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.
Explanation: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.
The voltage and current at the terminals of the circuit element in Fig. 1.5 are zero fort < 0. Fort 2 0 they areV =75 ~75e-1000t V,l = 50e -IOOOt mAa) Fund the maximum value of the power delivered to the circuit.b) Find the total energy delivered to the element.
The maximum power delivered to the circuit is found to be 3.75W, occurring at the start (t = 0). Calculating the total energy requires integrating the power over time, involving exponentials reducing over time to approach a finite total energy delivered.
Explanation:The question involves calculating the maximum power and the total energy delivered to a circuit element, where the voltage V and current I are given as time-dependent expressions.
Maximum Power Delivered
Power is calculated using P = IV. Inserting the given expressions for V and I,
P(t) = V(t) × I(t) = (75 - 75e^{-1000t}) × 50e^{-1000t} mA.
To find the maximum power, we would differentiate P(t) with respect to t and set the derivative equal to zero. However, because the setup includes exponential decay functions, their maximum product occurs at t = 0 for these specific functions.
Pmax = 75V × 50mA = 3.75W.
Total Energy Delivered
The total energy delivered can be found by integrating the power over time:
Energy = ∫ P(t) dt.
Considering the specific forms of V and I provided, this becomes an integral of the product of two exponentials, leading to an expression that evaluates the total energy consumed by the circuit over time.
Given the nature of exponential decay in V and I, the energy delivered approaches a finite value as t approaches infinity.
Your application must generate: - an array of thirteen random integers from 1-99, - then prompt the user to select a sorting option (Bubble Sort, Insertion Sort, Shell Sort, Merge Sort, or Quick Sort)
Answer:
The code is given which can be pasted in the Javascript file
Explanation:
The code is given as below
package Sorters;
public class javasort
{
private int[] arr=new int[13];
public void bubbleSort(int[] a){
int c,d,temp;
for (c = 0; c < ( 13 - 1 ); c++) {
for (d = 0; d < 13- c - 1; d++) {
if (a[d] > a[d+1]) /* For descending order use < */
{
temp = a[d];
a[d] = a[d+1];
a[d+1] = temp;
}
}
System.out.print("\n[");
for (int i = 0; i < 13; i++){
System.out.print(a[i]+" ");
}
System.out.print("]");
}
System.out.println("\nSorted list of numbers");
System.out.print("[");
for (int i = 0; i < 13; i++){
System.out.print(a[i]+" ");
}
System.out.print("]");
}
public void insertionSort(int[] a){
int temp;
for (int i = 1; i < 13; i++) {
for(int j = i ; j > 0 ; j--){
if(a[j] < a[j-1]){
temp = a[j];
a[j] = a[j-1];
a[j-1] = temp;
}
}
System.out.print("\n[");
for (int c = 0; c < 13; c++){
System.out.print(a[c]+" ");
}
System.out.print("]");
}
System.out.println("\nSorted list of numbers");
System.out.print("[");
for (int i = 0; i < 13; i++){
System.out.print(a[i]+" ");
}
System.out.print("]");
}
public void shellSort(int[] a){
int increment = a.length / 2;
while (increment > 0)
{
for (int i = increment; i < a.length; i++)
{
int j = i;
int temp = a[i];
while (j >= increment && a[j - increment] > temp)
{
a[j] = a[j - increment];
j = j - increment;
}
a[j] = temp;
}
if (increment == 2)
increment = 1;
else
increment *= (5.0 / 11);
System.out.print("\n[");
for (int c = 0; c < 13; c++){
System.out.print(a[c]+" ");
}
System.out.print("]");
}
System.out.println("\nSorted list of numbers");
System.out.print("[");
for (int i = 0; i < 13; i++){
System.out.print(a[i]+" ");
}
System.out.print("]");
}
public void MergeSort(int[] a, int low, int high){
int N = high - low;
if (N <= 1)
return;
int mid = low + N/2;
// recursively sort
