A Diesel cycle engine is analyzed using the air standard method. Given the conditions at state 1, compression ratio (r), and cutoff ratio (rc) determine the efficiency and other values listed below.
Note: The gas constant for air is R=0.287 kJ/kg-K.

--Given Values--
T1 (K) = 306
P1 (kPa) = 140
r = 10
rc = 1.75

a) Determine the specific internal energy (kJ/kg) at state 1.
Your Answer =
b) Determine the relative specific volume at state 1.
Your Answer =
c) Determine the relative specific volume at state 2.
Your Answer =
d) Determine the temperature (K) at state 2.
Your Answer =
e) Determine the pressure (kPa) at state 2.
Your Answer =
f) Determine the specific enthalpy (kJ/kg) at state 2.
Your Answer =
g) Determine the temperature (K) at state 3.
Your Answer =
h) Determine the pressure (kPa) at state 3.
Your Answer =
i) Determine the specific enthalpy (kJ/kg) at state 3.
Your Answer =
j) Determine the relative specific volume at state 3.
Your Answer =
k) Determine the relative specific volume at state 4.
Your Answer =
l) Determine the temperature (K) at state 4.
Your Answer =
m) Determine the pressure (kPa) at state 4.
Your Answer =
n) Determine the specific internal energy (kJ/kg) at state 4.
Your Answer =
o) Determine the net work per cycle (kJ/kg) of the engine.
Your Answer =
p) Determine the heat addition per cycle (kJ/kg) of the engine.
Your Answer =
q) Determine the efficiency (%) of the engine.
Your Answer =

help A through Q

Answer:

Code fixed below using Java

Explanation:

Error.java

import java.util.Random;

public class Error {

   public static void main(String[] args) {

       final int MAXCHEESE = 10;

       String[] names = new String[MAXCHEESE];

       double[] prices = new double[MAXCHEESE];

       double[] amounts = new double[MAXCHEESE];

       // Three Special Cheeses

       names[0] = "Humboldt Fog";

       prices[0] = 25.00;

       names[1] = "Red Hawk";

       prices[1] = 40.50;

       names[2] = "Teleme";

       prices[2] = 17.25;

       System.out.println("We sell " + MAXCHEESE + " kind of Cheese:");

       System.out.println(names[0] + ": $" + prices[0] + " per pound");

       System.out.println(names[1] + ": $" + prices[1] + " per pound");

       System.out.println(names[2] + ": $" + prices[2] + " per pound");

       Random ranGen = new Random(100);

       // error at initialising i

       // i should be from 0 to MAXCHEESE value

       for (int i = 0; i < MAXCHEESE; i++) {

           names[i] = "Cheese Type " + (char) ('A' + i);

           prices[i] = ranGen.nextInt(1000) / 100.0;

           amounts[i] = 0;

           System.out.println(names[i] + ": $" + prices[i] + " per pound");

       }        

   }

}

Answer:

The curve length (L) will be = 1218 ft

The elevations and stations for PVC and PVI

a. station of PVC = 103 + 91.00

b. station of PVI = 116 + 09.00

c. elevation of PVC = 432.18ft

d. elevation of PVI = 426.09ft

Explanation:

First calculate for the length (L)

To calculate the length, use the formula of "elevation at any point".

where, elevation at any point = 424.5.

and ∴ PVC Elevation = (420 + 0.01L)

Then, calculate for Station of PVC and PVI and elevation of PVC and PVI

Answer:

3-Drip edges prevent water from running underneath an overhang.

Explanation:

The only correct statement is in option 3. Drip edges are structures that are connected to the roof edges of buildings to ensure that the flow of water is properly controlled. They are typically used yo prevent water from getting to other parts of the building. They are made of non-corrosive and non-staining materials to make roofs of buildings beautiful.

Answer:

Drip edges prevent water from running underneath an overhang.

Explanation:

Answer:

Explanation:

Check attachment for solution

a. The value of R2 required for the desired no-load output voltage R2 = 10 kΩ

b. The most different output voltage is 5.8 V, which occurs when the load resistance is 300 kΩ.

c. The smallest resistor value that can be used for R1 is 45 kΩ.

Part A:

To find the value of R2 required for the desired no-load output voltage, we can use the following voltage divider equation:

vo = vs * R2 / (R1 + R2)

Substituting in the known values, we get:

6 V = 18 V * R2 / (50 kΩ + R2)

Solving for R2, we get:

R2 = 18 V * 6 V - 6 V * 50 kΩ / 6 V - 18 V

R2 = 10 kΩ

Part B:

To find the output voltage that is the most different from the design output voltage, we need to calculate the output voltage for each load resistance. We can use the following voltage divider equation:

vo = vs * (R2 / (R1 + R2) || RL)

Substituting in the known values for each load resistance, we get:

vo (RL = 300 kΩ) = 18 V * (10 kΩ / (50 kΩ + 10 kΩ) || 300 kΩ)

vo (RL = 300 kΩ) = 5.8 V

vo (RL = 200 kΩ) = 18 V * (10 kΩ / (50 kΩ + 10 kΩ) || 200 kΩ)

vo (RL = 200 kΩ) = 7.2 V

vo (RL = 100 kΩ) = 18 V * (10 kΩ / (50 kΩ + 10 kΩ) || 100 kΩ)

vo (RL = 100 kΩ) = 8.6 V

Therefore, the output voltage that is the most different from the design output voltage is 5.8 V, which occurs when the load resistance is 300 kΩ.

