Two balanced Y-connected loads in parallel, one drawing 15kW at 0.6 power factor lagging and the other drawing 10kVA at 0.8 power factor leading, are supplied by a balanced, three-phase, 480-volt source. (a) Draw the power triangle for each load and for the combined load. (b) Determine the power factor of the combined load and state whether lagging or leading. (c) Determine the magnitude of the line current from the source. (d) Δ-connected capacitors are now installed in parallel with the combined load. What value of capacitive reactance is needed in each leg of the A to make the source power factor unity? Give your answer in Ω. (e) Compute the magnitude of the current in each capacitor and the line current from the source.

Answer: See attachment below

Explanation:

To solve this problem we will apply the concepts related to real power in 3 phases, which is defined as the product between the phase voltage, the phase current and the power factor (Specifically given by the cosine of the phase angle). First we will find the phase voltage from the given voltage and proceed to find the current by clearing it from the previously mentioned formula. Our values are

[tex]V = 690V[/tex]

[tex]P_{real} = 2.3MW[/tex]

Real power in 3 phase

[tex]P_{real} = 3V_{ph}I_{ph} Cos\theta[/tex]

Now the Phase Voltage is,

[tex]V_{ph} = \frac{V}{\sqrt{3}}[/tex]

[tex]V_{ph} = \frac{690}{\sqrt{3}}[/tex]

[tex]V_{ph} = 398.37V[/tex]

The current phase would be,

[tex]P_{real} = 3V_{ph}I_{ph} Cos\theta[/tex]

Rearranging,

[tex]I_{ph}=\frac{P_{real}}{3V_{ph}Cos\theta}[/tex]

Replacing,

[tex]I_{ph}=\frac{2.3MW}{3( 398.37V)(0.85)}[/tex]

[tex]I_{ph}= 2.26kA/phase[/tex]

Therefore the current per phase is 2.26kA

You can use a series of commands to accomplish various tasks in GDB: 'print /x test' to view a variable in hex format, 'x/4wx $sp' to read the top four bytes on your stack by word, 'info registers' to check all register values, and 'break ece.c:391' to set a breakpoint at a specific line.

To achieve the tasks in GDB, you will need to use the following commands:

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The amount of water vapor present, we can use the relationship between specific volume, volume, and mass.

Specific volume= Volume/ Mass, (v) at a certain condition is 0.651 m³/kg and the volume (V) occupied by the water vapor is 1.2 m³, we can rearrange the formula to solve for mass (m):

Mass=  Specific volume/ Volume

           = 1.2/ 0.651

Mass= 1.2/ 0.651  

         = 1.844

Mass=1.844kg

Therefore,  the amount of water vapor present is approximately 1.844 kg.

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Answer:true

Explanation: I’m fram wokonda

The email message you provided is advising law firm employees to be alert for a new virus that can destroy files.

The virus is associated with .EDT file extensions, so the email message suggests that employees search their hard drives for these files and delete them.

This advice is generally sound. .EDT files are associated with Adobe Digital Editions, a software program that is used to read eBooks. If a .EDT file is infected with a virus, it could potentially damage or destroy the eBook that it is associated with.

However, it is important to note that not all .EDT files are infected with viruses. It is possible that an employee may have a legitimate .EDT file that they need to keep. Therefore, before deleting any .EDT files, it is important to make sure that they are not infected.

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Answer:

All Brake lights are dimmer than normal because high resistance in the brake switch could be the cause according to Technician B.

Explanation:

According to Technician A

When the bulb is faulty then no current will flow through bulb and it will be open circuit.So no light will produce in bulb .

According to Technician B

When a high resistance inserted in series  circuit the voltage across each resistance is reduced and this cause the light glow dimly.

Formula of resistance in series circuit

Rt=r1+r2+r3......

Answer:

Interface temperature = -7.27C

Rate of heat transfer = 10.91W/m2

Explanation:

The steps and the application of fourier's law of heat conduction is as shown in the attached file.

For the given composite wall, the temperature at the interface between the two layers is 17.275 °C and the rate of heat transfer through the wall is -0.0109 W/m².

To determine the temperature at the interface between the two layers, we can use the formula for 1-D steady-state heat conduction:

q = (T2 - T1) / ((L1 / k1) + (L2 / k2))

where q is the heat transfer rate per unit area, T1 and T2 are the temperatures of the inner and outer surfaces, L1 and L2 are the thicknesses of the insulation and siding layers, and k1 and k2 are the thermal conductivities of the insulation and siding layers, respectively.