MergeSort(a, low, mid);
MergeSort(a, mid, high);
// merge two sorted subarrays
int[] temp = new int[N];
int i = low, j = mid;
for (int k = 0; k < N; k++)
{
if (i == mid)
temp[k] = a[j++];
else if (j == high)
temp[k] = a[i++];
else if (a[j]<a[i])
temp[k] = a[j++];
else
temp[k] = a[i++];
}
for (int k = 0; k < N; k++)
a[low + k] = temp[k];
System.out.print("\n[");
for (int c = 0; c < 13; c++){
System.out.print(a[c]+" ");
}
System.out.print("]");
printM(a);
}
public void quickSort(int[] a,int low,int high){
System.out.print("\n[");
for (int c = 0; c < 13; c++){
System.out.print(a[c]+" ");
}
System.out.print("]");
int i =low, j = high;
int temp;
int pivot = a[(low + high) / 2];
/** partition **/
while (i <= j)
{
while (a[i] < pivot)
i++;
while (a[j] > pivot)
j--;
if (i <= j)
{
/** swap **/
temp = a[i];
a[i] = a[j];
a[j] = temp;
i++;
j--;
}
}
/** recursively sort lower half **/
if (low < j)
quickSort(a, low, j);
/** recursively sort upper half **/
if (i < high)
quickSort(a, i, high);
printM(a);
}
public void printM(int[] a){
arr=a;
}
public void fPrint(){
System.out.println("\nSorted list:");
System.out.print("\n[");
for (int c = 0; c < 13; c++){
System.out.print(arr[c]+" ");
}
System.out.print("]");
}
}
package mani;
import java.util.Random;
import java.util.Scanner;
public class javasorttest
{
public static void main(String[] args){
int[] a=new int[13];
Random r=new Random();
for(int i=0;i<13;i++){
a[i]=r.nextInt(99)+1;
}
System.out.print("[");
for (int c = 0; c < 13; c++){
System.out.print(a[c]+" ");
}
System.out.print("]");
javasort j=new javasort();
System.out.println("\nSelect the sorting algo.\n1.bubbleSort\n2.insertionSort\n3.shellSort\n4.MergeSort\n5.QuickSort.");
Scanner s=new Scanner(System.in);
int opt=s.nextInt();
switch(opt){
case 1:
j.bubbleSort(a);
break;
case 2:
j.insertionSort(a);
break;
case 3:
j.shellSort(a);
break;
case 4:
j.MergeSort(a, 0, 13);
j.fPrint();
break;
case 5:
j.quickSort(a ,0, 12);
j.fPrint();
break;
}
}
}
Two technicians are discussing hand tool use. Technician A says that a 6-point wrench is easier to use in tight places than a 12-point. Technician B says that a ratchet is used to loosen fasteners that are very tight. Who is correct?
Answer:
Technician B says that a ratchet is used to loosen fasteners that are very tight.
Explanation:
A ratchet is a common wrench device with a fastener component. A ratchet wrench is an essential tool that is used to fasten or loosen nuts and bolts.
Answer:
A ratchet wrench is usually used to loosen and tighten parts like steering linkages, tie rod end clamps and muffler clamps. Basically when nut is used on long thread a ratchet wrench is being used.
hence the technician B (option b) is correct.
Explanation:
9.43 An ideal air-standard Brayton cycle operates at steady state with compressor inlet conditions of 300 K and 100 kPa and a fixed turbine inlet temperature of 1700 K. For the cycle, (a) determine the net work developed per unit mass flowing, in kJ/kg, and the thermal efficiency for a compressor pressure ratio of 8. (b) plot the net work developed per unit mass flowing, in kJ/kg, and the thermal efficiency, each versus compressor pressure ratio ranging from 2 to 50.
Answer:
The net work output from the cycle = 520.67 [tex]\frac{KJ}{kg}[/tex]
The efficiency of Brayton cycle = 0.448
Explanation:
Compressor inlet temperature [tex]T_{1}[/tex] = 300 K
Turbine inlet temperature [tex]T_{3}[/tex] = 1700 K
Pressure ratio [tex]r_{p}[/tex] = 8
For the compressor the temperature - pressure relation is given by the formula,
⇒ [tex]\frac{T_{2} }{T_{1} }[/tex] = [tex]r_{p}^{\frac{\gamma - 1}{\gamma} }[/tex]
⇒ [tex]\frac{T_{2} }{300} = 8^{\frac{1.4 - 1}{1.4} }[/tex]
⇒ [tex]T_{2}[/tex] = 543.42 K
This is the temperature at compressor outlet.
