Why does the the diffusion capacitance fall off at high frequencies?


1. At high frequencies the dopants are vibrated across the metallurgical junction and the doping profiles smooths out.

2. Since Gd increases with frequency Cd must decrease with frequency and the parallel combination of Gd and Cd is constant.

3. When the applied bias frequency is higher than the speed which the majority carriers can respond the diode starts to disappear electrically.

4. It takes time to store and remove minority charge, so when the frequency is higher than the inverse of the minority carrier lifetime the carriers can't respond as well.

Answers

Answer 1

Answer:

The answer 1 is correct.

Explanation:

When a junction called the p-n is forward biased, the capacitance of diffusion will form or structured across depletion layers.

When at a higher frequency or frequencies, the dopants, are vibrated across the metallurgical junction and the doping profiles smooths out.


Related Questions

Problem 32.3 The telescoping arm ABC is used to provide an elevated platform for construction workers. The workers and the platform together have a msss of 250 kg and have a combined center of gravity located directly above C. For the position when θ = 25°, determine (a) the force exerted at B by the single hydraulic cylinder BD, (b) the force exerted on the supporting carriage at A.

Answers

Final answer:

The question involves applying principles of static equilibrium and mechanical advantage in physics to calculate forces exerted by a hydraulic cylinder and at a support point in a construction telescoping arm scenario.

Explanation:

The question relates to the application of static equilibrium and mechanical advantage concepts in physics to solve for forces in a telescoping arm used in construction. Specifically, it involves calculating force exerted by a hydraulic cylinder and the force at a support point when a mass is placed at a specific location on the arm.

In order to solve for the force exerted at point B by the hydraulic cylinder (part a) and the force exerted on the supporting carriage at A (part b) when the arm makes an angle  heta = 25 degrees with the horizontal, one would typically use the principles of torques and static equilibrium. Summation of torques around a point, often the pivot, and summation of forces in horizontal and vertical directions are standard procedures. This requires knowledge of the geometry of the construction, the location of the center of gravity, and the mass of the workers and platform.

The information given about the forces in a wheelbarrow and cranes are analogous examples that illustrate similar principles of physics applied to different scenarios. These examples demonstrate how to calculate the mechanical advantage and forces based on lever arms and mass distribution.

The force exerted at B by the hydraulic cylinder BD and the force exerted on the supporting carriage at A can be calculated using principles of moments and equilibrium of forces, respectively.

For part (a): The force exerted at B by the hydraulic cylinder BD can be calculated using the principle of moments. By summing the moments about point A, you can find the force at B.

For part (b): The force exerted on the supporting carriage at A can be determined by considering.

Derive an expression for the axial thrust exerted by a propeller if the thrust depends only on forward speed, angular speed, size, and viscosity and density ofthe fluid. How would the expression change if gravity were a relevant variable in the case of a ship propeller?

Answers

Answer:

Find the given attachment

Consider the problem of oxygen transfer from the interior lung cavity, across the lung tissue, to the network of blood vessels on the opposite side. The lung tissue (species B) may be approximated as a plane wall of thickness L. The inhalation process may be assumed to maintain a constant molar concentration CA(0) of oxygen (species A) in the tissue at its inner surface (x = 0), and assimilation of oxygen by the blood may be assumed to maintain a constant molar concentration CA(L) of oxygen in the tissue at its outer surface (x = L). There is oxygen consumption in the tissue due to metabolic processes, and the reaction is zero order, with N_A=-k_0.

Obtain expressions for the distribution of the oxygen concentration in the tissue and for the rate of assimilation of oxygen by the blood per unit tissue surface area.

Answers

Answer:

See attached images

The clock period of a ripple counter must be longer than the total propagation
delays through all of the flip-flops. For proper operation, the period / frequency
should be:
Tmin = N x tpd where tpd is the propagation delay fmax = 1/ Tmin = 1/ (N x tpd).
If the propagation delay is 15ns for a J-K FF, what is the largest
MOD counter that can be constructed to ensure the counter will operate for
frequencies > 12 MHz?

Answers

Answer:

See explaination

Explanation:

Please kindly check attachment for the step by step solution of the given problem

Create a JavaFX application that lets the user enter the food charge for a meal at a restaurant. When a button is clicked, the application should calculate and display the amount of an 18 percent tip on the total food charge, 7 percent sales tax, and the total of all three amounts. For example, if $20 is entered as a food charge for a meal then $3.6 should be displayed for the tip, $1.4 should be displayed for sales tax, and $25 should be displayed as a total of all three amounts.

