Compute the number of kilo- grams of hydrogen that pass per hour through a 6-mm-thick sheet of palladium having an area of 0.25 m^2 at 600°C. Assume a diffusion coefficient of 1.7 x 10^8 m^2/s, that the concentrations at the high- and low-pressure sides of the plate are 2.0 and 0.4 kg of hydrogen per cubic meter of palladium, and that steady-state conditions have been attained.

Properties of Carpenter's hammer possess

Explanation:

1.The head of a carpenter's hammer should possess the impact resistance, so that the chips do not peel off the striking face while working.

2.The hammer head should also be very hard, so that it does not deform while driving or eradicate any nails in wood.

3.Carpenter's hammer is used to impact smaller areas of an object.It can drive nails in the wood,can crush  the rock and shape the metal.It is not suitable for heavy work.

How hammer head is manufactured :

1.Hammer head is produced by metal forging process.

2.In this process metal is heated and this molten metal is placed in the cavities said to be dies.

3.One die is fixed and another die is movable.Ram forces the two dies under the forces which gives the metal desired shape.

4.The third process is repeated for several times.

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:

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:

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

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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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

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:

The answers to the question are

Explanation:

The number of employees in the firm = 62

Number of first-aid cases = 7

Number of medical treatment injuries = 3

Injury resulting restricted work activity = 1

Illness resulting in one week loss work day = 1

Illness resulting in loss of 6 weeks of work =1

Incident resulting in fatality = 1

Total hours worked = 40×62×50  = 124000 Hrs

Where 200,000 is the number of hours worked by Full time employees numbering 100 that  work for 40 Hours a week for 50 weeks in a year

Number of recordable incident =  Those incident that results in lost work days, death, restricted ability to work, or transfer to another task or more severe injury treatment beyond first aid

Therefore number recordabe incident = 7

Recordable Incident Rate is calculated by

IR = (Number of recordable incident Cases X 200,000)÷(Number of Employee labor hours worked)

= (7×200000)/124000= 11.29

The Lost Workday Rate is given by

Lost Workday Rate = (Total number of days lost to injury or illness)÷( Cumulative number of hours employees worked) × 200000

Total number of days lost to injury or illness = 42 days

Lost Workday Rate = (42/124000) × 200000 = 67.74

LWDI = (Number of incident resulting in lost workdays and restricted activity) ×200000 /(Total number of hours worked by all employees in one year)

LWDI = LWD×200000/ EH = 4×200000/124000 = 6.45

LWDI  = 6.45

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:

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:

attached below

Explanation:

Answer:

Power of the motor that kept the wheel rotating at 12 rev/hr despite friction is 342.79W.

Explanation:

Pls refer to the attached file. The explanation is long to pen down here.

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:

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: 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: 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:

The motor inertia is 7.958 X 10⁻² kg.m²

Explanation:

To determine the motor inertia, the following formula applies.

Neglecting the damping effect,

[tex]T = \frac{J}{9.55}.\frac{\delta n}{\delta T}[/tex]

Where;

T is the constant torque applied to the motor = 5Nm

J is the motor inertia = ?

δn is the change in angular speed of the motor = 1800 r/min

δT is change in time of the unloaded motor from rest = 3 sec

[tex]J = \frac{9.55* \delta T* T}{\delta n}[/tex]

[tex]J = \frac{9.55* 3* 5}{1800}[/tex] = 0.07958 kg.m² = 7.958 X 10⁻² kg.m²

Therefore, the motor inertia is 7.958 X 10⁻² kg.m²

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: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:

(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:

- 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:

z=60.32°, i=0.32°, Beam Radiation = 1097.2 W/m²,  Id = 94.2 W/m², Ir=14.1W/m², total radiation = 1205.4 W/m², Local time=1:21PM

Explanation:

A. Zenith Angle:

As we know that,

Zenith angle=z=90⁰-α=L(latitude)=29.68⁰

Another way to do it is to find α first,

At solar time hour angle is 0⁰. So, solar altitude becomes equal to latitude which could be written as

sinα=cosL

α=sin⁻¹(cosL)=sin⁻¹(cos29.68⁰)=60.32°

B. Angle of incidence:

angle of incidence= cosi=sin(α+β)=sin(60.32°+32°)=sin92.32°

i=cos⁻¹(sin92.32°)=0.32°

C. Beam Radiation:

First we need to calculate extra terrestrial radiations

Iext.=1353[1+0.034cos(360n/365)]

where n=264

=1345 W/m²

Now,

Beam Radiation=CIext⁻ⁿ

where n=0.1/sin60.32°

Beam Radiation = 1097.2 W/m²

D. Diffude Radiation:

difuse radiation = Id = 0.0921ₙcos²(β/2)

where β=30°

Id = 94.2 W/m²

E. Reflected Radiations:

Ir=pIn(sinα+0.092)sin²(β/2)