Same relation for turbine we can write this as,
⇒ [tex]\frac{T_{3} }{T_{4} }[/tex] = [tex]r_{p}^{\frac{\gamma - 1}{\gamma} }[/tex]
⇒[tex]\frac{1700 }{T_{4} }[/tex] = [tex]8^{0.2857}[/tex]
⇒ [tex]T_{4}[/tex] = 938.5 K
This is the temperature at turbine outlet.
Now the work output from the turbine [tex]W_{T}[/tex] = [tex]m C_{p} (T_{3} - T_{4} )[/tex]
Put all the values in above formula we get,
⇒ [tex]W_{T}[/tex] = 1 × 1.005 × ( 1700 - 938.5 )
[tex]W_{T} = 765.3 \frac{KJ}{kg}[/tex]
This is the work output from the turbine.
Now the work input to the compressor is [tex]W_{C}[/tex] = [tex]m C_{p} (T_{2} - T_{1} )[/tex]
Put all the values in above formula we get,
⇒ [tex]W_{C}[/tex] = 1 × 1.005 × ( 543.42 - 300 )
⇒ [tex]W_{C}[/tex] = 244.63 [tex]\frac{KJ}{kg}[/tex]
This is the work input to the compressor.
Net work output from the cycle [tex]W_{net} = W_{T} - W_{C}[/tex]
⇒ [tex]W_{net}[/tex] = 765.3 - 244.63
[tex]W_{net} = 520.67\frac{KJ}{kg}[/tex]
This is the net work output from the cycle.
The thermal efficiency is given by
[tex]E_{cycle} =1 - \frac{1}{r_{p}^{\frac{\gamma - 1}{\gamma} } }[/tex]
[tex]E_{cycle} =1 - \frac{1}{8^{\frac{1.4 - 1}{1.4} } }[/tex]
[tex]E_{cycle} = 0.448[/tex]
This is the efficiency of Brayton cycle.
(b). the graph between plot the net work developed per unit mass flowing and the thermal efficiency, each versus compressor pressure ratio ranging from 2 to 50 is shown in the image below.
- Draw the internal representation for the following lisp list(s). • (cons '( (Blue) (Yellow () Red) () Orange) '() ) • (cons '(Red (Yellow Blue)) '(() Yellow Orange) ) • (cons '(()(Green Blue)) '(Red (Yellow ()) Blue) )
Answer:
Explanation: see attachment
A load P is applied horizontally while the other end is fixed to a structure. where P = 185 N You have already examined the axial stresses inside the plates. In this problem, the bolted connections will be analyzed. Failure means that the pieces come apart because the bolts cannot hold the force. If two bolts are connecting the two plates, how much transverse force V must each bolt resist at each sheare joint?
Answer:
Explanation:
The bolt is under double shear because it connects 3 plates.
Now the shear force resisted by each shear joint(V):
V=P/(No. of shear area)=185/2=92.5N.
A 1.5-m-long aluminum rod must not stretch more than 1 mm andthe normal stress must not exceed 40 MPa when the rod is subjectedto a 3-kN axial load. Knowing that E = 70 GPa, determine therequired diameter of the rod.
Using the maximum stress limit and elongation criteria, the required diameter of the aluminum rod to withstand a 3-kN axial load without exceeding a 1 mm stretch and keeping the normal stress below 40 MPa is calculated to be approximately 9.8 mm.
Explanation:Determining the Required Diameter of an Aluminum Rod
To ensure a 1.5-m-long aluminum rod does not stretch more than 1 mm (0.001 m) under a 3-kN (3000 N) axial load while keeping the normal stress below 40 MPa (40×106 N/m2), we first calculate the cross-sectional area required using the formula for stress (σ) which is σ = F/A, where F is the force applied and A is the cross-sectional area. Given that the maximum allowable stress σ is 40 MPa, we can reorganize the formula to solve for A, the required cross-sectional area of the rod. This gives us A = F/σ.