Answers

Answer:

See explaination

Explanation:

package sample;

import javafx.application.Application;

import javafx.fxml.FXMLLoader;

import javafx.geometry.*;

import javafx.scene.Parent;

import javafx.scene.Scene;

import javafx.scene.control.Label;

import javafx.geometry.Insets;

import javafx.geometry.Pos;

import javafx.scene.Scene;

import javafx.scene.control.Button;

import javafx.scene.control.Label;

import javafx.scene.control.TextField;

import javafx.scene.layout.GridPane;

import javafx.stage.Stage;

public class Main extends Application {

atOverride // Replace the at with at symbol

public void start(Stage primaryStage) throws Exception{

primaryStage.setTitle("Calculator");

GridPane rootNode = new GridPane();

rootNode.setPadding(new Insets(15));

rootNode.setHgap(5);

rootNode.setVgap(5);

rootNode.setAlignment(Pos.CENTER);

Scene myScene = new Scene(rootNode, 300, 200);

rootNode.add(new Label("Amount:"), 0, 0);

TextField firstValue = new TextField();

rootNode.add(firstValue, 1, 0);

rootNode.add(new Label("Toatal is:"), 0, 5);

Button aButton = new Button("Calculate");

rootNode.add(aButton, 1, 2);

GridPane.setHalignment(aButton, HPos.LEFT);

TextField result = new TextField();

result.setEditable(false);

rootNode.add(result, 1, 5);

TextField tax = new TextField();

rootNode.add(new Label("Tax:"), 0, 3);

tax.setEditable(false);

rootNode.add(tax,1,3);

TextField tip = new TextField();

rootNode.add(new Label("Tip:"), 0, 4);

tip.setEditable(false);

rootNode.add(tip,1,4);

aButton.setOnAction(e -> {

Float value1 = Float.valueOf(firstValue.getText());

Float value2 =(value1*18)/100;

Float value3 = (value1*7)/100;

Float r = value1+value2+value3 ;

tax.setText(value3.toString());

tip.setText(value2.toString());

result.setText(r.toString());

});

primaryStage.setScene(myScene);

primaryStage.show();

}

public static void main(String[] args) {

launch(args);

}

}

Final answer:

The question is about creating a JavaFX application that calculates and displays the cost of a restaurant meal, including an 18% tip and 7% sales tax. To create the application, an interface is needed with a text field for user input, a button for the calculation, and labels to display the results. The calculation logic is added to the button event handler.

Explanation:

The subject of this question is in the area of Computers and Technology, specifically application development using JavaFX. To create the desired application, we need to first create an interface, typically using FXML, with a text field for user input (food charge), a button for the calculation, and three labels to display the tip, sales tax, and total respectively.

On the button event handler, we would include the logic to calculate the 18% tip, 7% sales tax, and add all three (food charge, tip, sales tax) to get the total amount. This is done using the following formulas: tip = food charge * 0.18, sales tax = food charge * 0.07, and total = food charge + tip + sales tax. Finally, we would display the calculated amounts on the respective labels.

The code is written in JavaFX, a software platform used to create and deliver desktop applications, along with rich internet applications that can run across a wide variety of devices.

Learn more about JavaFX application here:

https://brainly.com/question/31593283

#SPJ3

Determine the convention heat transfer coefficient inside the for the flow of (a) air and (b) water at a velocity of 2 m/s in an 8-cm- diameter and 7-m-long tube when the tube is subjected to uniform heat flux from all surfaces. Use fluid properties at 25oC.

Answers

Answer:

Find the attachments sequence wise for complete solution.

An extruder barrel has a diameter of 4.22 inches and a length of 75 inches. The screw rotates at 65 revolutions per minute. The screw channel depth = 0.23 in, and the flight angle = 21.4 degrees. The head pressure at the die end of the barrel is 705 lb/in2. The viscosity of the polymer melt is given as 145 x 10-4 lb·sec/in2. Calculate the volume flow rate in in3/sec of the plastic through the barrel.