Plugging in the given values:

q = (-10 - 20) / ((125 / 0.05) + (25 / 0.10)) = -30 / (2500 + 250) = -30 / 2750 = -0.0109 W/m².

The negative sign indicates heat transfer from the siding to the insulation.

To find the temperature at the interface, we can use the formula:

T_interface = T1 + (q * (L1 / k1))

Plugging in the values:

T_interface = 20 + (-0.0109 * (125 / 0.05)) = 20 - 2.725 = 17.275 °C.

Therefore, the temperature at the interface between the two layers is 17.275 °C and the rate of heat transfer through the wall is -0.0109 W/m².

Answer:

vec(a) = 16 i + 16 j

mag(a) = 22.63 ft/s^2

Explanation:

Given,

- The two components of velocity are given for fluid flow:

                                       u = 4*y ft/s

                                       v = 4*x ft/s

Find:

What is the time rate of change of the velocity vector V (i.e., the acceleration vector) for a fluid particle at x = 1 ft. and y = 1 ft. at time t = 1 second?

Solution:

- The rate of change of velocity is given to be acceleration. We will take derivative of each components of velocity with respect to time t:

                                      a_x = du / dt

                                      a_x = 4*dy/dt

                                      a_y = dv/dt

                                      a_y = 4*dx/dt

- The expressions dx/dt is the velocity component u and dy/dt is the velocity component v:

                                     a_x = 4*(4*y) = 16y

                                     a_y = 4*(4*x) = 16x

- The acceleration vector can be expressed by:

                                     vec(a) = 16y i + 16x j

- Evaluate vector (a) at x = 1 and y = 1:

                                     vec(a) = 16*1 i + 16*1 j = 16 i + 16 j

- The magnitude of acceleration is given by:

                                     mag(a) = sqrt ( a^2_x + a^2_y )

                                     mag(a) = sqrt ( 16^2 + 16^2 )

                                     mag(a) = 22.63 ft/s^2

Answer:

Explanation:

The concept of differential in rate of change is applied to solve the problem.

First is to compare the equation given V=ai+bxj  , where u = a and v = bx.

the detailed steps and appropriate integration and substitution is as shown in the attached file.

The streamline equation and the coordinates of the point are

A) [tex]x^2-4y+16[/tex]

B) (4, 12)

C) (3, 11.25)

Given that, [tex]V=ai+bxj[/tex], [tex]a=\frac{2m}{s}[/tex] and [tex]b=1s^{-1}[/tex]

Putting the value, we get

[tex]V=2i+1.xj[/tex]

From this equation, we get

[tex]u=2\frac{m}{s}[/tex] and [tex]v=x\frac{m}{s}[/tex]

Obtain the equation for the streamline passing through the point (2,5), streamline equation

[tex]\frac{dx}{u}=\frac{dy}{v}[/tex]

Putting the value, we get

[tex]\frac{dx}{2}=\frac{dy}{x}[/tex]

rearrange it and do integration:

[tex]\int xdx=\int 2dy[/tex]

[tex]\frac{x^2}{2}=2y+c -------(i)[/tex]

C is the integration constant.

[tex]\frac{2^2}{2}=(2\times5)+C[/tex]

[tex]C=-8[/tex]

Put the value of C in equation (i), we get

[tex]\frac{x^2}{2}=2y-8[/tex]

[tex]y=\frac{x^2}{4}+4[/tex]

[tex]x^2-4y+16=0[/tex]

b) [tex]u=2\frac{m}{s}[/tex]

[tex]\frac{dx}{dt}=2[/tex]

Rearrange and integrate

[tex]\int dx=\int 2dt[/tex]

[tex]x=2t+C_1[/tex]

Put x=0 and t=0

[tex]0=0+C_1[/tex]

[tex]C_1=0[/tex]

Put the value of constant we get

[tex]x=2t[/tex]

at [tex]t=2s[/tex]

[tex]x=2\times2= 4 --------(2)[/tex]

Now, [tex]v=x \frac{m}{s}[/tex]

[tex]\frac{dy}{dt}=x[/tex]

Rearrange and integrate:

[tex]\int dy=\int xdt[/tex]

[tex]y=xt+C_2[/tex]

Putting the value of x, y and t we get the value of constant.

