5. A driver is traveling at 90 km/h on a wet road. An object is spotted on the road 140m ahead and the driver is able to come to a stop just before hitting the object. Assuming standard reaction time and using the practical-stopping distance equation, determine the grade of the road.

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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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.

Full Question

Write a static method called quadrant that takes as parameters a pair of real numbers representing an (x, y) point and that returns the quadrant number for that point.

Recall that quadrants are numbered as integers from 1 to 4 with the upper-right quadrant numbered 1 and the subsequent quadrants numbered in a counter-clockwise fashion. Notice that the quadrant is determined by whether the x and y coordinates are positive or negative numbers. If a point falls on the x-axis or the y-axis, then the method should return 0.

Answer

public int quadrant(double x, double y) // Line 1

{

if(x > 0 && y > 0){ //Line 2

return 1;

}

if(x < 0 && y > 0){ // Line 3

return 2;

}

if(x < 0 && y < 0) {// Line 4

return 3;

}

if(x > 0 && y < 0) {// Line 5

return 4;

}

return 0;

}

Explanation:

Line 1 defines the static method.

The method is named quadrant and its of type integer.

Along with the method definition, two variables x and y of type double are also declared

Line 2 checks if x and y are greater than 0.

If yes then the program returns 1

Line 3 checks if x is less than 0 and y is greater than 0

If yes then the program returns 2

Line 4 checks if x and y are less than 0

If yes then the program returns 3

Line 5 checks is x is greater than 0 and y is less than 0

If yes then the program returns 4

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:

- 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:

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: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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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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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

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.

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:

Explanation:

[tex]\dot x (t) - ax(t)=bu(t)\\\\e^{-at}\dot x = e^{-at}ax = \frac{d}{dt} (e^{-at})=e^{-at}bu\\\\\int\limits^t_0 \frac{d}{d\tau}(e^{-a\tau}x(\tau))d\tau=e^{-at}x(t)-x(0)=\int\limits^t_0 e^{-a\tau}bu(\tau))d\tau\\\\x(t)=e^{at}x(0)+ \int\limits^t_0 e^{-a(t-\tau)}bu(\tau))d\tau[/tex]

Similarly,

[tex]e^{-At}\dot x(t) -e^{-At}Ax(t) = \frac{d}{dt} (e^{-At}x(t))=e^{-At}Bu(t)\\\\\int\limits^t_0 \frac{d}{d\tau}(e^{-A\tau}x(\tau))d\tau=e^{-At}x(t)-e^{-A.0}x(0)=\int\limits^t_0 e^{-A\tau}Bu(\tau))d\tau\\\\since\:\:\:e^{-A.0}=I \:\:\: and\:\:\:[e^{-At}]^{-1}=e^{At}\\\\x(t)=e^{At}x(0)+ e^{At}\int\limits^t_0 e^{-A\tau}Bu(\tau))d\tau\\\\x(t)=e^{At}x(0)+ \int\limits^t_0 e^{A(t-\tau)}Bu(\tau))d\tau[/tex]

For initial condition x(0) = 0

[tex]x(t)= \int\limits^t_0 e^{A(t-\tau)}Bu(\tau))d\tau[/tex]

Answer: See attachment below

Explanation:

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.

Hi, you haven't provided the programing language in which you need the code, I'll just explain how to do it using Python, and you can apply a similar method for any programming language.

Answer:

1. def pyramid_volume(base_length, base_width, pyramid_height):

2.     volume = base_length*base_width*pyramid_height/3

3.     return(volume)

Explanation step by step:

In the image below you can see the result of calling the function with input 4.5, 2.1, 3.0.

The function 'pyramid_volume' can be used to compute the volume of a pyramid with a rectangular base. Its calculation uses the equation: V = (base_area * height) / 3, where base_area = base_length * base_width. Given the inputs 4.5, 2.1, and 3.0 for base_length, base_width, and pyramid_height respectively, the volume would be 9.45

The function pyramid_volume is best defined in the context of Mathematics, particularly, geometric formulas. To compute the volume of a pyramid with a rectangular base, you can utilize the formula: V = (base_area * height) / 3. Here, base_area is equal to base_length times base_width. Hence, the function pyramid_volume could be defined as follows:

Volume = lambda base_length, base_width, pyramid_height: (base_length * base_width * pyramid_height) / 3.

Thus, if you input 4.5, 2.1, and 3.0 as your base_length, base_width, pyramid_height respectively into the function, it will return the volume of the pyramid, which in this case is 9.45.

