the sum of the resistance and reactance.

The inductive reactance is ...

  X_L=j\omega L=j2\pi fL=j100\pi\cdot 15.915\times10^{-3}\approx j4.99984\,\Omega

The capacitive reactance is ...

  X_C=\dfrac{1}{j\omega C}=\dfrac{-j}{100\pi\cdot 318.31\times10^{-6}F}\approx -j10.00000\,\Omega

<u>Branch 1</u>

The impedance of branch 1 is ...

  Z1 = 8 +j4.99984 Ω

so the current is ...

  I1 = V/Z = 240/(8 +j4.99984) ≈ 25.440∠-32.005°

The power factor is cos(-32.005°) ≈ 0.848 (lagging)

<u>Branch 2</u>

The impedance of branch 2 is ...

  Z2 = 5 -j10 Ω

so the current is ...

  I2 = 240/(5 +j10) ≈ 21.466∠63.435°

The power factor is cos(63.436°) ≈ 0.447 (leading)

<u>Total current</u>

The total current is the sum of the branch currents. A suitable calculator can add these vectors without first converting them to rectangular form.

  It = I1 +I2 = (21.573 -j13.483) +(9.6 +j19.2)

  It ≈ 31.173 +j5.717 ≈ 31.693∠10.392°

The power factor for the circuit is cos(10.392°) ≈ 0.984 (leading)

__

The phasor diagram of the currents is attached.

_____

<em>Additional comment</em>

Given two vectors, their sum can be computed several ways. One way to compute the sum is to use the Law of Cosines. In this application, the angle between the vectors is the supplement of the difference of the vector angles: 84.560°.

3 0
2 years ago
Pick a subjectarea/field/topic that you are interested in. For each of the following Bonham- Carver uses of GIS give an example
Vanyuwa [196]

Answer:

I hope following attachment will help you a lot!

Explanation:

3 0
3 years ago
How to design a solar panel<br>​
artcher [175]

Answer:

#1) Find out how much power you need

#2 Calculate the amount of batteries you need.

#3 Calculate the number of solar panels needed for your location and time of year.

#4 Select a solar charge controller.

#5 Select an inverter.

#6 Balance of system

Explanation: To design solar panel, consider the following steps

1.) Find the power consumption demands

The first step in designing a solar PV system is to find out the total power and energy consumption of all loads that need to be supplied by the solar PV system as follows:

Calculate total Watt-hours per day for each appliance used.

 Add the Watt-hours needed for all appliances together to get the total Watt-hours per day which must be delivered to the appliances.

Calculate total Watt-hours per day needed from the PV modules.

Multiply the total appliances Watt-hours per day times 1.3 (the energy lost in the system) to get the total Watt-hours per day which must be provided by the panels.

2. Size the PV modules

Different size of PV modules will produce different amount of power. To find out the sizing of PV module, the total peak watt produced needs. The peak watt (Wp) produced depends on size of the PV module and climate of site location. We have to consider panel generation factor which is different in each site location. For Thailand, the panel generation factor is 3.43. To determine the sizing of PV modules, calculate as follows:

2.1 Calculate the total Watt-peak rating needed for PV modules

Divide the total Watt-hours per day needed from the PV modules (from item 1.2) by 3.43 to get the total Watt-peak rating needed for the PV panels needed to operate the appliances.

Calculate the number of PV panels for the system

Divide the answer obtained in item 2.1 by the rated output Watt-peak of the PV modules available to you. Increase any fractional part of result to the next highest full number and that will be the 

number of PV modules required.

Result of the calculation is the minimum number of PV panels. If more PV modules are installed, the system will perform better and battery life will be improved. If fewer PV modules are used, the system may not work at all during cloudy periods and battery life will be shortened.

3. Inverter sizing

An inverter is used in the system where AC power output is needed. The input rating of the inverter should never be lower than the total watt of appliances. The inverter must have the same nominal voltage as your battery.

For stand-alone systems, the inverter must be large enough to handle the total amount of Watts you will be using at one time. The inverter size should be 25-30% bigger than total Watts of appliances. In case of appliance type is motor or compressor then inverter size should be minimum 3 times the capacity of those appliances and must be added to the inverter capacity to handle surge current during starting.

For grid tie systems or grid connected systems, the input rating of the inverter should be same as PV array rating to allow for safe and efficient operation.

4. Battery sizing

The battery type recommended for using in solar PV system is deep cycle battery. Deep cycle battery is specifically designed for to be discharged to low energy level and rapid recharged or cycle charged and discharged day after day for years. The battery should be large enough to store sufficient energy to operate the appliances at night and cloudy days. To find out the size of battery, calculate as follows:

     4.1 Calculate total Watt-hours per day used by appliances.

     4.2 Divide the total Watt-hours per day used by 0.85 for battery loss.

     4.3 Divide the answer obtained in item 4.2 by 0.6 for depth of discharge.

     4.4 Divide the answer obtained in item 4.3 by the nominal battery voltage.

     4.5 Multiply the answer obtained in item 4.4 with days of autonomy (the number of days that you need the system to operate when there is no power produced by PV panels) to get the required Ampere-hour capacity of deep-cycle battery.

Battery Capacity (Ah) = Total Watt-hours per day used by appliancesx Days of autonomy

(0.85 x 0.6 x nominal battery voltage)

5. Solar charge controller sizing

The solar charge controller is typically rated against Amperage and Voltage capacities. Select the solar charge controller to match the voltage of PV array and batteries and then identify which type of solar charge controller is right for your application. Make sure that solar charge controller has enough capacity to handle the current from PV array.

For the series charge controller type, the sizing of controller depends on the total PV input current which is delivered to the controller and also depends on PV panel configuration (series or parallel configuration).

According to standard practice, the sizing of solar charge controller is to take the short circuit current (Isc) of the PV array, and multiply it by 1.3

Solar charge controller rating = Total short circuit current of PV array x 1.3

5 0
3 years ago
Eutectic product in Fe-C system is called A. Pearlite B. Bainite C. Ledeburite D. Spheroidite
zimovet [89]

Answer:

Eutectic product in Fe-C system is called Ledeburite-C.

6 0
3 years ago
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