Substituting the given values, A = 3000 N / (40×106 N/m2) = 7.5×10-5 m2. To ensure the rod does not exceed the maximum stretch limit when this force is applied, we must also consider the modulus of elasticity (E) for aluminum, which is given as 70 GPa (70×109 N/m2). The formula for elongation (ΔL) under a force is ΔL = (FL)/(AE), where L is the original length of the rod. Given the requirements, the diameter can be calculated from the cross-sectional area (A = πd2/4), where d is the diameter of the rod.
From the area calculated earlier, we can determine the diameter is required to be sufficiently large to maintain stress and elongation within specified limits. Rearranging A = πd2/4 to solve for d, we find d to be approximately 9.8 mm, considering the area necessary to keep stress below 40 MPa while allowing for the specified elongation limit.
Write a program with two inputs, current price and last month's price (both integers). Then, output a summary listing the price, the change since last month, and the estimated monthly mortgage computed as (currentPrice * 0.045) / 12. in coral
Complete Question
Sites like Zillow get input about house prices from a database and provide nice summaries for readers. Write a program with two inputs, current price and last month's price (both integers). Then, output a summary listing the price, the change since last month, and the estimated monthly mortgage computed as (currentPrice * 0.045) / 12. Ex: If the input is 200000 210000, the output is: This house is $200000. The change is $-10000 since last month. The estimated monthly mortgage is $750.
Use Coral Programming Language
Answer:
// Program is written in Coral Programming Language
// Comments are used for explanatory purpose
// Program starts here
// Variable declaration
int currentprice
int prevprice
int change
float mortgage
Put "Enter current price to output" to output
currentprice = Get next input
Put "Enter last month price to output" to output
prevprice = Get next input
// Calculate Change since last month
change = currentprice - prevprice
// Calculate Monthly Mortgage
mortgage = currentprice * 0.045 / 12
// Print Results
Put "This house is $" to output
Put currentprice to output
Put "\n" to output
Put "This change is $" to output
Put change to output
Put "\n" to output
Put "This house is $" to output
Put currentprice to output
Put "since last month\n" to output
Put "This estimated monthly mortgage is $" to output
Put mortgage to output
// End of Program
What is the average distance in microns an electron can travel with a diffusion coefficient of 25 cm^2/s if the electron lifetime is 7.7 microseconds. Three significant digits and fixed point notation.
Answer: The average distance the electron can travel in microns is 1.387um/s
Explanation: The average distance the electron can travel is the distance an exited electron can travel before it joins together. It is also called the diffusion length of that electron.
It is gotten, using the formula below
Ld = √DLt
Ld = diffusion length
D = Diffusion coefficient
Lt = life time
Where
D = 25cm2/s
Lt = 7.7
CONVERT cm2/s to um2/s
1cm2/s = 100000000um2/s
Therefore D is
25cm2/s = 2500000000um2/s = 2.5e9um2/s
Ld = √(2.5e9 × 7.7) = 138744.37um/s
Ld = 1.387e5um/s
This is the average distance the excited electron can travel before it recombine
Determine if each of the following statements is true or false:
1. Larger atoms are better nucleophiles due to polarizability.
2. The identity of the nucleophile affects the rate of an SN1 reaction.
3. SN2 reactions proceed via frontside attack.
4. Bimolecular reactions tend to be stereoselective.
5. SN2 reactions invert all stereocenters in a haloalkane.
6. Cl-, OH-, and H- are good all leaving groups.
7. Good bases tend to be good nucleophiles
8. Branching adjacent to a reacting carbon slows SN2 reactions due to steric hindrance.
9. SN1 reactions are slowed by hyperconjugation of the carbocation intermediate.
10. The rate determining step for SN1 reactions is the same as the rate determining step for E1 reactions.
Answer:
Explanation:
1) true ( reason is due to polarizability given in statement)
2)false ( nucleophile does not involve in rate determining step)
3)false ( it is an inverse attack , attacks from backside)
4) true
5) false ( it inverts stereo centres which are being attacked)
6) false (not H-)
7) true ( strong bases are good nucleophiles)
8) true ( because of steric hindrance)
9)false ( ther are stabilized by hyper conjugation)
10) false
Answer:
1) True
2) False
3) False
4)
5)
6) True
7)True
8) True
9) False
10) True
Explanation:
1) Generally, polarization increases down the column of the periodic table.
