For each of the characteristic equations of feedback control systems given, determine the range of K so that the system is asymptotically stable. Determine the value of Kso that the system is marginally stable and the frequency of sustained oscillation if applicable. s4 + Ks3 + 2s2 + (K + 1)s + 10 = 0

Answer:

total weight of the aggregate = 3594878.28 lbs

Explanation:

given data

density = 121.8 lb/ft³

area =  1000 ft x 60 ft x 6 in = 1000 ft x 60 ft x 0.5 ft

moisture = 3.5 %

compaction = 95%

solution

we get here first volume of the space that is filled with the aggregate that is

volume = 1000 ft x 60 ft x 0.5 ft  = 30,000 cu ft

now we get fill space with aggregate that compact to 95% of dry density.

so we fill space with aggregate of density that is = 95% of 121.8

= 115.71 lb/ cu ft

so now dry weight of aggregate is

dry weight of aggregate = 30,000  × 115.71 = 3471300 lb

when we assume that moisture percentage is by weight

then weight of the moisture in aggregate will be

weight of the moisture in aggregate = 3.56 % of 3471300 lb  

weight of the moisture in aggregate = 123578.28 lbs

and

we get total weight of the aggregate to fill space that is

total weight of the aggregate = 3471300 lb +123578.28 lb

total weight of the aggregate = 3594878.28 lbs

Answer:

Depth in the contracted section = 2.896m

Velocity in the contracted section = 2.072m/s

Explanation:

Please see that attachment for the solving.

Assumptions:

1. Negligible head losses

2. Horizontal channel bottom

Answer:

0.34

Explanation:

See the attached picture.

Answer:

time = 4.89 min

Explanation:

given data

volume = 15,000 ft³

ventilation rate of oven = 2,100 cubic feet per minute

safety factor (K) = 3

evaporates at a rate = 0.6 cubic feet per minute

solution

we get here first solvent additional rate in oven that is

solvent additional rate = [tex]\frac{0.6}{1500}[/tex]

solvent additional rate = 4 × [tex]10^{-5}[/tex] min

solvent additional rate == 4 × [tex]10^{-5}[/tex] × [tex]10^{6}[/tex]

solvent additional rate =  40 ppm/min

and

solvent removal rate due to ventilation will be

removal rate = [tex]\frac{2100}{1500}[/tex] × concentration in ppm

removal rate = 0.14 C  ppm/min

and

net additional rate is

net additional rate (c) = m - r

so net additional rate (c) is = 40 - 0.14C

so here

[tex]\frac{dc}{dt}[/tex] = 40 - 0.14C

so take integrate from 0 to t

[tex]\int\limits^t_o {dt}[/tex] = [tex]\int\limits^{425/3}_0 \frac{dc}{40-0.14C}[/tex]    ....................1

here factor of safety is 3 so time taken is [tex]\frac{425}{3}[/tex]  

solve it we get

time = [tex][\frac{-50}{7} \ log(40-\frac{7c}{50} ]^{425/3} _0[/tex]

time = 4.89 min

Answer: V = 208514.156 ft/hr

Explanation:

we will begin by giving a step by step order of answering.

given that

A = area available for convection which will be same for both cases

h (still) = heat transfer coefficient in still air

h(blowing) = heat transfer coefficient in blowing air.

Therefore,

h(still) A [40-(-30)] = h(blowing) A (40-15)

canceling out we have

h(still) (70) = h(blowing) (25)

where h(still) = 4 Btu/hr.ft².°F

4 × 70 =  h(blowing) (25)

h (blowing) = 11.2 Btu/hr.ft².°F

Also, we have that NUD = 0.0239 ReD 0805

h (blowing) D / k = 0.0239 (ρVD/μ)˄0.805

where from our data,

D = diameter = 1 ft

ρ = density = 0.0845 lbm/ft

μ = viscosity = 3.996×10-2 bm/ft hr

K = 0.0134 Btu/hr.ft²°F

So from

h (blowing) D / k = 0.0239 (ρVD/μ)˄0.805

we have;

11.2 × 1 / K =  0.0239 (0.0845×V×1 / 3.996×10-2 )˄0.805

where K = 0.0134

V = 208514.156 ft/hr

cheers i hope this helps

Answer:

(a)

(i) pseudo code :-

current = i

// assuming parent of root is -1

while A[parent] < A[current] && parent != -1 do,

if A[parent] < A[current] // if current element is bigger than parent then shift it up

swap(A[current],A[parent])

current = parent

(ii) In heap we create a complete binary tree which has height of log(n). In shift up we will take maximum steps equal to the height of tree so number of comparison will be in term of O(log(n))

(b)

(i) There are two cases while comparing with grandparent. If grandparent is less than current node then surely parent node also will be less than current node so swap current node with parent and then swap parent node with grandparent.