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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:

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

Answer:

(a) attached below

(b) [tex]pf_{C}=0.85[/tex] [tex]lagging[/tex]

(c) [tex]I_{C} =32.37 A[/tex]

(d) [tex]X_{C} =49.37[/tex] Ω

(e) [tex]I_{cap} =9.72 A[/tex] and [tex]I_{line} =27.66 A[/tex]

Explanation:

Given data:

[tex]P_{1}=15 kW[/tex]

[tex]S_{2} =10 kVA[/tex]

[tex]pf_{1} =0.6[/tex] [tex]lagging[/tex]

[tex]pf_{2}=0.8[/tex] [tex]leading[/tex]

[tex]V=480[/tex] [tex]Volts[/tex]

(a) Draw the power triangle for each load and for the combined load.

[tex]\alpha_{1}=cos^{-1} (0.6)=53.13[/tex]°

[tex]\alpha_{2}=cos^{-1} (0.8)=36.86[/tex]°

[tex]S_{1}=P_{1} /pf_{1} =15/0.6=25 kVA[/tex]

[tex]Q_{1}=P_{1} tan(\alpha_{1} )=15*tan(53.13)=19.99[/tex] ≅ [tex]20kVAR[/tex]

[tex]P_{2} =S_{2}*pf_{2} =10*0.8=8 kW[/tex]

[tex]Q_{2} =P_{2} tan(\alpha_{2} )=8*tan(-36.86)=-5.99[/tex] ≅ [tex]-6 kVAR[/tex]

The negative sign means that the load 2 is providing reactive power rather than consuming  

Then the combined load will be

[tex]P_{c} =P_{1} +P_{2} =15+8=23 kW[/tex]

[tex]Q_{c} =Q_{1} +Q_{2} =20-6=14 kVAR[/tex]

(b) Determine the power factor of the combined load and state whether lagging or leading.

[tex]S_{c} =P_{c} +jQ_{c} =23+14j[/tex]

or in the polar form

[tex]S_{c} =26.92<31.32[/tex]°

[tex]pf_{C}=cos(31.32) =0.85[/tex] [tex]lagging[/tex]

The relationship between Apparent power S and Current I is

[tex]S=VI^{*}[/tex]

Since there is conjugate of current I therefore, the angle will become negative and hence power factor will be lagging.

(c) Determine the magnitude of the line current from the source.

Current of the combined load can be found by

[tex]I_{C} =S_{C}/\sqrt{3}*V[/tex]

[tex]I_{C} =26.92*10^3/\sqrt{3}*480=32.37 A[/tex]

(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

[tex]Q_{C} =3*V^2/X_{C}[/tex]

[tex]X_{C} =3*V^2/Q_{C}[/tex]

[tex]X_{C} =3*(480)^2/14*10^3[/tex] Ω

(e) Compute the magnitude of the current in each capacitor and the line current from the source.

Current flowing in the capacitor is  

[tex]I_{cap} =V/X_{C} =480/49.37=9.72 A[/tex]

Line current flowing from the source is

[tex]I_{line} =P_{C} /3*V=23*10^3/3*480=27.66 A[/tex]

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: 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:

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:

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

Answer:

Explanation: 100 and 110 Hz

Here fL=2kHz, fH=2.1kHz, B=0.1kHz, fL=2kHz,  

n≤fH/(fH-fL)

n≤(2.1*1000)/((2.1-2)*1000)

n≤20

2fH/n  ≤ fs

(2*2.1*1000)/20 ≤ fs

210≤ fs

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:

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.

Optimistic attitude, self-efficacy, stress inoculation, secure relationships, and spirituality are factors that help individuals cope with adversity. These elements foster hardiness and stress-related growth, allowing individuals to manage chronic and acute stressors more effectively and pursue personal development.

Maintaining an optimistic attitude, building self-efficacy, strengthening stress inoculation, experiencing secure personal relationships, and practicing spirituality or religiosity are factors to help individuals thrive in the face of adversity.

Adversity can include both chronic stressors, such as long-term unemployment, and acute stressors, like illnesses or loss. These stressors require coping mechanisms and personal strengths for an individual to manage effectively. Optimism is a general tendency to expect positive outcomes and it contributes to less stress and happier dispositions. Self-efficacy, defined by Albert Bandura, is the belief in one's abilities to reach goals, which empowers a proactive response to stressors. The concept of hardiness, which relates to optimism and self-efficacy, describes those who are less affected by life's stressors and use effective coping strategies.

Furthermore, stress-related growth or thriving describes the ability of individuals to exceed previous performances and show increased strength in the face of adversity. The nurturing of secure personal relationships provides social support that can aid in increasing positive affect and reducing the number of physical symptoms during stressful times.

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)

Learn more about the streamline passing through the point here:

https://brainly.com/question/34111016.

#SPJ3

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