Answers

Answer:

volume flow rate Q  = 53.23 in³/s

Explanation:

given data

diameter = 4.22 inches

length = 75 inches

screw rotates = 65 revolutions per minute

depth = 0.23 in

flight angle = 21.4 degrees

head pressure = 705 lb/in²

viscosity = 145 x [tex]10^{-4}[/tex] lb·sec/in²

solution

we get here volume flow rate of platstic in barrel that is express as

volume flow rate Q = volume flow rate of die - volume flow of extruder barrel   ................1

here

volume flow rate extruder barrel is

flow rate   = [tex]\frac{\pi \times 705 \times 4.22 \times 0.23\times sin21.4 }{12\times 145 \times 10^{-4}\times 75 }[/tex]    

flow rate = 60.10  

and

volume flow rate of die is express as

volume flow rate = 0.5 × π² × D² × Ndc ×  sinA × cosA   .............2

put here value and w eget

volume flow rate = 0.5 × π² × 4.22² × 0.23 × 1 × sin21.4 × cos21.4

volume flow rate = 6.866

so put value in equation 1 we get

volume flow rate Q  = 60.10 - 6.866  

volume flow rate Q  = 53.23

Compare the heat transfer coefficients for laminar forced and free convection over vertical flat plates. Develop an approximate relation between the Reynolds and Rayleigh numbers such that the heat transfer coefficients for pure forced convection and pure free convection are equal.

Answers

Answer:

Check the attached images below.

Explanation:

It Is required to develop an approximate relation between . Reynolds and Grashor numbers such that the heat-transfer coefficients for pure forced convection and pure .e convection are equal, assuming laminar flow, by comparing the heat-transfer coefficients for forced or free convection over vertical hat plates.  

write the equation or heat transfer coefficient Mr (arced comedian.

Kindly check the attached images below.

An airplane starts from rest, 6050 ft down a runway at uniform accelerationthen takes off with a speed of 150mi / h . It then climbs in a straight line with a uniform acceleration of 2 ft/s^ 2 until it reaches a constant speed of 195mi / h . How far has the plane traveled when it reaches this constant speed?

Answers

Given Information:  

distance = s₁ = 6050 ft

velocity = v₁ = 0 mi/hr

velocity = v₂ = 150 mi/hr

velocity = v₃ = 195  mi/hr

Acceleration = a = 2 ft/s²

Required Information:  

distance = s₂ = ?

Answer:

distance = s₂ = 14,399 ft

Explanation:

We know from the equations of motion,

v₃² = v₂² + 2a(s₂ - s₁)

We want to find out the distance s₂

2a(s₂ - s₁) = v₃² - v₂²

s₂ - s₁ = (v₃² - v₂²)/2a

s₂ = (v₃² - v₂²)/2a + s₁

First convert given velocities from mi/hr to ft/s

1 mile has 5280 feet and 1 hour has 3600 seconds

velocity = v₂ = 150*(5280/3600) = 220 ft/s

velocity = v₃ = 195*(5280/3600) = 286 ft/s

s₂ = (v₃² - v₂²)/2a + s₁

s₂ = (286² - 220²)/2*2 + 6050

s₂ = 33396/4 + 6050

s₂ = 8349 + 6050

s₂ = 14,399 ft

Therefore, the plane would have traveled a distance of 14,399 ft when it reaches a constant speed of 286 ft/s

ou want to amplify a bio-potential signal that varies between 2.5 V and 2.6 V. Design an amplifier circuit for this signal such that the output spans 0 V to +10 V. The signal cannot be inverted. You can use any number of op amps and any number of resistors (with any values). But you can use only one +10 V DC voltage source (for powering the op amps as well as for any other needs). Clearly draw the complete circuit and show all component values.

Answers

Answer:

See attachment

Explanation:

Gain= Vo/Vin

If we set Vout=9.62V corresponding to Vin=2.6V, then gain will be 3.7

Using above value of gain, let's design non-inverting op-amp configuration

Gain= 1+Rf/Rin

3.7= 1= Rf/Rin

2.7= Rf/Rin

If Rin=100Ω then Rf= 270Ω

Air enters the first compressor stage of a cold air-standard Brayton cycle with regeneration and intercooling at 100 kPa, 300 K, with a mass flow rate of 6 kg/s. The overall compressor pressure ratio is 10, and the pressure ratios are the same across each compressor stage. The temperature at the inlet to the second compressor stage is 300 K. The temperature at the inlet to the turbine is 1400 K. The compressor stages and turbine each have isentropic efficiencies of 80% and the regenerator effectiveness is 80%.
For k = 1.4, calculate:
(A) the thermal efficiency of the cycle.
(B) the back work ratio.
(C) the net power developed, in kW.
(D) the rates of exergy destruction in each compressor stage and the turbine stage as well as the regenerator, in kW, for T0 = 300 K.