If above condition is not true then we will check for parent node and if it is less than current node then swap these.

pseudo code :-

current = i

// assuming parent of root is -1

parent is parent node of current node

while A[parent] < A[current] && parent != -1 do,

if A[grandparent] < A[current] // if current element is bigger than parent then shift it up

swap(A[current],A[parent])

swap(A[grandparent],A[parent])

current = grandparent

else if A[parent] < A[current]

swap(A[parent],A[current])

current = parent

(ii) Here we are skipping the one level so max we can make our comparison half from last approach, that would be (height/2)

so order would be log(n)/2

(iii) C++ code :-

#include<bits/stdc++.h>

using namespace std;

// function to return index of parent node

int parent(int i)

{

if(i == 0)

return -1;

return (i-1)/2;

}

// function to return index of grandparent node

int grandparent(int i)

{

int p = parent(i);

if(p == -1)

return -1;

else

return parent(p);

}

void shift_up(int A[], int n, int ind)

{

int curr = ind-1; // because array is 0-indexed

while(parent(curr) != -1 && A[parent(curr)] < A[curr])

{

int g = grandparent(curr);

int p = parent(curr);

if(g != -1 && A[g] < A[curr])

{

swap(A[curr],A[p]);

swap(A[p],A[g]);

curr = g;

}

else if(A[p] < A[curr])

{

swap(A[p],A[curr]);

curr = p;

}

}

}

int main()

{

int n;

cout<<"enter the number of elements :-\n";

cin>>n;

int A[n];

cout<<"enter the elements of array :-\n";

for(int i=0;i<n;i++)

cin>>A[i];

int ind;

cout<<"enter the index value :- \n";

cin>>ind;

shift_up(A,n,ind);

cout<<"array after shift up :-\n";

for(int i=0;i<n;i++)

cout<<A[i]<<" ";

cout<<endl;

}

Explanation:

Answer and Explanation:

The answer is attached below

Answer:

See image attached.

Explanation:

This device features priority encoding of the inputs

to ensure that only the highest order data line is en-

coded. Nine input lines are encoded to a four line

BCD output. The implied decimal zero condition re-

quires no input condition as zero is encoded when

all nine datalinesare athigh logic level. Alldata input

and outputs are active at the low logic level. All in-

puts are equipped with protection circuits against

static discharge and transient excess voltage.

A 10-to-4 encoder is requested to be designed for mapping 1-out-of-10 input code to a modified BCD output, where the digits 8 and 9 are encoded as 'E' and 'F'. The implementation involves the use of digital logic gates and could also reference quantum gates depending on the context of the course.

The student is asking to design a 10-to-4 encoder with a specific bit pattern for the numbers 8 and 9. This encoder takes a l-out-of-10 input code and produces a BCD-like output with special cases for inputs 8 ('E') and 9 ('F'). The design would necessitate the use of digital logic gates to map each of the 10 inputs to correspondent 4-bit BCD outputs, with the additional requirement to encode the inputs representing the decimal numbers 8 and 9 into the hexadecimal digits 'E' and 'F', respectively.

In terms of circuitry, the encoder might use a combination of logical gates such as AND, OR, and NOT to create the desired output for each input. For example, when the ninth input is active (representing the number 8), the output should be '1110', which signifies 'E' in hexadecimal notation. Similarly, an active tenth input (representing the number 9) should produce '1111', corresponding to 'F' in hexadecimal.

The encoding and decoding elements are described using terms such as CnNOT, CNOT, I (Identity), and Toffoli gates, which suggests a more sophisticated setup, possibly involving quantum computing principles, as these terms relate to quantum gates.