Part C:

To change the values of R1 and R2 so that the design output voltage vo = 6 V is achieved when the load resistance is RL = 200 kΩ rather than at no-load, we can use the following voltage divider equation:

vo = vs * R2 / (R1 + R2)

Substituting in the known values, we get:

6 V = 18 V * R2 / (R1 + 200 kΩ)

Solving for R2, we get:

R2 = 30 kΩ

Next, we need to calculate the value of R1 to ensure that the actual output voltage does not drop below 5.4 V when RL = 100 kΩ. We can use the following voltage divider equation:

vo = vs * (R2 / (R1 + R2) || RL)

Substituting in the known values, we get:

5.4 V = 18 V * (30 kΩ / (R1 + 30 kΩ) || 100 kΩ)

Solving for R1, we get:

R1 = 45 kΩ

Therefore, the smallest resistor value that can be used for R1 is 45 kΩ.

For such more question on voltage

https://brainly.com/question/28632127

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Complete Question

The complete question is shown on the first uploaded image

Answer:

(a) the pressure drop is [tex]\Delta P_L = 100.185\ kPa[/tex]

(b) the head loss is the [tex]h_L =10.22\ m[/tex]

(c) the pumping power requirement to overcome this pressure drop [tex]W_p = 901.665\ W[/tex]

Explanation:

So the question we are told that the water is at a temperature of 15°C

 And also we are told that the density of the water is [tex]\rho=999.1\ kg/m^3[/tex]

 We are also told that the dynamic viscosity of the water is                                        [tex]\mu = 1.138 *10^{-3} kg/m \cdot s[/tex]

   From the diagram the length of the pipe is [tex]L=[/tex] 30 m    

  The diameter is given as [tex]D=[/tex] 5 cm [tex]\frac{5}{100} m = 0.05m[/tex]

   The volumetric flow rate is given as  [tex]Q=[/tex] 9 L/s [tex]= \frac{9}{1000} m^3/ s = 0.009\ m^3/s[/tex]

Now the objective of this solution is to obtain

      i the pressure drop,  ii the head loss iii  the pumping power requirement to overcome this pressure drop

to obtain this we need to get the cross-sectional area of the pipe which will help when looking at the flow analysis

       So mathematically the cross-sectional area is

                               [tex]A_s =\frac{\pi}{4} D^2[/tex]

                                    [tex]= \frac{\pi}{4} (5*10^{-2})[/tex]

                                   [tex]=1.963*10^{-3} m^2[/tex]

Next thing to do is to obtain average flow velocity which is mathematically represented as

                 [tex]v = \frac{Q}{A_s}[/tex]

                 [tex]v =\frac{9*10^{-3}}{1.963*10^{-3}}[/tex]

                  [tex]=4.585 \ m/s[/tex]

 Now to determine the type of flow we have i.e to know whether it a laminar flow , a turbulent flow or an intermediary flow

  We use the Reynolds number

if it is below [tex]4000[/tex] then it is a laminar flow but if it is higher then it is a turbulent flow ,now when it is exactly the value then it is an intermediary flow

     This Reynolds number is mathematically represented as

                           [tex]Re = \frac{\rho vD}{\mu}[/tex]

                                [tex]=\frac{(999.1)(4.585)(0.05)}{1.138*10^{-3}}[/tex]

                               [tex]=2.0127*10^5[/tex]

Now since this number is greater than 4000 the flow is turbulent

So we are going to be analyse the flow using the Colebrook's  equation which is mathematically represented as

                [tex]\frac{1}{\sqrt{f} } =-2.0\ log [\frac{\epsilon/D_h}{3.7} +\frac{2.51}{Re\sqrt{f} } ][/tex]

 Where f is the friction  factor , [tex]\epsilon[/tex] is the surface roughness ,

Now generally the surface roughness for stainless steel is

                                             [tex]\epsilon = 0.002 mm = 2*10^{-6} m[/tex]

Now substituting the values into the equation we have

                                   [tex]\frac{1}{\sqrt{f} } =-2.0\ log [\frac{2*10^{-6}/5*10^{-2}}{3.7} +\frac{2.51}{(2.0127*10^{5})\sqrt{f} } ][/tex]

So solving to obtain f we have

                                  [tex]\frac{1}{\sqrt{f} } =-2.0\ log [\frac{4*10^{-5}}{3.7} +\frac{2.51}{(2.0127*10^{5})\sqrt{f} } ][/tex]

                                 [tex]= -2.0log[1.0811 *10^{-5} +\frac{1.2421*10^{-5}}{\sqrt{f} } ][/tex]

                              [tex]f = 0.0159[/tex]

Generally the pressure drop is mathematically represented as

                                            [tex]\Delta P_L =f\ \frac{L}{D} \ \frac{\rho v^2}{2}[/tex]

Now substituting values into the equation

                                           [tex]= 0.0159 [\frac{30}{5*10^{-5}} ][\frac{(999.1)(4.585)}{2} ][/tex]

                                           [tex]=100.185 *10^3 Pa[/tex]

                                           [tex]=100.185 kPa[/tex]  