[tex]4(0\times0)+C_2[/tex]

[tex]C_2=4[/tex]

Putting the value of constant we get

[tex]y=xt+4[/tex]

Put the value of [tex]t=2[/tex] and [tex]x=4[/tex] we get

[tex]y=(4\times2)+4[/tex]

[tex]= 8+4[/tex]

[tex]y= 12 -------(3)[/tex]

From (2) and (3)

we get the coordinates of the particle that passed through point [tex](0,4)[/tex] at [tex]t=0[/tex].

After [tex]t=2s[/tex] as [tex](4, 12)[/tex]

c) first we have to find the integration constant through initial condition then put the value of time to get the coordinates.

From above we have:

(1) [tex]x=2t+C_1[/tex] and

(2) [tex]y=xt+C_2[/tex]

Given [tex]t=2s[/tex] and [tex]P(1, 4.25)[/tex]

Putting the value in first equation, we get

[tex]1=(2\times2)+C_1[/tex]

[tex]C_1 =-3[/tex]

Put the value of time [tex]t=3s[/tex] we get

[tex]x=(2\times3)-3=3 ---------(a)[/tex]

Now, putting the value in second equation  we get

[tex]4.25=(1\times2)+C_2[/tex]

[tex]C_2 =2.25[/tex]

Then [tex]y=xt+2.25[/tex]

Put the value of time [tex]t=3s[/tex] we get

[tex]y=(3\times3)+2.25=11.25 -------(b)[/tex]

From (a) and (b) we get the coordinate of the point after [tex]t=3s[/tex] is [tex](3, 11.25)[/tex].

Therefore, streamline equation and the coordinates of the point are

A) [tex]x^2-4y+16[/tex]

B) (4, 12)

C) (3, 11.25)

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Answer:

slope of the equation = -0.006778 and the intercept = 27.11

Explanation:

The following details were also given ; They predict the monthly T shirts to be 1750 shirts, if you price them at $15.25 per shirt or 791 shirts if you price them at $21.75 per shirt.

The detailed steps and calculation is as shown in the attached file.

Answer:

- Provides an overview of your points

Provides a review of your key points near the end of your presentation

Provides transitions as you move from point to point

Explanation:

What does a blueprint slide do? Check all that apply. Provides a review of your key points near the end of your presentation Provides transitions as you move from point to point Provides an overview of your points Provides a clear explanation of one main point Provides a template for your presentation outline

- Provides an overview of your points

Provides a review of your key points near the end of your presentation

Provides transitions as you move from point to point

. A blueprint slide would not be useful for a clear explanation of one point or as a template for your outline. You would have to use a different slide layouts.

Blueprints slide establishes how you are going to work through your presentation. it is on the introductory part of your presentation.

Contains a description of your statements, a recap of your important points after your presentations, and transitioning as you get further from source to destination.

Thus the response above is correct.

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Answer:

The cathode ray oscilloscope (CRO) consists of a set of blocks. Those are vertical amplifier, delay line, trip circuit, time base generator, horizontal amplifier, cathode ray tube (CRT) and power supply. The CRO block diagram is shown in attached figure.

The function of each CRO block is mentioned below,

Vertical amplifier amplifies the input signal, which will be displayed on the CRT screen.

Delay line provides a certain amount of delay to the signal, which is obtained at the output of the vertical amplifier. This delayed signal is then applied to the CRT vertical deflection plates.

Trigger circuit produces a trigger signal to synchronize the horizontal and vertical deviations of the electron beam.

Time base generator produces a sawtooth signal, which is useful for horizontal deviation of the electron beam.

Horizontal amplifier amplifies the sawtooth signal and then connects it to the CRT horizontal deflection plates.

Power supply produces high and low voltages. The high negative voltage and the low positive voltage apply to CRT and other circuits respectively.

Cathode ray tube (CRT)

it is the main important block of CRO and consists mainly of four parts. Those are electronic guns, vertical deflection plates, horizontal deflection plates and fluorescent display.

The electron beam, which is produced by an electron gun, is deflected both vertically and horizontally by a pair of vertical deflection plates and a pair of horizontal deflection plates, respectively. Finally, the deflected beam will appear as a point on the fluorescent screen.