Answers

the answer would be d because it only makes logical sense

A flexible pavement has a SN of 3.8 (all drainage coefficients are equal to 1.0). The initial PSI is 4.7 and the terminal serviceability is 2.5. The soil has a CBR of 9. The overall standard deviation is 0.40 and the reliability is 95%. The pavement is currently designed for 1800 equivalent 18-kip single-axle loads per day. If the number of 18-kip single-axle loads were to increase by 30%, by how many years would the pavement design life be reduced

Answers

Answer:

2.83 years

Explanation:

Please kindly see the attached file at thw attachment area for a detailed and step by step solution to the given problem.

Logical variables: Running late? Complete the tunction RunningLate such that the logical variable on Time is true if no Traffic is true and gasEmpty is false. Ex >> n°Traffic = true ; >> gasEmpty-false; >onTime - Runninglate(noTraffic,gasEmpty) logical Your Function B Save C Reset EE MATLAB Documentation 1 function onTime - RunningLate (noTraffic,gasEmpty) 2 % complete the logical expression to the right of . using the variables n°Traffic and gasEmpty 4 onTime - 6 end Code to call your function C Reset 1 noTraffic true; gasEmpty true; 2 onTime-Runninglate(noTraffic, gasEmpty)

Answers

Final answer:

The function RunningLate returns true for the variable onTime when noTraffic is true and gasEmpty is false, which is achieved by the logical expression noTraffic && ~gasEmpty in MATLAB.

Explanation:

The task is to complete the function RunningLate in MATLAB, which returns a logical variable onTime that is true if noTraffic is true and gasEmpty is false. The finished MATLAB code inside the function should be:

onTime = noTraffic && ~gasEmpty;

This code uses the logical AND operator (&&) to check that there is no traffic, and the NOT operator (~) to check that the gas tank is not empty.

Select the statement that is false.

A. If two graphs G and H are isomorphic, then they have the same total degree.
B. If two graphs G and H have the same degree sequence, then G and H are isomorphic.
C. If two graphs G and H have the same degree sequence, then G and H must have the same number of edges.
D. If two graphs G and H have the same number of edges then G and H must have the same total degree.

Answers

Answer:

D

Explanation:

the way vertices are connected may be different so having same number of edges do not mean that total degree will also be same.

A 3-phase stepping motor is to be used to drive a linear axis for a robot. The motor output shaft will be connected to a screw thread with a screw pitch of 1 mm. We want to be able to have a spatial control of at least 0.05 mm. a. How many poles should the motor have? b.How many pulses are needed from the controller every second to move the linear axis at a rate of 90 mm/sec?

Answers

Final answer:

The number of poles required for a 3-phase stepping motor to achieve 0.05 mm spatial control depends on the increments per revolution, which was not provided. To move the axis at 90 mm/sec, the pulses per second from the controller will be calculated based on the screw pitch and the number of increments per revolution. Additional information on the stepper motor is needed for precise calculations.

Explanation:

To answer the student's questions regarding a 3-phase stepping motor for spatial control in a robotics application:

Poles of the motor: The number of poles in a stepper motor determines the resolution of movement. To achieve a spatial control of at least 0.05 mm with a screw pitch of 1 mm, the motor needs to have enough poles to allow for a fractional turn equal to that precision. Given that stepper motors can have a range of 5,000 to 10,000 increments in a 90-degree rotation, the exact number of poles required can be calculated based on the needed increments per mm.Pulses needed for motion: To move the axis at a speed of 90 mm/sec, the number of pulses required from the controller per second will depend on the revolutions per minute (rpm) the screw thread needs to turn, which is determined by the number of poles or increments per revolution of the motor. Generally, to calculate this, we would use the formula (speed in mm/sec) / (screw pitch in mm) * (number of increments per revolution), giving us the pulses per second.

Without the specific number of increments per revolution of the stepping motor, it's impossible to provide an exact answer. However, typically stepper motors with more poles can provide finer control, and thus would be preferred in this application to meet the 0.05 mm spatial control requirement.