The Poisson's ratio is 0.36

Explanation:

Given-

Diameter = 10 mm = 10 X 10⁻³m

Force, F = 15000 N

Diameter reduction = 7 X 10⁻3 mm = 7 X 10⁻⁶ mm

The equation for Poisson's ratio:

ν = [tex]\frac{-E_{x} }{E_{z} }[/tex]

and

εₓ = Δd / d₀

εz = σ / E

    = F / AE

We know,

Area of circle = π (d₀/2)²

[tex]E_{z}[/tex]= F / π (d₀/2)² E

εz = 4F / πd₀²E

Therefore, Poisson's ratio is

ν = - Δd ÷ d / 4F ÷ πd₀²E

ν =  -d₀ΔDπE / 4F

ν = (10 X 10⁻³) (-7 X 10⁻⁶) (3.14) (100 X 10⁻⁹) / 4 (15000)

ν = 0.36

Therefore, the Poisson's ratio is 0.36

Answer:

note:

solution is attached in word form due to error in mathematical equation. furthermore i also attach Screenshot of solution in word due to different version of MS Office please find the attachment

(a) The RTL specification for an arithmetic shift right on an eight-bit cell can be described as follows:

1. Take the eight-bit input value and assign it to a variable, let's call it "input".

2. Create a temporary variable, let's call it "shifted".

3. Assign the most significant bit (MSB) of the "input" to the least significant bit (LSB) of the "shifted" variable.

4. Shift the remaining seven bits of the "input" one position to the right, discarding the least significant bit (LSB) and shifting in the sign bit to the most significant bit (MSB).

5. Repeat step 3 and step 4 seven more times to shift all the bits in the "input" variable to the right.

6. Output the final value of the "shifted" variable, which represents the result of the arithmetic shift right.

Here's an example to illustrate the RTL specification:

Input: 10111010

Shifted: 11011101 (after one shift)

Shifted: 11101110 (after two shifts)

Shifted: 11110111 (after three shifts)

Shifted: 11111011 (after four shifts)

Shifted: 11111101 (after five shifts)

Shifted: 11111110 (after six shifts)

Shifted: 11111111 (after seven shifts)

Shifted: 11111111 (after eight shifts)

(b) The RTL specification for an arithmetic shift left on an eight-bit cell can be described as follows:

1. Take the eight-bit input value and assign it to a variable, let's call it "input".

2. Create a temporary variable, let's call it "shifted".

3. Assign the least significant bit (LSB) of the "input" to the most significant bit (MSB) of the "shifted" variable.

4. Shift the remaining seven bits of the "input" one position to the left, discarding the most significant bit (MSB) and shifting in a zero to the least significant bit (LSB).

5. Repeat step 3 and step 4 seven more times to shift all the bits in the "input" variable to the left.

6. Output the final value of the "shifted" variable, which represents the result of the arithmetic shift left.

Here's an example to illustrate the RTL specification:

Input: 10111010

Shifted: 01110100 (after one shift)

Shifted: 11101000 (after two shifts)

Shifted: 11010000 (after three shifts)

Shifted: 10100000 (after four shifts)

Shifted: 01000000 (after five shifts)

Shifted: 10000000 (after six shifts)

Shifted: 00000000 (after seven shifts)

Shifted: 00000000 (after eight shifts)

Know more about RTL specifications,

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Answer:

Explanation:

The answer to the above question is given in attached files.

Answer:

solution in the picture attached

Explanation:

Foreign keys help define the relationship between tables, which is what puts the "relational" in "relational database." They enable database developers to preserve referential integrity across their system.

To Learn more About Foreign keys refer to:

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Comments on Question

Your questions is incomplete.

I'll provide a general answer to create a checkbox programmatically

Answer:

// Comments are used for explanatory purpose

// Function to create a checkbox programmatically

public class CreateCheckBox extends Application {

// launch the application

public void ClubObject(Stage chk)

{

// Title

chk.setTitle("CheckBox Title");

// Set Tile pane

TilePane tp = new TilePane();

// Add Labels

Label lbl = new Label("Creating a Checkbox Object");

// Create an array to hold 2 checkbox labels

String chklbl[] = { "Checkbox 1", "Checkbox 2" };

// Add checkbox labels

tp.getChildren().add(lbl);

// Add checkbox items

for (int i = 0; i < chklbl.length; i++) {

// Create checkbox object

CheckBox cbk = new CheckBox(chklbl[i]);

// add label

tp.getChildren().add(cbk);

// set IndeterMinate to true

cbk.setIndeterminate(true);

}

// create a scene to accommodate the title pane

Scene sc = new Scene(tp, 150, 200);

// set the scene

chk.setScene(sc);

chk.show();

}

Answer:

The given grammar is :

S = T V ;

V = C X

X = , V | ε

T = float | double

C = z | w

1.