Generally the head loss in the pipe is mathematically represented as

                        [tex]h_L =\frac{\Delta P_L}{\rho g}[/tex]

                            [tex]= \frac{100.185 *10^3}{(999.1)(9.81)}[/tex]

                             [tex]=10.22m[/tex]

Generally the power input required to overcome this pressure drop is mathematically represented as

                     [tex]W_p = Q \Delta P_L[/tex]

                         [tex]=(9*10^{-3}(100.185*10^3))[/tex]

                         [tex]= 901.665\ W[/tex]

Answer:

0.028 kg per Seconds; 8.4

Explanation:

N-S, 4 x 100 = 400 m^2, concentration 5/400= 0.0125 kg/s

E-W 4 x 100 =400 m^2 , => 10/400 =0.025 kg/s

(0.0125^2 + 0.025^2)^(1/2) =0.028 kg/s

in 5 minutes = 5 x 60 x 0.028 =8.4

Answer:

Explanation:

It is generally known that the thermal stresses developed during thermal cycle results into cracking, and these thermal stresses are due to temperature gradients .

Stresses will be equivalently lower for a particular temperature gradient when the thermal expansion is low.

It also known that there will be a reduction in the temperature gradient if the thermal conductivity is high, as heat is dissipated faster and more equally and  with it, as well as  when deformation takes place due to thermal stresses, cracking occurs but if the ductility is high, more deformation will be allowed without cracking and thus reduces the tendency for cracking.

Thermal expansion and conductivity influence cracking during thermal cycling because materials expand or contract at different rates causing stress. Differences in these properties between bonded materials or different parts of a structure can lead to cracks. Understanding these properties is important in designing materials to minimize thermal stress.

The two physical properties that have a major influence on the cracking of workpieces, tools, or dies during thermal cycling are thermal conductivity and thermal expansion. Thermal expansion occurs due to the tendency of a material to change in volume in response to a change in temperature. When different parts of a workpiece, or different materials bonded together, have different thermal expansion coefficients or varied thermal conductivities, thermal stress can result as the materials expand or contract at different rates. This stress leads to the formation of cracks, especially if the material is rigid and cannot accommodate the stress through deformation.

For example, metal implants in the body may need replacement because of the lack of bonding between metal and bone due to different expansion coefficients. Similarly, the expansion of fillings in teeth can be different from that of tooth enamel, causing discomfort or damage. The understanding of thermal expansion and conductivity is critical in designing materials and structures, such as railroad tracks and roadways with adequate expansion joints to prevent buckling, or using materials like Pyrex for cooking pans to minimize cracking from thermal stress.

The calculated resulting strain of the metal is 0.0002329

How to determine the resulting strain

To calculate the resulting strain [tex](\(\epsilon\))[/tex], we can use Hooke's Law

This is represented as

[tex]\[\epsilon = \frac{\sigma}{E}\][/tex]

Where:

-[tex]\(\sigma\)[/tex] is the stress

- E is the elastic modulus

From the question, we have

Force F = 2650 N

The cross-sectional area is calclated as

[tex]\(A\) = 10.5 mm x 13.7 mm[/tex]

This gives

[tex]A = \(143.85 \times 10^{-6}\) m\(^2\)[/tex] (converted to [tex]m\(^2\)[/tex])

Calculating the stress [tex](\(\sigma\))[/tex] using the formula:

[tex]\[\sigma = \frac{F}{A}\][/tex]

Substitute the given values:

[tex]\[\sigma = \frac{2650}{143.85 \times 10^{-6}} = \frac{2650}{143.85 \times 10^{-6}} = 18422118.06 \text{ Pa}\][/tex]

Next, we have

[tex]\[\epsilon = \frac{\sigma}{E} = \frac{18422118.06}{79 \times 10^9} = \frac{18422118.06}{79 \times 10^9} = 0.0002329\][/tex]

Hence, the resulting strain is 0.0002329

The power that must be supplied to the motor is 136 hp

Explanation:

Given-

weight of the elevator, m = 1000 lb

Force on the table, F = 500 lb

Distance, s = 27 ft

Efficiency, ε = 0.65

Power  = ?

According to the equation of motion:

F = ma

[tex]3(500) - 1000 = \frac{1000}{32.2} * a[/tex]

a = 16.1 ft/s²

We know,

[tex]v^2 - u^2 = 2a (S - So)\\\\v^2 - (0)^2 = 2 * 16.1 (27-0)\\\\v = 29.48m/s[/tex]

To calculate the output power:

Pout = F. v

Pout = 3 (500) * 29.48

Pout = 44220 lb.ft/s

As efficiency is given and output power is known, we can calculate the input power.

ε = Pout / Pin

0.65 = 44220 / Pin

Pin = 68030.8 lb.ft/s

Pin = 68030.8 / 500 hp

     = 136 hp

Therefore, the power that must be supplied to the motor is 136 hp

To calculate the power supplied to the motor, we use the work done by the motor and the motor efficiency. However, without the time, it takes to lift the elevator, a crucial piece of information is missing, thus preventing an exact calculation.

To determine the power that must be supplied to the motor at the instant the 1000-lb elevator has been hoisted 27 ft from rest, we must first calculate the useful power output. Since the motor has an efficiency of 0.65 and exerts a constant force of 500 lb, we can use the work-energy principle along with the efficiency to find the required power.