Consider a circular grill whose diameter is 0.3 m. The bottom of the grill is covered with hot coal bricks at 961 K, while the wire mesh on top of the grill is covered with steaks initially at 274 K. The distance between the coal bricks and the steaks is 0.20 m. Treating both the steaks and the coal bricks as blackbodies, determine the initial rate of radiation heat transfer from the coal bricks to the steaks. Also, determine the initial rate of radiation heat transfer to the steaks if the side opening of the grill is covered by aluminum foil, which can be approximated as a reradiating surface.

Answers

Answer:

Step 1

Given

Diameter of circular grill,   D = 0.3m

Distance between the coal bricks and the steaks,  L = 0.2m

Temperatures of the hot coal bricks,  T₁ = 950k

Temperatures of the steaks, T₂ = 5°c

Explanation:

See attached images for steps 2, 3, 4 and 5

The journals in a high speed oil engine are 80 mm in diameter. and 40 mm long. The radial clearance is 0.060mm. Each supports a load of 9 kN when the shaft is rotating at 3600 rpm. The bearing is lubricated with SAE 40 oil supplied at atmospheric pressure and average operating temperature is about 65oC. Using Raimondi- Boyd charts analyze the bearing under steady state operation.

Answers

Answer: S = 0.284

Explanation:

Data: d = 80 mm;

l =40 mm;

c = 0.06 mm;

F = 9kN;

n = 3600rpm = 60 rps

SAE 40 oil

T= 65°C

Therefore:

p = F / ld

= 9 x1000 /40 x 80

= 2.813 MPa

μ = 30 cp at 65°C for SAE 40 oil

S = r^2 x μ x n / c^2 x p

S= ( 40 )^2 x 30*10^-3 x 60 / (0.06)^2 x 2.813*10^6

S = 2880 / 10,126.6

S = 0.284

l/d = ½,

h o /c = 0.38

ε = e /c = 0.62

h o = 0.38 x C

= 0.382 x 0.06

=0.023mm

= 23µm

e = 0.62 x C

= 0.62 x 0.06

= 0.037 mm

Viscosity temperature curves of SAE graded oils

(r /c) f = 7.5,

S = 0.284

l /d = ½

f = 7.5 x (c / r)

= 7.5x (0.06/40)

= 0.0113

The mean of hours that the average person watches television each day is 4.18 hours with a standard deviation of 1.19 hours. Find probability that someone watches between 3 and 5 hours a day

Answers

Answer:

[tex] z = \frac{3-4.18}{1.19}=-0.992[/tex]

[tex] z = \frac{5-4.18}{1.19}=0.689[/tex]

And we can find this probability with this difference:

[tex] P(-0.992<z<0.689) = P(z<0.689) -P(z<-0.992) =0.752 -0.161=0.591[/tex]

And then we can conclude that the probability that someone watches between 3 and 5 hours a day is approximately 0.591 using a normal distribution

Explanation:

For this case we can define the random variable X as "hours that a person watches television". For this case we don't have the distribution for X but we have the following parameters:

[tex]\mu = 4.18,\sigma =1.19[/tex]

We can assume that the distribution for X is normal

[tex] X \sim N(\mu = 4.18 , \sigma =1.19)[/tex]

And we want to find this probability:

[tex] P(3 <X<5)[/tex]

And we can use the z score formula given by:

[tex] z=\frac[X- \mu}{\sigma}[/tex]

And we can find the z score for each limit and we got:

[tex] z = \frac{3-4.18}{1.19}=-0.992[/tex]

[tex] z = \frac{5-4.18}{1.19}=0.689[/tex]

And we can find this probability with this difference:

[tex] P(-0.992<z<0.689) = P(z<0.689) -P(z<-0.992) =0.752 -0.161=0.591[/tex]

And then we can conclude that the probability that someone watches between 3 and 5 hours a day is approximately 0.591 using a normal distribution

Convert 580,000 kW to [GW] with correct number of significant digits.

Answers

Answer:

[tex]x = 0.58\,GW[/tex]

Explanation:

The conversion is:

[tex]x = (580000\,kW)\cdot \left(\frac{1\,GW}{1000000\,kW} \right)[/tex]

[tex]x = 0.58\,GW[/tex]

Insulated Gas Turbine Air enters an adiabatic gas turbine at 1050 K and 1 MPa and leaves at 400 kPa. Kinetic and potential energy changes can be ignored. Treat the air as an ideal gas with constant specific heats. Let k = 1.4 a. Determine the theoretical exit temperature - corresponding to part "b" below, the maximum theoretical power output. b. Determine the maximum theoretical work output for the gas turbine in kJ/kg. c. If the isentropic turbine efficiency is 0.8, what is the actual work output of the turbine in kJ/kg?