In this way, CRO will display the input signal applied on the CRT screen. So, we can analyze the signals in the time domain using CRO.

Explanation:

The oscilloscopes which is widely used for analysis purpose of circuits is divided into four main groups: the horizontal and vertical controls, the input controls and the activation controls.

Found in the front panel section marked Horizontal, the oscilloscope's horizontal controls allow users to adjust the horizontal scale of the screen. This section includes the control of the horizontal delay (displacement), as well as the control that indicates the time per division on the x-axis. The first control allows users to scan through a time range, while the latter allows users to approach a particular time range by decreasing the time per division.

Meanwhile, the oscilloscope's vertical controls are usually found in a section specifically marked as Vertical. The controls found in this section allow users to adjust the vertical appearance of the screen and include the control that indicates the number of volts per division on the axis and the grid of the screen. Also in this section is the control of the vertical displacement of the waveform, which translates the waveform up or down on the screen.

Signal activation helps provide a usable and stable display and allows users to synchronize the oscilloscope acquisition in the waveform of interest. The oscilloscope trigger controls allow users to choose the vertical trigger level, as well as the desired trigger capability. Common types of activation include fault activation, edge activation and pulse width activation.

Useful for identifying random errors or failures, the activation of faults allows users to fire at a pulse or event whose width is less than or greater than a specific period of time. This activation mode allows users to capture errors or technical problems that do not occur very frequently, which makes them very difficult to see.

The most famous trigger mode, edge tripping occurs when the voltage exceeds a set threshold value. This mode allows users to choose between shooting on a falling or rising edge.

Although pulse width activation is comparable to fault activation when users search for pulse width, it is, however, more general since it allows users to fire pulses of specified width. Users can also select the polarity of the pulses to be activated and set the horizontal position of the trigger. This allows users to see what happened during pre-shot or post-shot.

The input panels of an oscilloscope usually include two or four analog channels. They are usually numbered and have a button associated with each channel that allows users to activate and deactivate them. This section may also include a selection that allows users to specify the DC or AC coupling. Selecting the DC coupling implies that the entire signal will be input. The AC pairing, on the other hand, blocks the DC component and focuses the waveform around zero volts. Operators can also identify the probe impedance of the channels through a selection button. In adding, the input panels permit users to select the type of sampling to be used.

Answer: No, the mining process isn't faster when you know in advance that the land must be restored.

Explanation:

The mining process would be slower when mining companies know in advance that the land must be restored because they have to be more careful about their impact on the environment & avoid taking unnecessary damage to ruin the land which they know they must restore.

So, it's evident how this would slow down the mining process. If there were no obligations to restore land, the mining companies would just come at the mining space and run their mining process fast without big concerns for the consequences.

But the caution definitely slows them down.

People often have different points of view. If I think mining process is faster when you know in advance that the land must be restored, My answer is No.

This is because even without miners been aware that the land had to be restored before mining, they still carelessly remove minerals without taking into cognizant the damage it will do to the land.

There are different processes of mining. They include;

The type of mining method used is usually based on the kind of resource that is needed, the deposit's location etc.

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The particle's position at t = 2 s is (11 ft, 2 ft, 21 ft).

To determine the particle's position at t = 2 s, we need to integrate the acceleration function twice. First, we'll find the velocity function:

[tex]\[ \mathbf{v}(t) =\int \mathbf{a}(t) \, dt \][/tex]

[tex]\[ \int (6t\mathbf{i} + 12t^2 \mathbf{k}) \, dt \\= 3t^2 \mathbf{i} + 4t^3 \mathbf{k} + \mathbf{C} \][/tex]

where C₁ is the constant of integration.

Next, we'll find the position function:

[tex]\[ \mathbf{v}(t) = \int (3t^2 \mathbf{i} + 4t^3 \mathbf{k} + \mathbf{C}_1) \, dt \\= t^3 \mathbf{i} + t^4 \mathbf{k} + \mathbf{C}_1 t + \mathbf{C}_2 \][/tex]

where C₂ is the constant of integration.