The tangent function is defined as tan(theta) = sin(theta)/cos(theta). This expression can be evaluated to solve for the tangent as long as the magnitude of cos(theta) is no too near to 0. Assume that theta is given in degrees, and write the MATLAB statements to evaluate tan(theta) as long as the magnitude of cos(theta) is greater than or equal to 10e-2. If the magnitude of cos(theta) is less than 10e-2, write out an error message instead.
Answer:
The code is as attached here.
Explanation:
The code is as given below
theta = input(' Enter the value of theta?');
y = sin(theta*pi()/180);
z = cos(theta*pi()/180);
if z < 0.01
fprintf('The value of theta is very low')
else
t=round(y/z,2);
disp(['The value of tangent theta is ',num2str(t)]);
end
In MATLAB, evaluate tangent of theta if the magnitude of cosine is
[tex]> = 10[/tex]⁻²; otherwise, display an error message.
Here are the MATLAB statements to evaluate [tex]\( \tan(\theta) \)[/tex] as long as the magnitude of [tex]\( \cos(\theta) \)[/tex] is greater than or equal to [tex]\( 10^{-2} \)[/tex], and display an error message if the magnitude of [tex]\( \cos(\theta) \)[/tex] is less than [tex]\( 10^{-2} \)[/tex]:
```matlab
% Define theta in degrees
theta_deg = input('Enter the value of theta in degrees: ');
% Convert theta to radians
theta_rad = deg2rad(theta_deg);
% Calculate cosine of theta
cos_theta = cos(theta_rad);
% Check if the magnitude of cos(theta) is greater than or equal to 10^-2
if abs(cos_theta) >= 1e-2
% Evaluate tangent of theta
tan_theta = sin(theta_rad) / cos_theta;
disp(['tan(theta) = ', num2str(tan_theta)]);
else
% Display error message
disp('Error: The magnitude of cos(theta) is too small. Cannot evaluate tan(theta).');
end
```
This script prompts the user to enter the value of [tex]\( \theta \)[/tex] in degrees. It then converts [tex]\( \theta \)[/tex] to radians and calculates [tex]\( \cos(\theta) \)[/tex]. If the magnitude of [tex]\( \cos(\theta) \)[/tex] is greater than or equal to [tex]\( 10^{-2} \)[/tex], it evaluates [tex]\( \tan(\theta) \)[/tex] using the given formula and displays the result. Otherwise, it displays an error message indicating that the magnitude of [tex]\( \cos(\theta) \)[/tex] is too small to evaluate [tex]\( \tan(\theta) \).[/tex]
g For this project you are required to perform Matrix operations (Addition, Subtraction and Multiplication). For each of the operations mentioned above you have to make a function in addition to two other functions for ‘Inputting’ and ‘Displaying’ the Matrices in Row Order (rows are populated left to right, in sequence). In total there will be 5 functions: 3 for the operations and 2 functions for inputting and displaying the data.