Nullable variables are the variables which generate ε ( epsilon ) after one or more steps.

From the given grammar,

Nullable variable is X as it generates ε ( epsilon ) in the production rule : X -> ε.

No other variables generate variable X or ε.

So, only variable X is nullable.

2.

First of nullable variable X is First (X ) = , and ε (epsilon).

L.H.S.

The first of other varibles are :

First (S) = {float, double }

First (T) = {float, double }

First (V) = {z, w}

First (C) = {z, w}

R.H.S.

First (T V ; ) = {float, double }

First ( C X ) = {z, w}

First (, V) = ,

First ( ε ) = ε

First (float) = float

First (double) = double

First (z) = z

First (w) = w

3.

Follow of nullable variable X is Follow (V).

Follow (S) = $

Follow (T) = {z, w}

Follow (V) = ;

Follow (X) = Follow (V) = ;

Follow (C) = , and ;

Explanation:

Answer: 2.26x10^-4 v

Explanation:

Lenght of the selonoid = 24x10^-2m

Diameter of the selonoid = 2.3cm

The radius will then be = 1.15cm = 1.15x10^-2m

The area of the selonoid = ¶r^2 = 3.142 x (1.15x10^-2)^2 = 0.000415m^2.

Number of turns on selonoid N1 is 750

For the small center coil, number of turns N2 is 19.

There is a change in current dI/dt from 0 to 5.1 in 0.7s, dI/dt = (5.1-0)/0.7

dI/dt = 7.29A/s.

Induced EMF on selonoid due to magnetic Flux due to changing current in small coil is given as;

E = -M(dI/dt), where M is the mutual inductance of the coils.

but M = (u°AN1N2)/L, where u°= 4¶x10^-7,

A = area of selonoid,

L = Lenght of selonoid.

M = (4¶X10^-7X0.000415X750X19)/(24X10^-2)

M = 3.096X10^-5H

Induced EMF E = 3.096X10^-5 x 7.29

E = 2.26x10^-4V

Answer / Explanation:

To answer this question, we first define the parameters which are,

Volume: This can be defined or refereed to the quantity of three-dimensional space enclosed by a closed surface. For the purpose of illustration, we can say that the space that a substance or shape occupies or contains. The SI used in measuring volume is mostly in cubic metre.

Therefore,

Volume, where Volume = base area x height x 1/3

where,

Base area = base length x base width

However, we also watch out for the division integer.

So moving forward to write the code solving the question, we have:

double Pyramid Volume (double baseLength, double baseWidth, double pyramid Height)

{

double baseArea = baseLength * baseWidth;

double vol = ((baseArea * pyramidHeight) * 1/3);

return vol;

}

int main() {

  cout << "Volume for 1.0, 1.0, 1.0 is: " << PyramidVolume(1.0, 1.0, 1.0) <<

endl;

  return 0;

}

Answer:

The problem demands a function PyramidVolume() in C language since printf() command exists in C language. Complete code, output and explanation is provided below.

Code in C Language:

#include <stdio.h>

double PyramidVolume(double baseLength, double baseWidth, double pyramidHeight)

{

return baseLength*baseWidth*pyramidHeight/3;

}

int main()

{

  printf("Volume for 1.0, 1.0, 1.0 is: %.2f\n",PyramidVolume(1, 1, 1));

  printf("Volume for 2.5, 5, 2.0 is: %.2f\n",PyramidVolume(2.5, 5, 2.0));

return 0;

}

Output:

Please also refer to the attached output results

Volume for 1.0, 1.0, 1.0 is: 0.33

Volume for 2.5, 5, 2.0 is: 8.33

Explanation:

A function PyramidVolume of type double is created which takes three inputs arguments of type double length, width and height and returns the volume of the pyramid as per the formula given in the question.

Then in the main function we called the PyramidVolume() function with inputs 1.0, 1.0, 1.0 and it returned correct output.

We again tested it with different inputs and again it returned correct output.

Answer:

Average access time for a hard disk = 11.38 ms

Explanation:

See attached pictures for step by step explanation.

To answer the student's question on the exit temperature and rate of entropy generation in an adiabatic nozzle, we need more context-specific information or the assumption that steam behaves as an ideal gas.