Given the initial conditions, we know that at t = 0, the particle is at rest (v = 0) and located at point (3 ft, 2 ft, 5 ft). We can use these conditions to find the values of C₁ and C₂:

[tex]c_{1} = (0 i + 0 j + 0 k) - (3(0^{2} i + 4(0^{4} k) = 0[/tex]

[tex]c_{2}[/tex][tex]= (3 i + 2 j + 5 k)[/tex] - ([tex]i + 0^{4}k[/tex] )

[tex]= (3 i + 2 j + 5 k)[/tex]

Now we can find the position at t = 2 s:

= [tex]2^{3}[/tex][tex]i[/tex] + [tex]2^{4}[/tex][tex]k + (0)(2) + (3 i + 2 j + 5 k)[/tex]

[tex]= (8 i + 16 k) + (3 i + 2 j + 5 k)[/tex]

[tex]= (11 i + 2 j + 21 k)[/tex]

So the particle's position at t = 2 s is (11 ft, 2 ft, 21 ft).

Answer:

import java.util.Scanner;

  public class PeopleWeights {

    public static void main(String[] args) {

    Scanner reader = new Scanner(System.in);  

    double weightOne = reader.nextDouble();

    System.out.println("Enter 1st weight:");

    double weightTwo = reader.nextDouble();

    System.out.println("Enter 2nd weight :");

    double weightThree = reader.nextDouble();

    System.out.println("Enter 3rd weight :");

    double weightFour = reader.nextDouble();

    System.out.println("Enter 4th weight :");

    double weightFive = reader.nextDouble();

    System.out.println("Enter 5th weight :");

     double sum = weightOne + weightTwo + weightThree + weightFour + weightFive;

     double[] MyArr = new double[5];

     MyArr[0] = weightOne;

     MyArr[1] = weightTwo;

     MyArr[2] = weightThree;

     MyArr[3] = weightFour;

     MyArr[4] = weightFive;

     System.out.printf("You entered: " + "%.1f %.1f %.1f %.1f %.1f ", weightOne, weightTwo, weightThree, weightFour, weightFive);

     double average = sum / 5;

     System.out.println();

     System.out.println();

     System.out.println("Total weight: " + sum);

     System.out.println("Average weight: " + average);

     double max = MyArr[0];

     for (int counter = 1; counter < MyArr.length; counter++){

        if (MyArr[counter] > max){

           max = MyArr[counter];

        }

     }

     System.out.println("Max weight: " + max);

  }

import java.util.Scanner;

  public class PeopleWeights {

    public static void main(String[] args) {

    Scanner reader = new Scanner(System.in);  

    double weightOne = reader.nextDouble();

    System.out.println("Enter 1st weight:");

    double weightTwo = reader.nextDouble();

    System.out.println("Enter 2nd weight :");

    double weightThree = reader.nextDouble();

    System.out.println("Enter 3rd weight :");

    double weightFour = reader.nextDouble();

    System.out.println("Enter 4th weight :");

    double weightFive = reader.nextDouble();

    System.out.println("Enter 5th weight :");

     double sum = weightOne + weightTwo + weightThree + weightFour + weightFive;

     double[] MyArr = new double[5];

     MyArr[0] = weightOne;

     MyArr[1] = weightTwo;

     MyArr[2] = weightThree;

     MyArr[3] = weightFour;

     MyArr[4] = weightFive;

     System.out.printf("You entered: " + "%.1f %.1f %.1f %.1f %.1f ", weightOne, weightTwo, weightThree, weightFour, weightFive);

     double average = sum / 5;

     System.out.println();

     System.out.println();

     System.out.println("Total weight: " + sum);

     System.out.println("Average weight: " + average);

     double max = MyArr[0];

     for (int counter = 1; counter < MyArr.length; counter++){

        if (MyArr[counter] > max){

           max = MyArr[counter];

        }

     }

     System.out.println("Max weight: " + max);

  }

Answer:

The source code file to this question has been attached to this response. Please download it and go through it. The file contains comments explaining significant sections of the code. Please go through the comments in the code.

Answer: First pond= 8.0 mg/L

second pond = 4.2 mg/L

Explanation:

Cn/Co = [ 1/ [ 1 + (k*t/n)]]

for the first pond

where Co into the first pond is 20mg/L,

Cn = 20*[ 1/ [ 1+ ((1 x 5)/0.3)]]

Cn = 8 mg/L

into the second pond we calculate the BOD leaving the pond of volume 3million liters

we have Cn = 4.2 mg/L

Answer: BULK RELATIVE DENSITY.

WHY?

BULK RELATIVE DENSITY GIVES A BETTER UNDERSTANDING OF THE QUALITY OF THE MATERIAL.