The work done by the motor on the elevator is the force applied times the distance moved. The useful work is W = force × distance = 500 lb × 27 ft.

The useful power output, or work done per unit time, can be found by P = W / t. However, the time is not provided directly in this scenario; we assume this process occurs instantaneously, which is not realistic for real-world scenarios.

Since power supplied is equal to the useful power output divided by the efficiency, we have Power supplied = Useful power / ε.

Due to missing time information, the exact numerical answer cannot be provided without assumptions or additional data concerning how quickly the elevator is lifted.

To find the work done during the polytropic expansion of water in a piston-cylinder assembly, the initial and final volumes are needed to apply the polytropic process work formula. Without this information, it's not possible to calculate the energy transfer by work.

The question asks to determine the energy transfer by work during a polytropic process where pressure p and specific volume v are related by the equation pv1.2 = constant. To find the work done by the water when it expands in a piston-cylinder assembly, you would typically use the polytropic process work formula:

W = (P1V1 - P2V2) / (n - 1)

However, to use this formula, you would need the initial and final volumes, V1 and V2, which are not provided in the question. Without these volumes or additional information such as tables or diagrams that could help find these values through the relationship of states, it is impossible to provide a numerical answer to the question asked.

Answer:

Explanation: see attachment below

Answer:

2. The decrease in pressure will be 8psia

3. The NWS will report the pressure to be 0.0657inchHg

Explanation:

2. This is an isothermal system because it was only the volume of the gas that was increased, and temperature was constant. To find the pressure, use the Boyle's law equation below.

P1V1 ÷ P2V2..............(1)

make P2 the subject of the formula

P2 = P1V1 ÷ V2 ..............(2)

P1 = 12psia

V1 = 10ft3

V2 = 15ft3

Using equation 2

P2 = (12 × 10) ÷ 15 = 8psia

Therefore, the resulting pressure of the gas will decrease to 8psia, which obeys Boyle's law that states that pressure is inversely proportion to volume in a given mass of gas. That means, as volume increases pressure should decrease.

3. The National weather Service report, are the national agency that forecast the weather, to report how the climate will change. To report the pressure of a weather, their use a pressure barometer, and reports the reading in the pressure barometer, which is the hight the Mercury in the barometer increased to.

Using a pressure barometer.

Patm = pgh..............(3)

Make h the subject of formula

h = Patm ÷ pg ..............(4)

Patm is the Atmospheric pressure = 4.3psi

p is the density of Mercury = 13.6

g is the force of gravity = 9.8

h is the hight of the Mercury in the barometer, which is the pressure that is measured.

Using equation 4

h = 4.3 ÷ (13.6 × 9.8) = 0.0322629psi

To convert Pound- force per square inch to inch of mecury (that's is psi to inchHg)

1psi = 2.03602inchHg

Therefore

0.0322629psi = 0.065689399inchHg

The NWS will report the pressure to be 0.0657inchHg

The codes for the files are:

File: Artist.py

This file will contain the Artist class with a constructor initializing the artist's name, birth year, and death year.

# Artist. py

class Artist:

   def __init__(self, name="None", birth_year=0, death_year=0):

       self.name = name

       self. birth_year = birth_year

       self. death_year = death_year

File: Artwork.py

This file will define the Artwork class and import the Artist class from Artist.py. It initializes the artwork's title, year of creation, and artist.

# Artwork. py

from Artist import Artist

class Artwork:

   def __init__(self, title="None", year_created=0, artist=Artist()):

       self. title = title

       self. year_created = year_created

       self. artist = artist

File: main. py

This file will import both Artist and Artwork classes and create instances of them to demonstrate their usage.

# main. py

from Artist import Artist

from Artwork import Artwork

# Creating an instance of Artist

artist1 = Artist("Vincent van Gogh", 1853, 1890)

# Creating an instance of Artwork using the artist instance

artwork1 = Artwork("Starry Night", 1889, artist1)

# Printing artist and artwork details to verify

print(f"Artist: {artwork1. artist. name}, Born: {artwork1. artist. birth_year}, Died: {artwork1. artist. death_year}")

print(f"Artwork: {artwork1. title}, Created in: {artwork1. year_created}")

Execution and Output

When you run the main.py script, it will create an Artist object for Vincent van Gogh and an Artwork object for one of his most famous pieces, "Starry Night". The script will then print the details of both the artist and the artwork.

Expected Output:

Artist: Vincent van Gogh, Born: 1853, Died: 1890

Artwork: Starry Night, Created in: 1889

Answer:

import java.awt.*;

import java.util.*;

import javax.swing.*;

/**

An icon that contains a moveable shape.

*/

public class ShapeIcon implements Icon

{

private int width;

private int height;

private MoveableShape shape;

private ArrayList <MoveableShape> cars;

public ShapeIcon(MoveableShape shape,

int width, int height)

{

this.shape = shape;

this.width = width;

this.height = height;

cars = new ArrayList<MoveableShape>();

}

public int getIconWidth()

{

return width;

}

public int getIconHeight()

{

return height;

}

public void paintIcon(Component c, Graphics g, int x, int y)

{

Graphics2D g2 = (Graphics2D) g;

shape.draw(g2);

}

}

import java.awt.*;

import java.awt.event.*;

import javax.swing.*;

/**

This program implements an animation that moves

a car shape.