The student's question involves determining the exit temperature and the rate of entropy generation in an adiabatic nozzle process, which falls under the subject of thermodynamics, a branch of Physics. Calculating these properties requires knowledge of the first and second laws of thermodynamics, as well as the ideal gas law and flow equations.

For part (a), the exit temperature can be estimated using the ideal gas law and the conservation of energy. For part (b), the Clausius–Clapeyron relation and the concept of entropy would be applied. However, to accurately determine these values, more information would be needed, such as specific heat capacities, or the assumption that the steam behaves as an ideal gas.

Given the provided problem statements refer to various examples, which could lead to confusion, the correct answer would be provided if the precise context and details specific to the student's actual question were known.

Answer:

The lift coefficient is 0.3192 while that of the moment about the leading edge is-0.1306.

Explanation:

The Upper Surface Cp is given as

[tex]Cp_u=-0.8 *0.6 +0.1 \int\limits^1_{0.6} \, dx =-0.8*0.6+0.4*0.1[/tex]

The Lower Surface Cp is given as

[tex]Cp_l=-0.4 *0.6 +0.1 \int\limits^1_{0.6} \, dx =-0.4*0.6+0.4*0.1[/tex]

The difference of the Cp over the airfoil is given as

[tex]\Delta Cp=Cp_l-Cp_u\\\Delta Cp=-0.4*0.6+0.4*0.1-(-0.8*0.6-0.4*0.1)\\\Delta Cp=-0.4*0.6+0.4*0.1+0.8*0.6+0.4*0.1\\\Delta Cp=0.4*0.6+0.4*0.2\\\Delta Cp=0.32[/tex]

Now the Lift Coefficient is given as

[tex]C_L=\Delta C_p cos(\alpha_i)\\C_L=0.32\times cos(4*\frac{\pi}{180})\\C_L=0.3192[/tex]

Now the coefficient of moment about the leading edge is given as

[tex]C_M=-0.3*0.4*0.6-(0.6+\dfrac{0.4}{3})*0.2*0.4\\C_M=-0.1306[/tex]

So the lift coefficient is 0.3192 while that of the moment about the leading edge is-0.1306.

The lift coefficient and the pitching moment coefficient about the leading edge due to lift are respectively; 0.3192 and -0.13

We are given;

Distance of upper surface from leading edge to percentage of chord = -0.8

Percentage of chord for both surfaces = 60% = 0.6

Rate of increase at trailing edge for both surfaces = +0.1

Distance of lower surface from leading edge to percentage of chord = -0.6

angle of incidence; α_i = 4° = 4π/180 rad

Let us first calculate the Cp constant for both the upper and lower surface.

Cp for upper surface is;

Cp_u = (-0.8 × 0.6) - 0.1∫¹₀.₆ dx

Solving this integral gives;

Cp_u =  (-0.8 × 0.6) - (0.1 × 0.4)

Cp_u =  -0.52

Cp for lower surface is;

Cp_l = (-0.4 × 0.6) + 0.1∫¹₀.₆ dx

Solving this integral gives;

Cp_l =  (-0.4 × 0.6) + (0.1 × 0.4)

Cp_l =  -0.2

Change in Cp across the foil is;

ΔCp = Cp_l - Cp_u

ΔCp = -0.2 - (-0.52)

ΔCp = 0.32

Formula for the lift coefficient is;

C_L = ΔCp * cosα_i

C_L = 0.32 * cos (4π/180)

C_L = 0.3192

Formula for the pitching moment coefficient is;

(-0.3 * 0.4 * 0.6) - ((0.6 + (0.4/3)) * 0.2 * 0.4)

C_m,p =  -0.072 - 0.059

C_mp ≈ -0.13

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Answer:

Explanation:using the product immediately in the next step

Answer:

a. true

Explanation:

Firstly, we need to understand what takes places during the compression process in a quasi-equilibrium process. A quasi-equilibrium process is a process in during which the system remains very close to a state of equilibrium at all times.  When a compression process is quasi-equilibrium, the work done during the compression is returned to the surroundings during expansion, no exchange of heat, and then the system and the surroundings return to their initial states. Thus a reversible process.

While for a non-quasi equilibrium process, it takes more work to move the piston against this high-pressure region.

Answer:

True

Explanation:

Because in quasi static process, process occurs very slowly so it is treated as reversible process.