Explanation:Bulk relative density is a type of Specific gravity which is often used in determining the volume occupied by the aggregates in various mixtures containing aggregate including Portland cement concrete, bituminous concrete etc this mixtures are proportioned based on an absolute volume basis. Bulk relative density is considered because of its ability to give a better understanding of the materials which makes up the mixture.

Answer:

Explanation:

Given

Temperature of solid [tex]T=500\ K[/tex]

Einstein Temperature [tex]T_E=300\ K[/tex]

Heat Capacity in the Einstein model is given by

[tex]C_v=3R\left [ \frac{T_E}{T}\right ]^2\frac{e^{\frac{T_E}{T}}}{\left ( e^{\frac{T_E}{T}}-1\right )^2}[/tex]

[tex]e^{\frac{3}{5}}=1.822[/tex]

Substitute the values

[tex]C_v=3R\times (\frac{300}{500})^2\times (\frac{1.822}{(1.822-1)^2})[/tex]

[tex]C_v=3R\times \frac{9}{25}\times \frac{1.822}{(0.822)^2}[/tex]

[tex]C_v=0.97\times (3R)[/tex]            

Answer:

Gage pressure at the bottom of the tank = 46.1kP

Explanation:

Note that it is required to expressed the gage pressure ,pressure at the free surface=0 gage

Pressure due to the oil layer= (3m)*((800kg/m^3)(9.81m/s^2))/1000=25.9kpa

Pressure due to the brine layer=(2m)*((1030kg/m^3)(9.81m/s^2))/1000=20.2ka

Therefore, the pressure at the bottom=25.9+20.2=46.1kP

Answer:

a) the two-way concrete joist framing system

Explanation:

A waffle slab is also known as ribbed slab, it is a slab which as waffle like appearance with holes beneath. It is adopted in construction projects that has long length, length more than 12m. The waffle slab is rigid, therefore it is used in building that needs minimal vibration.

Answer:

True

Explanation:

The passageway between the riser and the main cavity must have a small cross-sectional area to reduce waste of material in metal casting.

Answer:

View Image

Explanation:

You didn't provide me a picture of the opamp.

I'm gonna assume that this is an ideal opamp, therefore the input impedance can be assumed to be ∞ . This basically implies that...

Since no current is going into the + and - side of the opamp, then

i₁ = i₂

Since V₊ is connected to ground (0V) then V₋ must also be 0V.

V₊ = V₋  = 0

Use whatever method you want to solve for v_out and v_in then divide them. There's so many different ways of solving this circuit.

You didn't give me what the input voltage was so I can't give you the entire answer. I'll just give you the equations needed to plug in your values to get your answers.

Answer: a) 0.77 and 0.39 b) 3.9 and 0.20

Explanation:

We have to find two things here i.e link parameter and channel utilization efficiency.

a)

We are given

Bitrate= 12 Mbps,

Distance= 6 miles

Propagation Velocity= 3 x 10⁸ m/s

Length of frame= 500 bits

We know that The link parameter is related to propagation time and frame time

Where Propagation time= Bitrate x Distance= 12 Mbps x 6 miles

and Frame time= Propagation velocity x length of frame= 3 x 10⁸ x 500

So,

Link parameter= [tex]\frac{12*10^6*6*1609.34}{3*10^8*500}[/tex]

Link parameter= 0.77

Channel utilization efficiency= [tex]\frac{1}{1+2* link parameter}[/tex]

Channel utilization efficiency= 0.39

b)

We are given

Bitrate= 10 Mbps,

Distance= 15 km

Propagation Velocity= 3 x 10⁸ m/s

Length of frame= 256 bits

We know that The link parameter is related to propagation time and frame time, Here QPSK is used so bitrate is multiplied by 2,

Where Propagation time= Bitrate x Distance= 2 x 10 Mbps x 15 km

and Frame time= Propagation velocity x length of frame= 3 x 10⁸ x 256

So,

Link parameter= [tex]\frac{2*10*10^6*15*1000}{3*10^8*256}[/tex]

Link parameter= 3.9

Channel utilization efficiency= [tex]\frac{1}{1+2* link parameter}[/tex]

Channel utilization efficiency= 0.20

Answer:

The area of solar panel needed in Buffalo is 22.83 m²

The area of solar panel needed in Phoenix is 31 m²

Explanation:

FOR BUFFALO:

First we, calculate the the annual power requirement by dividing the energy by the time of 1 year.