*/

public class AnimationTester

{

private static final int ICON_WIDTH = 400;

private static final int ICON_HEIGHT = 100;

private static final int CAR_WIDTH = 100;

public static void main(String[] args)

{

JFrame frame = new JFrame();

final MoveableShape shape

= new CarShape(0, 0, CAR_WIDTH);

ShapeIcon icon = new ShapeIcon(shape,

ICON_WIDTH, ICON_HEIGHT);

final JLabel label = new JLabel(icon);

frame.setLayout(new FlowLayout());

frame.add(label);

frame.setDefaultCloseOperation(JFrame.EXIT_ON_CLOSE);

frame.pack();

frame.setVisible(true);

final int DELAY = 10; //100

// Milliseconds between timer ticks

Timer t = new Timer(DELAY, new

ActionListener()

{

public void actionPerformed(ActionEvent event)

{

shape.translate(1, 0);

label.repaint();

}

});

t.start();

}

}

import java.awt.*;

import java.awt.geom.*;

import java.util.*;

/**

A car that can be moved around.

*/

public class CarShape implements MoveableShape

{

private int x;

private int y;

private int width;

/**

Constructs a car item.

@param x the left of the bounding rectangle

@param y the top of the bounding rectangle

@param width the width of the bounding rectangle

*/

public CarShape(int x, int y, int width)

{

this.x = x;

this.y = y;

this.width = width;

}

public void translate(int dx, int dy)

{

// x += dx;

// y += dy;

if (x <= 400)

x += dx;

else

x = 0;

y += dy;

}

public void draw(Graphics2D g2)

{

Rectangle2D.Double body

= new Rectangle2D.Double(x, y + width / 6,

width - 1, width / 6);

Ellipse2D.Double frontTire

= new Ellipse2D.Double(x + width / 6, y + width / 3,

width / 6, width / 6);

Ellipse2D.Double rearTire

= new Ellipse2D.Double(x + width * 2 / 3, y + width / 3,

width / 6, width / 6);

// The bottom of the front windshield

Point2D.Double r1

= new Point2D.Double(x + width / 6, y + width / 6);

// The front of the roof

Point2D.Double r2

= new Point2D.Double(x + width / 3, y);

// The rear of the roof

Point2D.Double r3

= new Point2D.Double(x + width * 2 / 3, y);

// The bottom of the rear windshield

Point2D.Double r4

= new Point2D.Double(x + width * 5 / 6, y + width / 6);

Line2D.Double frontWindshield

= new Line2D.Double(r1, r2);

Line2D.Double roofTop

= new Line2D.Double(r2, r3);

Line2D.Double rearWindshield

= new Line2D.Double(r3, r4);

g2.draw(body);

g2.draw(frontTire);

g2.draw(rearTire);

g2.draw(frontWindshield);

g2.draw(roofTop);

g2.draw(rearWindshield);

Explanation:

Answer:

The original length of the specimen is found to be 76.093 mm.

Explanation:

From the conservation of mass principal, we know that the volume of the specimen must remain constant. Therefore, comparing the volumes of both initial and final state as state 1 and state 2:

Initial Volume = Final Volume

πd1²L1/4 = πd2²L2/4

d1²L1 = d2²L2

L1 = d2²L2/d1²

where,

d1 = initial diameter = 19.636 mm

d2 = final diameter = 19.661 mm

L1 = Initial Length = Original Length = ?

L2 = Final Length = 75.9 mm

Therefore, using values:

L1 = (19.661 mm)²(75.9 mm)/(19.636 mm)²

L1 = 76.093 mm

Answer:

potential energy=ε Qq/r. Putting values of charge along with sign and calculating the potential energy will tell whether the resultant force is attractive or repulsive.

Explanation:

potential energy=ε Qq/r

q and Q are charges on each particle

So in case the particles have attractive force between them, one particle will be negative and the other will be positiive. The result will be negative.

In case the particles have repulsive forces, the charge on both particles will be either positive or negative. The result will be positive

It appears that your answer contains either a link or inappropriate words. Please correct and submit again! error

Had to screenshot the solution check attached

Answer:

The surface energy of a single crystal depends on the planar density (i.e., degree of atomic packing) of the exposed surface plane because of the number of unsatisfied bonds. As the planar density increases, the number of nearest atoms in the plane increases, which results in an increase in the number of satisfied atomic bonds in the plane, and a decrease in the number of unsatisfied bonds. Since the number of unsatisfied bonds diminishes, so also does the surface energy decrease. (That is, surface energy decreases with an increase in planar density.)

Explanation:

Answer:

A) C = 1/120

B) P( x ≥ 5) = 1/3

Explanation:

I've attached explanations to this.

The value of C is: C = 1/126

The probability that X will be at least 5 is 1/21.