Answers

Answer:A certain vehicle loses 3.5% of its value each year. If the vehicle has an initial value of $11,168, construct a model that represents the value of the vehicle after a certain number of years. Use your model to compute the value of the vehicle at the end of 6 years.

A certain vehicle loses 3.5% of its value each year. If the vehicle has an initial value of $11,168, construct a model that represents the value of the vehicle after a certain number of years. Use your model to compute the value of the vehicle at the end of 6 years.

Explanation:A certain vehicle loses 3.5% of its value each year. If the vehicle has an initial value of $11,168, construct a model that represents the value of the vehicle after a certain number of years. Use your model to compute the value of the vehicle at the end of 6 years.

"From the earth to the moon". In Jules Verne’s 1865 story with this title, three men went to the moon in a shell fired from a giant cannon sunk in the earth in Florida.


(a) Find the minimum muzzle speed needed to shoot a shell straight up to an altitude equal to the 2 times earth’s radius RE.


(b) Find the minimum muzzle speed that would allow a shell to reach the height of the moon, 385,000 km, center of the earth to center of the moon

Answers

Answer:

Check the explanation

Explanation:

Kindly check the attached image below to get the step by step explanation to the above question.

A 10 wt % aqueous solution of sodium chloride is fed to an evaporative crystallizer which operates at 80o C. The product is a slurry of solute crystals suspended in a saturated solution. The unit is to produce 1000 kg/crystals per hour. If the pump that handles the slurry cannot handle more than 40 wt % solids, find the feed rate to the crystallizer and the evaporation rate of the water. (20 pts)

Answers

Answer: The feed rate is

17,020kg/he and the rate is 13,520kg/h

Check the attachment for step by step explanation

Consider a 50 x 106 m3 lake fed by a polluted stream with a flow rate of 75 m3 /s and a pollutant concentration of 25.5 mg/L. There is also a sewage outfall that discharges 3.0 m3 /s of wastewater with a pollutant concentration of 125 mg/L. Stream and sewage wastes have a reaction rate coefficient of 5% per day. Find the steady-state (i.e., effluent) pollutant concentration and flow rate

Answers

Answer:

24.78 mg/L

Explanation:

The step-to-step explanation is written legibly with clear explanation in the diagram attached below.

Answer:

Answer: Input rate = 2.288 x 10⁶mg/s

             Output rate =78 x 10³Cmg/s

            Decay rate = 28.94 x 10 ⁵C mg/s

Explanation:

Assuming that complete and instantaneous mixing occurs in the lake, this implies that the concentration in the lake C is the same as the concentration of the mix leaving the lake Cm

   Input rate = Output rate  + KCV

Input rate = Q₁C₁ + QwCw

                = (75.0m³/s x 25.5mg/L + 3.0m³/s x 125.0mg/L) x 10³L/m³

                = 2.288 x 10⁶mg/s

Output rate = QmCm = (Q₁ +Qw)C

                   =(75 + 3.0)m³/s x Cmg/L x 10³L/m³ = 78 x 10³Cmg/s

Decay rate = KCV = 5/d x Cmg/L x 50 x 10⁶ x 10³L/m³  

                                           24 hr/d x 3600s/hr

                     = 28.94 x 10 ⁵C mg/s

So,

                      =2.288 x 10⁶ = 78 x 10³C + 28.94 x 10 ⁵C  = 29.72 X 10⁵C

                      = 2.288 x 10⁶          

                        29.72 X 10⁵         = 0.77 mg/l

                   C =

Consider schedules S3, S4, and S5 below. Determine whether each schedule is
strict, cascadeless, recoverable, or nonrecoverable. (Determine the strictest
recoverability condition that each schedule satisfies.)
S3: r1 (X); r2 (Z); r1 (Z); r3 (X); r3 (Y); w1 (X); c1; w3 (Y); c3; r2 (Y); w2 (Z);
w2 (Y); c2;
S4: r1 (X); r2 (Z); r1 (Z); r3 (X); r3 (Y); w1 (X); w3 (Y); r2 (Y); w2 (Z); w2 (Y); c1;
c2; c3;
S5: r1 (X); r2 (Z); r3 (X); r1 (Z); r2 (Y); r3 (Y); w1 (X); c1; w2 (Z); w3 (Y); w2 (Y);
c3; c2;

Answers

Answer:

A schedule is strict if it satisfies the following conditions:

Tj reads a data item X after Ti has written to X and Ti is terminated means aborted or committed.