Therefore,

Power Requirement = P = 6000 KWhr/(1 yr)(365 days/yr)(24 hr/day)

P = 684.93 W

Now, to calculate the area we use the formula:

Area = A = P/(Annual Insolation)(Conversion Factor)

A = 684.93 W/(200 W/m²)(0.15)

A = 22.83 m²

FOR PHOENIX:

First we, calculate the the annual power requirement by dividing the energy by the time of 1 year.

Therefore,

Power Requirement = P = 11000 KWhr/(1 yr)(365 days/yr)(24 hr/day)

P = 1255.7 W

Now, to calculate the area we use the formula:

Area = A = P/(Annual Insolation)(Conversion Factor)

A = 1255.7 W/(270 W/m²)(0.15)

A = 31 m²

To calculate the mechanical power used to overcome friction in piping for the described pump operation, calculate the hydraulic power needed, then subtract the pump's output power (based on its efficiency) from its total input.

The question involves determining the mechanical power used to overcome frictional effects in piping for an 80-percent-efficient pump with a power input of 20 hp, which is pumping water at a rate of 1.5 ft3/s from a lake to a pool 80 ft above. Firstly, calculate the total hydraulic power required to move the water to the desired height, then calculate the power output of the pump based on its efficiency. Finally, subtract this value from the total input power to find the power used to overcome friction.

Through such calculations, one can find that the mechanical power used to overcome frictional effects in piping is 2.37 hp.

Answer:

Explanation:

The concept of Hooke's law was applied as it relates to deformation.

The detailed steps and appropriate substitution is as shown in the attached file.

Answer:

estimated actual endurance strength = 183.22 MPa

Explanation:

given data

actual endurance limit for SAE 4130 WQT 1300

cross section = 20.0 mm by 60 mm

reliability = 99%

solution

we use here table for get some value for AISI 4130 WQT 1300 steel

Sut = 676 MPa

Sn = 260 MPa

and here

stress factor Cst = 1

and wrought steel Cm = 1

and reliability factor Cr is = 0.81

so here equivalent diameter will be

equivalent diameter = 0.808 [tex]\sqrt{bh}[/tex]    ...........1

equivalent diameter = 0. 808 [tex]\sqrt{20*60}[/tex]  

equivalent diameter = 28 mm

so here size factor will be = 0.87 by the table

so now we can get estimated actual endurance strength will be

actual endurance strength = Sn × Cm × Cst × Cr × Cs ...............2

put here value

actual endurance strength = 260 × 1 × 1 × 0.81 × 0.87

actual endurance strength = 183.22 MPa

Answer:

Space mean speed = 44 mi/h

Explanation:

Using Greenshield's linear model

q = Uf ( D - [tex]D^{2}[/tex]/Dj )

qcap = capacity flow that gives Dcap

Dcap = Dj/2

qcap = Uf. Dj/4

Where

U = space mean speed

Uf =  free flow speed

D = density

Dj = jam density

now,

Dj = 4 × 3300/55

    = 240v/h

q = Dj ( U - [tex]U^{2}[/tex]/Uf)

2100 = 240 ( U - [tex]U^{2}[/tex]/55)

Solve for U

U = 44m/h

Answer:

t = 30.1 sec

Explanation:

If the ant is moving at a constant speed, the velocity vector will have the same magnitude at any point, and can be decomposed in two vectors, along directions perpendicular each other.

If we choose these directions coincident with the long edge of the paper, and the other perpendicular to it, the components of the velocity vector, along these axes, can be calculated as the projections of this vector along these axes.

We are only interested in the component of the velocity across the paper, that can be calculated as follows:

vₓ = v* sin θ, where v is the magnitude of the velocity, and θ the angle that forms v with the long edge.

We know that v= 1.3 cm/s, and θ = 61º, so we can find vₓ as follows:

vₓ = 1.3 cm/s * sin 61º = 1.3 cm/s * 0.875 = 1.14 cm/s

Applying the definition of average velocity, we can solve for t:

t =[tex]\frac{x}{vx}[/tex] = [tex]\frac{34.2 cm}{1.14 cm/s} =30.1 sec[/tex]

⇒ t = 30.1 sec