To determine the value of C, we need to make sure that the sum of the probabilities of all possible values of X is equal to 1. This means that the following equation must hold:

\sum_{x=0}^7 P(X = x) = 1

Substituting in the given expression for P(X = x), we get:

\sum_{x=0}^7 C(x + 1)(8 - x) = 1

Expanding the summation and simplifying, we get:

7C + 21C + 35C + 35C + 21C + 7C = 1

126C = 1

Therefore, the value of C is:

C = 1/126

To find the probability that X will be at least 5, we need to calculate the following sum:

P(X \ge 5) = P(X = 5) + P(X = 6) + P(X = 7)

Substituting in the expression for P(X = x), we get:

P(X \ge 5) = C(5 + 1)(8 - 5) + C(6 + 1)(8 - 6) + C(7 + 1)(8 - 7)

P(X \ge 5) = C(6)(3) + C(7)(2) + C(8)(1)

Substituting in the value of C, we get:

P(X \ge 5) = (1/126)(6)(3) + (1/126)(7)(2) + (1/126)(8)(1)

P(X \ge 5) = 1/21

Therefore, the probability that X will be at least 5 is 1/21.

For such more question on probability

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Answer:

Vaporization is the process by which a substance changes from its solid or liquid state to a gaseous state.

Since both liquids are of the same volume and are placed under the same temperature condition, for them to not to vaporize at the same time, they must have been in different containers.

For vaporization to take place, the volume of liquid, amount of air exposure and area of the surface must be considered.

Maybe the first liquid was in a dish which has a large opening, thereby exposing a large amount which can make water to evaporate faster, whereas the second liquid was somehow enclosed (in a deeper dish).

Answer:

The output will be (3, 4) becomes (8, 10)

Explanation:

#include <stdio.h>

//If you send a pointer to a int, you are allowing the contents of that int to change.

void CoordTransform(int xVal,int yVal,int* xNew,int* yNew){

*xNew = (xVal+1)*2;

*yNew = (yVal+1)*2;

}

int main(void) {

int xValNew = 0;

int yValNew = 0;

CoordTransform(3, 4, &xValNew, &yValNew);

printf("(3, 4) becomes (%d, %d)\n", xValNew, yValNew);

return 0;

}

Answer:

A. [tex]9\pi h^2 - \pi h^3 -90 = 0\\\\[/tex]

B. h1 = 2.0813772719

    h2 = 2.0272465228

    h3 = 2.0269057423

Explanation:

V = πh^2 x [3R − h]/3

Given v=30, R=3

[tex]30 = \pi h^2 [\frac{9-h}{3} ]\\\\9\pi h^2 - \pi h^3 -90 = 0\\\\[/tex]

B. An initial guess within the interval [0; 6] is selected,

initial guess h0 = 1.5

Newton method  x₂ = x₁ - f'(x₁)/f(x₁)

           h₂ = h₁ - f'(x₁)/f(x₁)

            [tex]h_2 = h_1 - f'(x_1)/f(x_1)\\\\h_2 = h_1 - \frac{18\pi h-3\pi h^2}{9\pi h^2 - \pi h^3 -90}[/tex]            

h1 = 2.0813772719

h2 = 2.0272465228

h3 = 2.0269057423

Answer:

Th = 40.91 C

Explanation:

We must use the equation for heat current in conduction

[tex]H = \frac{kA(Th-Tc)}{L}[/tex]

                         Where k is the thermal conductivity of the material

                                     A is the cross-sectional area

                                     (Th -Tc) is the temperature difference

                                  and L is the length of the material

Thus

[tex]A = 2\pi r*0.1 = 0.00157079 m^{2}[/tex]

[tex]50 = \frac{5*0.00157079(Th-25)}{0.0025}[/tex]

Isolating Th, we have

[tex]Th = \frac{50*0.0025}{5*0.00157079}+25[/tex]

Th = 40.91 C

The question relates to estimating the temperature of a heater based on heat conduction, but it lacks the necessary details about the heater's material properties to provide a numerical answer.

The student is asking about the temperature reached by an electrical heater when dissipating power. To estimate this temperature, one must consider the concept of heat conduction in which the rate of heat transfer (Q/t) is influenced by the thermal conductivity (k) of the material, surface area (A), thickness (d), and temperature difference across the material. In this problem, the heater dissipates 50 W of power, the length of the heater is 100 mm, the diameter is 5 mm, and the thermal conductivity of the surrounding material is 5 W/m·K. To calculate the temperature reached by the heater, we would typically use the equation for steady-state heat transfer. However, the question lacks sufficient details such as the properties of the heater material, which would be necessary to determine the temperature distribution and to estimate the temperature reached by the heater. Therefore, with the information provided, we can only discuss the factors involved in calculating the temperature reached by the heater but cannot provide a numerical answer.

Answer:

a The probability Needed is 0.424

b1 The probability Needed is 0.567

b2 The median of the life time distribution is less than 28 cause its probability is less than the probability of 28

c The Needed life time is 70 weeks

d The needed lifetime is t = 77 weeks

Explanation:

Let A represent the life time of the transistor in weeks and follows a gamma distribution with parameters ([tex]\alpha,\beta[/tex]).