Answer:
C++ code is explained below
Explanation:
#include<iostream>
using namespace std;
//Function Declarations
void add();
void sub();
void mul();
//Main Code Displays Menu And Take User Input
int main()
{
int choice;
cout << "\nMenu";
cout << "\nChoice 1:addition";
cout << "\nChoice 2:subtraction";
cout << "\nChoice 3:multiplication";
cout << "\nChoice 0:exit";
cout << "\n\nEnter your choice: ";
cin >> choice;
cout << "\n";
switch(choice)
{
case 1: add();
break;
case 2: sub();
break;
case 3: mul();
break;
case 0: cout << "Exited";
exit(1);
default: cout << "Invalid";
}
main();
}
//Addition Of Matrix
void add()
{
int rows1,cols1,i,j,rows2,cols2;
cout << "\nmatrix1 # of rows: ";
cin >> rows1;
cout << "\nmatrix1 # of columns: ";
cin >> cols1;
int m1[rows1][cols1];
//Taking First Matrix
for(i=0;i<rows1;i++)
for(j=0;j<cols1;j++)
{
cout << "\nEnter element (" << i << "," << j << "): ";
cin >> m1[i][j];
cout << "\n";
}
//Printing 1st Matrix
for(i=0;i<rows1;i++)
{
for(j=0;j<cols1;j++)
cout << m1[i][j] << " ";
cout << "\n";
}
cout << "\nmatrix2 # of rows: ";
cin >> rows2;
cout << "\nmatrix2 # of columns: ";
cin >> cols2;
int m2[rows2][cols2];
//Taking Second Matrix
for(i=0;i<rows2;i++)
for(j=0;j<cols2;j++)
{
cout << "\nEnter element (" << i << "," << j << "): ";
cin >> m2[i][j];
cout << "\n";
}
//Displaying second Matrix
cout << "\n";
for(i=0;i<rows2;i++)
{
for(j=0;j<cols2;j++)
cout << m2[i][j] << " ";
cout << "\n";
}
//Displaying Sum of m1 & m2
if(rows1 == rows2 && cols1 == cols2)
{
cout << "\n";
for(i=0;i<rows1;i++)
{
for(j=0;j<cols1;j++)
cout << m1[i][j]+m2[i][j] << " ";
cout << "\n";
}
}
else
cout << "operation is not supported";
main();
}
void sub()
{
int rows1,cols1,i,j,k,rows2,cols2;
cout << "\nmatrix1 # of rows: ";
cin >> rows1;
cout << "\nmatrix1 # of columns: ";
cin >> cols1;
int m1[rows1][cols1];
for(i=0;i<rows1;i++)
for(j=0;j<cols1;j++)
{
cout << "\nEnter element (" << i << "," << j << "): ";
cin >> m1[i][j];
cout << "\n";
}
for(i=0;i<rows1;i++)
{
for(j=0;j<cols1;j++)
cout << m1[i][j] << " ";
cout << "\n";
}
cout << "\nmatrix2 # of rows: ";
cin >> rows2;
cout << "\nmatrix2 # of columns: ";
cin >> cols2;
int m2[rows2][cols2];
for(i=0;i<rows2;i++)
for(j=0;j<cols2;j++)
{
cout << "\nEnter element (" << i << "," << j << "): ";
cin >> m2[i][j];
cout << "\n";
}
for(i=0;i<rows2;i++)
{
for(j=0;j<cols2;j++)
cout << m1[i][j] << " ";
cout << "\n";
}
cout << "\n";
//Displaying Subtraction of m1 & m2
if(rows1 == rows2 && cols1 == cols2)
{
for(i=0;i<rows1;i++)
{
for(j=0;j<cols1;j++)
cout << m1[i][j]-m2[i][j] << " ";
cout << "\n";
}
}
else
cout << "operation is not supported";
main();
}
void mul()
{
int rows1,cols1,i,j,k,rows2,cols2,mul[10][10];
cout << "\nmatrix1 # of rows: ";
cin >> rows1;
cout << "\nmatrix1 # of columns: ";
cin >> cols1;
int m1[rows1][cols1];
for(i=0;i<rows1;i++)
for(j=0;j<cols1;j++)
{
cout << "\nEnter element (" << i << "," << j << "): ";
cin >> m1[i][j];
cout << "\n";
}
cout << "\n";
for(i=0;i<rows1;i++)
{
for(j=0;j<cols1;j++)
cout << m1[i][j] << " ";
cout << "\n";
}
cout << "\nmatrix2 # of rows: ";
cin >> rows2;
cout << "\nmatrix2 # of columns: ";
cin >> cols2;
int m2[rows2][cols2];
for(i=0;i<rows2;i++)
for(j=0;j<cols2;j++)
{
cout << "\nEnter element (" << i << "," << j << "): ";
cin >> m2[i][j];
cout << "\n";
}
cout << "\n";
//Displaying Matrix 2
for(i=0;i<rows2;i++)
{
for(j=0;j<cols2;j++)
cout << m2[i][j] << " ";
cout << "\n";
}
if(cols1!=rows2)
cout << "operation is not supported";
else
{
//Initializing results as 0
for(i = 0; i < rows1; ++i)
for(j = 0; j < cols2; ++j)
mul[i][j]=0;
// Multiplying matrix m1 and m2 and storing in array mul.