Tj writes a data item X after Ti has written to X and Ti is terminated means aborted or committed.

Explanation:

See attached image

(a) Determine the temperature of the insulated walls. (b) Determine the net radiation heat rate from surface 2 per unit conduit length. 13.48 A long conduit is constructed with diffuse, gray walls 0.5 m wide. The top and bottom of the conduit are insulated. The emissivities of the walls are ε 1 = 0.45, ε 2 = 0.65, and ε 3 = 0.15, respectively, while the temperatures of walls 1 and 2 are 500 K and 700 K, respectively.

Answers

Answer:

Explanation:

the solution to the problem is given in the pictures attached. (b) is answered first then (a). I hope the explanation helps you.Thank you

2.(10 pts)A proposed engine cycle employs an ideal gas and consists of the following sequence of transformations; a) Isothermal compression at 300 o K from a pressure of 1bar to a pressure of 30bar b) Constant pressure heating to a temperature of 1600 o K. c) Isothermal expansion at 1600 o K to the original pressure of 1 bar. d) Constant pressure cooling to a temperature of 300 o K to complete the cycle An ideal regenerator connects d) to b) so that the heat given up in d) is used for the heating in b). For an engine using a kilamole of gas find the net work in kJ and the thermal efficiency. You may assume C p

Answers

Answer:

Check the explanation

Explanation:

For ideal regeneration heat loss in cooling aqual to heat gain in compression so temperature Tb=Td as  can be seen in the step by step solution in the attached images below.

An n- channel enhancement- mode MOSFET with 50 nm thick HfO2 high- k gate dielectric (Pr = 25) has a flat band voltage of 0.5 V, and substrate doping of 1018 cm-3. The intrinsic carrier concentration is 1011 cm-3, effective electron channel mobility is 250 cm2/V­s, and Pr = 15. What is the drive current for a 50 om wide and 2 om long device at VG = 3 V and VD = 0.05 V? What is the saturation current at this gate bias?

Answers

Answer:

Find the complete solution in the given attachments

(30 pts) A simply supported beam with a span L=20 ft and cross sectional dimensions: b=14 in; h=20 in; d=17.5 in. is reinforced with tension steel As=5 in2 . The beam supports a uniformly distributed dead load (including its own weight) DL=2.2 kips/ft and a uniformly distributed live load LL=1.8 kips/ft. The properties of the materials are as follows: f’c=4000 psi, steel fy=60,000 psi. Calculate the long-term deflections in the beam after five years.

Answers

Answer:

Zx = 176In³

Explanation:

See attached image file

"Design a sequential circuit with two T flip-flops A and B, and one input x. When x = 0, the circuit remains in the same state. When x = 1, the circuit goes through the state transitions from 00 to 01, to 10, to 11, back to 00, and repeats. What is the boolean equation of the input of flip-flop A (TA)? Type variables and operations without blanks."

Answers

Answer:

See attached images for the diagrams and tables

A piece of corroded steel plate was found in a submerged ocean vessel. It was estimated that the original area of the plate was 17 in.2 and that approximately 2.1 kg had corroded away during the submersion. Assuming a corrosion penetration rate of 200 mpy for this alloy in seawater, estimate the time of submersion in years. The density of steel is 7.9 g/cm3.

Answers

Answer:

4.8 years

Explanation:

The rate of corrosion (CPR) is defined as the rate at which a metal corrodes in a specific environment. It depends on the environmental condition and the type of metal. It is expressed in inches per year or melts per year. CPR is given by:

[tex]CPR=\frac{KW}{\rho At}[/tex]

Where K is a constant = 534, W is the weight corroded = 2.1 kg = 2.1 × 10⁶mg, A is the area = 17 in², ρ is the density = 7.9 g/cm³

From the CPR equation:

[tex]t=\frac{KW}{\rho A(CPR)}=\frac{534*2.1*10^6}{7.9*17*200}= 4.17*10^4 hrs=4.8years[/tex]

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