Looking at what we are give we can say that

                 E(A) = 28 weeks and  [tex]\sigma_A[/tex]  = 14 weeks

Hence the mean of the gamma distribution is E(A) = [tex]\alpha\beta ------(1)[/tex]

and the variance of the gamma distribution is V(A) = [tex]\alpha\beta^2 ------(2)[/tex]

Looking at Equation 2

              [tex]\alpha\beta^2 =V(A)[/tex]

              [tex](\alpha\beta)\beta = V(A)[/tex]

             [tex]E(A)\beta = V(A)[/tex]

                     [tex]\beta = \frac{V(A)}{E(A)}[/tex]

[tex]Note:\alpha\beta = E(A)\\\ and \ \sigma_A^2 = V(A)[/tex]

                    [tex]\beta = \frac{\sigma_A^2}{E(A)}[/tex]

                    [tex]\beta = \frac{14^2}{28} = \frac{196}{28} = 7[/tex]

Looking at Equation 1

               E(A)  = [tex]\alpha \beta[/tex]

                  28 = [tex]\alpha (7)[/tex]

          =>    [tex]\alpha = \frac{28}{7}[/tex]

         =>    [tex]\alpha = 4[/tex]

Now considering the first question

        i.e to find the probability that a transistor will last between 14 and 28 weeks

            P(14 < A < 28) = P(A ≤28) - P(A ≤ 14)

The general formula for probability of  gamma distribution

                                   = [tex]F[\frac{a}{\beta},\alpha ] - F[\frac{a}{\beta},\alpha ][/tex]

                                   [tex]= F[\frac{28}{7},4 ] - F[\frac{14}{7},4 ][/tex]

                                   [tex]= F(4,4) - F(2,4)[/tex]

                                   [tex]= 0.567 - 0.143[/tex]    (This is gotten from the gamma table)

                                  [tex]= 0.424[/tex]

Now considering the second question        

     i.e to find the probability that a transistor will last at most 28 weeks  

                           [tex]P(A \le 28)[/tex]    =    [tex]F[\frac{28}{7},4][/tex]

                                                [tex]=F(4,4)[/tex]

                                                [tex]= 0.567[/tex]     (This is obtained from the gamma Table)

 we need to determine the median of the life time distribution less than 28

        from what we are given [tex]P(X \le \mu ) = 0.5[/tex] we can see that [tex]\mu[/tex]<28 since [tex]P(A \le 28)[/tex]  = 0.567 and this is greater than 0.5

Now considering the Third  question

       i.e to find the  [tex]99^{th}[/tex] percentile of the life time distribution

       Generally

        F(a : [tex]\alpha[/tex]) =99%

        F(a : 4) = 0.99

Looking at the gamma table the tabulated values at [tex]\alpha =4[/tex] and the corresponding 0.99 percent value to a is 10

         Now , find the product of [tex]\beta =7[/tex] with a = 10 to obtain the [tex]99^{th}[/tex] percentile

                   [tex]99^{th} \ percentile = a\beta[/tex]

                                            [tex]= 10 * 7[/tex]

                                            [tex]=70[/tex]

Now considering the Fourth  question

    i.e to determine the number of weeks such that 0.5% of all transistor will still be operating

    The first thing to do is to find the value in the incomplete gamma function with [tex]\alpha = 4[/tex] in such a way that

              F(a:4) = 1 - 0.5%

              F(a: 4) = 1 - 0.005

              F(x: 4)  = 0.995

So we will now obtain the 99.5 percentile

 Looking at the gamma tabulated values at [tex]\alpha = 4[/tex] and the corresponding 0.995 percent value for a (x depending on what you want to denoted it with) is 11

   So we would the obtain the product of [tex]\beta =7[/tex] with  a = 11  which is the [tex]99.5^{th}[/tex] percentile

                              [tex]t = a\beta[/tex]

                                 [tex]= 11 *7[/tex]

                                [tex]= 77[/tex]

The probability will be:

(a) 0.4240

(b) 0.5670

(c) 70

(d) 77

According to the question,

→ [tex]\mu = \alpha \beta[/tex]

 [tex]28= \alpha \beta[/tex] ...(equation 1)

→ [tex]\sigma^2 = \alpha \beta^2[/tex]

 [tex]14^2 = \alpha \beta^2[/tex] ...(equation 2)

By putting "equation 1" in "equation 2", we get

→ [tex]14^2 = \alpha \beta\times \beta[/tex]

   [tex]14^2 = 28\times \beta[/tex]

→ [tex]196 = 28\times \beta[/tex]

      [tex]\beta = 7[/tex]

      [tex]\alpha = 4[/tex]

(a) P(14 and 28)

→ [tex]P(14< X< 28 ) = F(\frac{x}{\beta}, \alpha )- F (\frac{x}{\beta}, d )[/tex]

                             [tex]= F(\frac{28}{7}, 4 )- F(\frac{14}{7} ,4)[/tex]

                             [tex]= F(4,4)-F(2,4)[/tex]

By using gamma tables, we get

                             [tex]= 0.5670-0.1430[/tex]

                             [tex]= 0.4240[/tex]

(b) P(almost 28)

→ [tex]P(x \leq 28) = F(\frac{x}{\beta}, \alpha )[/tex]

                    [tex]= F (\frac{28}{7} ,4)[/tex]

                    [tex]= F(4,4)[/tex]

                    [tex]= 0.5670[/tex]

(c) 99 percentile

at x = 10,

99th percentile,

→ [tex]x \beta = 10\times 7[/tex]

        [tex]= 70[/tex]

(d)

x = 11

→ [tex]t = x \beta[/tex]

     [tex]= 11\times 7[/tex]

     [tex]= 77[/tex]

Thus the above answers are correct.

Learn more:

https://brainly.com/question/14281632

Answer:

Take any algorithm if that algorithm solves this problem it can be represented as a ternary decision tree. Therefore each question has at most three answers.