= (0.2)(1097.1)(sin60.32+0.092)sin²(30/2)

= 14.1W/m²

F. Total Radiation:

total radiation = beam radiation + diffuse radiation + reflected raddiation

= 1205.4 W/m²

G. Local Time:

LST= ST-ET-(lₓ-l(local))4min/₀

     = 12:00-7.9min-(75°-82.27°)4min/₀

     =12:21PM

Local time

LDT=LST+=12:21+1:00=1:21PM

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:

a.The current is cut in half

b.the current stays the same

c.the current doubles

d.the current becomes zero

Explanation:

The current doubles

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:

The question is incomplete, below is the complete question "The electric potential difference between the ground and a cloud in a particular thunderstorm is [tex]3.0*10^{9}V[/tex]. What is the magnitude of the change in the electric potential energy of an electron that moves between the ground and the cloud?"

Answer:

[tex]U=3.0*10^{9}eV[/tex]

Explanation:

data given,

Potential difference,V=3.0*10^9V

charge on an electron, q=1e.

Recall that the relationship between potential difference (v), charge(Q) and the potential energy(U) is expressed as

[tex]U=qV[/tex]

from the question, we asked to determine potential energy given the charge and the potential difference.

Hence if we substitute values into the equation, we arrive at

[tex]U=qV\\U=3.0*10^{9}*1e\\U=3.0*10^{9}eV[/tex]

Hence the magnitude of the change in the electric potential energy of an electron that moves between the ground and the cloud is [tex]3.0*10^{9}eV[/tex]

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:

Detailed Answer is given below,

Explanation:

(a) Elastic strain

Elastic deformation is defined as the limit for the deformation values up to which the object will bounce and return to the original shape when the load is removed.

When the object is subjected to an external load, it deforms. If the load exceeds the load corresponding to the elastic limit of the object, then, after removing the load, the object cannot return to its original geometric specification.

(b) Plastic strain

Plastic Strain is a deformation that cannot be recovered after eliminating the deforming force. In other words, if the applied tension is greater than the elastic limit of the material, there will be a permanent deformation.

(c) Creep strain

The deformation of the material at a higher temperature is much more than at the normal temperature known as creep.

Subsequently, the deformation produced in heated material is called creep deformation. Creep deformation is a function of temperature.

(d) Tensile viscosity

Viscosity is a main parameter when measuring flux fluids, such as liquids, semi-solids, gases and even solids. Brookfield deals with liquids and semi-solids. Viscosity measurements are made together with the quality and efficiency of the product. Any person involved in the characterization of the flow, in research or development, quality control or transfer of fluids, at one time or another is involved with some kind of viscosity measurement.

(e) Recovery

Recovery is a process whereby deformed grains can reduce their stored energy by eliminating or reorganizing defects in their crystalline structure. These defects, mainly dislocations, are introduced by plastic deformation of the material and act to increase the elastic limit of a material.

(f) Relaxation

Stress relaxation occurs in polymers when they remain tense for long periods of time.

These alloys have very good resistance to stress relaxation and, therefore, are used as spring materials.

Stress relaxation is a gradual reduction of stress over time in constant tension.

Answer:

Due to differences in the nature of the adhesive force and cohesive forces in the molecules of the individual substances and the glass tube.

Explanation:

To understand why this is so, a deep understanding of adhesive and cohesive force is required.

First, what is adhesive force?

Note: Adhesive and cohesive forces are best discussed under macroscopic level.

Adhesive force is the Intermolecular forces that exist between atoms of different molecules e g the forces explain when unlike charges stick together.

Cohesive force on the other hand is the Intermolecular force that exist between atoms of the same molecules e.g the force between hydrogen bonding.

Hence to explain the case scenario above, the adhesive force between the water molecules and the glass molecules is higher than the cohesive force between the water molecules. Hence the high rise.

For the case of the Hexane, the cohesive forces between the molecules hexane is far greater then the adhesive force between the glass molecules and the hexane molecules.