Answer:

therefore the exit pressure of air is 331.2 kPa

Explanation:

Answer:

The code is as below whereas the output is attached herewith

Explanation:

package brainly.priorityQueue;

class PriorityJobQueue {

   Job[] arr;

   int size;

   int count;

   PriorityJobQueue(int size){

       this.size = size;

       arr = new Job[size];

       count = 0;

   }

   // Function to insert an element into the priority queue

   void insert(Job value){

       if(count == size){

           System.out.println("Cannot insert the key");

           return;

       }

       arr[count++] = value;

       heapifyUpwards(count);

   }

   // Function to heapify an element upwards

   void heapifyUpwards(int x){

       if(x<=0)

           return;

       int par = (x-1)/2;

       Job temp;

       if(arr[x-1].getPriority() < arr[par].getPriority()){

           temp = arr[par];

           arr[par] = arr[x-1];

           arr[x-1] = temp;

           heapifyUpwards(par+1);

       }

   }

   // Function to extract the minimum value from the priority queue

   Job extractMin(){

       Job rvalue = null;

       try {

           rvalue = arr[0].clone();

       } catch (CloneNotSupportedException e) {

           e.printStackTrace();

       }

       arr[0].setPriority(Integer.MAX_VALUE);

       heapifyDownwards(0);

       return rvalue;

   }

   // Function to heapify an element downwards

   void heapifyDownwards(int index){

       if(index >=arr.length)

           return;

       Job temp;

       int min = index;

       int left,right;

       left = 2*index;

       right = left+1;

       if(left<arr.length && arr[index].getPriority() > arr[left].getPriority()){

           min =left;

       }

       if(right <arr.length && arr[min].getPriority() > arr[right].getPriority()){

           min = right;

       }

       if(min!=index) {

           temp = arr[min];

           arr[min] = arr[index];

           arr[index] = temp;

           heapifyDownwards(min);

       }

   }

   // Function to implement the heapsort using priority queue

   static void heapSort(Job[] array){

       PriorityJobQueue object = new PriorityJobQueue(array.length);

       int i;

       for(i=0; i<array.length; i++){

           object.insert(array[i]);

       }

       for(i=0; i<array.length; i++){

           array[i] = object.extractMin();

       }

   }

}

package brainly.priorityQueue;

import java.io.BufferedReader;

import java.io.IOException;

import java.io.InputStreamReader;

import java.util.Arrays;

public class PriorityJobQueueTest {

   // Function to read user input

   public static void main(String[] args) {

       BufferedReader br = new BufferedReader(new InputStreamReader(System.in));

       int n;

       System.out.println("Enter the number of elements in the array");

       try{

           n = Integer.parseInt(br.readLine());

       }catch (IOException e){

           System.out.println("An error occurred");

           return;

       }

       System.out.println("Enter array elements");

       Job[] array = new Job[n];

       int i;

       for(i=0; i<array.length; i++){

           Job job  =new Job();

           try{

               job.setJobId(i);

               System.out.println("Element "+i +"priority:");

               job.setJobName("Name"+i);

               job.setSubmitterName("SubmitterName"+i);

               job.setPriority(Integer.parseInt(br.readLine()));

               array[i] = job;

           }catch (IOException e){

               System.out.println("An error occurred");

           }

       }

       System.out.println("The initial array is");

       System.out.println(Arrays.toString(array));

       PriorityJobQueue.heapSort(array);

       System.out.println("The sorted array is");

       System.out.println(Arrays.toString(array));

       Job[] readyQueue =new Job[4];

   }

}

The student's question involves calculating the water level drop at the center of a rotating cylindrical container, which requires understanding of rotational motion and equilibrium in physics.

The student is asking about the change in water level in a rotating cylindrical container due to the centrifugal force. In a rotating frame, the water surface forms a parabolic shape and the level at the center drops. To solve this problem, principles of rotational motion and physics are applied. The task is to calculate the drop in the center of the water surface within a cylindrical container rotating at a constant angular speed. This can be determined by setting up the equilibrium of forces acting on a particle of the water in the radial direction and using the relationship between the angular speed, radius, and acceleration in a rotating system.