There are ten possible verdicts, the height of such kind of tree should satisfy

ℎ >= ⌈log3(10)⌉ = 3

Hence no such algorithm can ask less than three questions in the worst case.

---

b)

Each and every internal node represents a question asking whether m belongs to one of three possible subset of {0, 1, 2, 3, 4, 5, 6, 7, 8, 9} or not

For example 0123|456|789 represented the questionDoes “m: belongs to {0, 1, 2, 3}, to {4, 5, 6}, or to {7, 8, 9}?"

Verdicts are placed in brackets "[ ]"

Explanation:

decision tree is attached below

Answer:

The diffusion coefficient for magnesium in aluminum at the given conditions is 2.22x10⁻¹⁴m²/s.

Explanation:

The diffusion coefficient represents how easily a substance (solute, e.g.: magnesium) can move through another substance (solvent, e.g.: aluminum).

The formula to calculate the diffusion coefficient is:

D=D₀.exp(- EA/R.T)

Each term of the formula means:

D: diffusion coefficient

D₀: Pre-exponential factor (it depends on each particular system, in this case the magnesium-aluminium system)

EA: activation energy

R: gas constant

T: temperature

1st) It is necessary to make a unit conversion for the temperature from °C to K, because we need to solve the problem according to the units of the gas constant (R) to assure the units consistency, so:

Temperature conversion from °C to K: 500 + 273 = 773 K

2nd) Replace the terms in the formula and solve:

D=D₀.exp(- EA/R.T)

D=1.2x10⁻⁴m²/s . exp[- 144,000 J/mole / (8.314 J/mol K . 773K)]

D=1.2x10⁻⁴m²/s . exp(-22.41)

D=1.2x10⁻⁴m²/s . 1.85x10⁻¹⁰

D= 2.22x10⁻¹⁴m²/s

Finally, the result for the diffusion coefficient  is 2.22x10⁻¹⁴m²/s.

Answer:

A) Ductility = 11% EL

B) Radius after deformation = 4.27 mm

Explanation:

A) From equations in steel test,

Tensile Strength (Ts) = 3.45 x HB

Where HB is brinell hardness;

Thus, Ts = 3.45 x 250 = 862MPa

From image 1 attached below, for steel at Tensile strength of 862 MPa, %CW = 27%.

Also, from image 2,at CW of 27%,

Ductility is approximately, 11% EL

B) Now we know that formula for %CW is;

%CW = (Ao - Ad)/(Ao)

Where Ao is area with initial radius and Ad is area deformation.

Thus;

%CW = [[π(ro)² - π(rd)²] /π(ro)²] x 100

%CW = [1 - (rd)²/(ro)²]

1 - (%CW/100) = (rd)²/(ro)²

So;

(rd)²[1 - (%CW/100)] = (ro)²

So putting the values as gotten initially ;

(ro)² = 5²([1 - (27/100)]

(ro)² = 25 - 6.75

(ro) ² = 18.25

ro = √18.25

So ro = 4.27 mm

Answer:

the answer is attributes for each entity

Answer: boundary work is 1.33kJ and heat lost is 200kJ

Explanation: mass of water m = 5kg

Molar mass of water mm = 18kg/mol

No of moles n = 5/18 = 0.28

Pressure P = 500x10^3

Temperature T = 300°c = 573K

R = 8.314pa.m3/mol/k a constant.

Using PV = nRT

V = nRT/P

V = (0.28x8.314x573)÷(500x10^3)

V = 0.00266m3

Work on boundary w = PV

W = 500x10^3 x 0.00266 = 1330 J

= 1.33kJ

Heat lost on cooling is equal to electrical heat Q used to heat the water initially

Q =Ivt

I = current = 4A

v = voltage = 100

T = time = 500sec

Q = 4x100x500 = 200000 = 200kJ

The question is about determining cache hits and misses when performing a matrix transposition, taking into account the cache's specifications. Given the setup, each new cache block accessed will result in a miss, followed by hits if subsequent accesses fall within the same block. The non-sequential access pattern of matrix transposition is likely to incur more cache misses compared to sequential access.

The student is asking about cache behavior when transposing a matrix, which involves changing the positions of each element in the array so that rows become columns and vice versa. In a cache memory system that is direct-mapped, block size determines how many elements will fit within one block of cache. Because the array starts at a known memory address and each integer occupies 4 bytes (given by the property that sizeof(int) is 4), we can predict how the cache will behave during the execution of the transpose routine.

Given the cache specifics - direct-mapped, write-through, write-allocate, and block size of 16 bytes (which can hold 4 integers), and size of 32 bytes total - the access pattern will largely depend on the congruence of array indices and cache block boundaries. For example, when accessing the source array src at address 0 (src[0][0]), it will be a cache miss as the cache is initially empty. However, subsequent accesses like src[0][1], src[0][2], and src[0][3] may hit as they reside in the same cache block if the block can accommodate them. Similarly, writes to the destination array dst will initially miss and then follow a similar pattern of hits and misses based on cache boundaries and block allocation behavior after write misses (due to write-allocate policy).

Matrix transposition involves swapping elements such that element (i, j) becomes element (j, i). As a result of accessing elements in this non-sequential manner, there could be more cache misses compared to sequential access because the transpose operation accesses elements across different rows and hence, potentially different cache lines.