Answer: The approximate voltage would be the result from computing analogRead(A0)*3.3 V / 1023

Explanation:

Depending on the binary read of the function analogRead(A0) we would get a binary value between 0 to 1023, being 0 associated to 0V and 1023 to 3.3V, then we can use a three rule to get the X voltage corresponding to the binary readings as follows:

3.3 V ---------> 1023

X  V------------>analogRead(A0)

Then [tex]X=\frac{3.3 \times analogRead(A0)}{1023} Volts[/tex]

Thus depending on the valule analogRead(A0) has in bits we get an approximate value of the voltage at pin A0, with a precission of 3,2mV approximately (3.3v/1023).

Answer:

Explanation: see attachment

Answer:

a) A = 1667 and B = 2353

b) Oven A

c) Oven A

d) Below 13,333 pizza: Oven A

Above 13,334 pizza: Oven B

Explanation:

We have the following data:

                             Oven A:                             Oven B:

Capacity                 20 p/hr                              40p/hr

Fixed Cost            $20,000                            $30,000

Variable Cost        $2.00/p                             $1.25/p

Selling Price: $14

a) Break-even point  →   Cost = Revenue

([tex]x[/tex] refers to the number of pizza sold)

Oven A:

20000 + 2[tex]x[/tex] = 14[tex]x[/tex]

20000 = 14[tex]x[/tex] - 2[tex]x[/tex]

[tex]x[/tex] = 20000/ 12

[tex]x[/tex] = 1666.67 ≈ 1667 pizza

Oven B:

30000 + 1.25[tex]x[/tex] = 14[tex]x[/tex]

30000 = 14[tex]x[/tex] - 1.25[tex]x[/tex]

[tex]x[/tex] = 30000/ 12.75

[tex]x[/tex] = 2352.9 ≈ 2353 pizza

b) Comparing both oven for 9,000 pizza

Profit = Selling Price - Cost Price

Oven A:

Profit = (9000 x 14) - (20,000 +  2 x 9000)

Profit = 126000 - 38000

Profit = 88000

Oven B:

Profit = (9000 x 14) - (30,000 +  1.25 x 9000)

Profit = 126000 - 41250

Profit = 84750

Oven A is more profitable.

c)

Oven A:

Profit = (12000 x 14) - (20,000 +  2 x 12000)

Profit = 168000 - 44000

Profit = 124000

Oven B:

Profit = (12000 x 14) - (30,000 +  1.25 x 12000)

Profit = 168000 - 45000

Profit = 123000

Oven A is more profitable.

d) Using the equation formed in a):

20,000 - 12[tex]x[/tex] < 30,000 - 12.75[tex]x[/tex]

12.75[tex]x[/tex] - 12[tex]x[/tex] < 30000 - 20000

0.75[tex]x[/tex] < 10000

[tex]x[/tex] < 10000/0.75

[tex]x[/tex] < 13333.3

Hence, if the production is below 13,333 Oven A is beneficial.

For production of 13,334 and above, Oven B is beneficial.

Answer:

a.       Find the break even points in units for each oven.

Breakeven for type A pizza x =  = 1,666.6 units of pizza need to be sold in order to obtain breakeven for Type A

Breakeven for type B pizza x =  = 2,352.9 units of pizza need to be sold in order to obtain breakeven for Type B

b.      If the owner expects to sell 9000 pizzas, which oven should she purchase?

Type B: because the profit will be twice what will be obtainable from type A considering the fact that it produces pizza at the ration of TypeB:TypeA, 40:20 or 2:1

Profit for type a = 9000/20 x 14 = 6,300 – 1,666,6units ($23, 3332) = 4366.4 units

Profit for type B = 10,247.1 units of pizza - which makes it justifiable

Explanation:

Answer and Explanation:

The answer is attached below

Answer:

critical stress required is  18.92 MPa

Explanation:

given data

specific surface energy = 1.0 J/m²

modulus of elasticity = 225 GPa

internal crack of length = 0.8 mm

solution

we get here one half length of internal crack that is

2a = 0.8 mm

so a = 0.4 mm = 0.4 × [tex]10^{-3}[/tex] m

so we get here critical stress that is

[tex]\sigma _c = \sqrt{\frac{2E \gamma }{\pi a}}[/tex]     ...............1

put here value we get

[tex]\sigma _c[/tex] =   [tex]\sqrt{\frac{2\times 225\times 10^9 \times 1 }{\pi \times 0.4\times 10^{-3}}}[/tex]

[tex]\sigma _c[/tex] =  18923493.9151 N/m²

[tex]\sigma _c[/tex] =   18.92 MPa