for(i = 0; i < rows1; i++)
for(j = 0; j < cols2; j++)
for(k = 0; k < cols1; k++)
mul[i][j] += m1[i][k] * m2[k][j];
// Displaying the result.
cout << "\n";
for(i = 0; i < rows1; ++i)
for(j = 0; j < cols2; ++j)
{
cout << " " << mul[i][j];
if(j == cols2-1)
cout << endl;
}
}
main();
}
Assume that a phase winding of the synchronous machine of Problem 4.11 consists of one 5-turn, full-pitch coil per pole pair, with the coils connected in series to form the phase winding. If the machine is operating at rated speed and under the operating conditions of Problem 4.11, calculate the rms generated voltage per phase.
Answer:
The solution and complete explanation for the above question and mentioned conditions is given below in the attached document.i hope my explanation will help you in understanding this particular question.
Explanation:
A piston–cylinder assembly contains propane, initially at 27°C, 1 bar, and a volume of 0.2 m3. The propane undergoes a process to a final pressure of 4 bar, during which the pressure–volume relationship is pV1.1 = constant. For the propane, evaluate the work and heat transfer, each in kJ. Kinetic and potential energy effects can be ignored.
The question involves applying thermodynamics to calculate work and heat transfer for propane in a piston-cylinder assembly undergoing a process with a specific pressure-volume relationship. It requires integrating over the process path for work and applying the first law of thermodynamics for heat transfer, considering the neglect of kinetic and potential energy effects.
Explanation:A piston–cylinder assembly containing propane undergoes a process where the pressure-volume relationship is given as pV1.1 = constant. To evaluate the work and heat transfer for the propane, we apply the principles of thermodynamics, specifically the first law of thermodynamics, and the properties of processes adhering to specific equations of state. The work done in such processes can be calculated using the integral of p dV, considering the pressure-volume relationship provided. The heat transfer can then be inferred by applying the first law of thermodynamics, which equates the change in internal energy to the net heat added to the system minus the work done by the system.
The initial and final states of the propane provide the necessary boundary conditions to evaluate these quantities. However, without specific values for the molar mass or specific heat capacities of propane at constant pressure and volume, exact numerical answers cannot be provided. Generally, for processes described by a polytropic equation (pVn = constant), the work done is W = (p2V2 - p1V1)/(1-n) for an ideal gas, where p1, V1 are the initial pressure and volume, and p2, V2 are the final conditions. Heat transfer, Q, requires specific thermal properties of propane and can be approached via Q = ΔU + W, with ΔU denoting the change in internal energy of the gas.
For a precise evaluation, one would typically reference thermodynamic tables for propane or apply real gas equations of state considering the polytropic process specifics. It is essential to note that kinetic and potential energy changes are negligible, focusing the analysis solely on the thermodynamic work and heat transfer.
A worker is asked to move 30 boxes from a desk onto a shelf within 3 minutes (assume there is enough space and no other lifting work within 8 hours). The shelf height is 55 inches, and the desk height is 30 inches. The initial horizontal distance from the box to the body is 7 inches. Assume that all the boxes are in the same size (8 inches edge, cube), each weigh 20 lb., and have well designed handles. Is there any lifting risk according NIOSH lifting guide
Answer:
LI = Lifting Index = 0.71
No lifting risk is involved
Explanation:
NIOSH Lifting Index
LI = Load Weight / Recommended Weight Limit
LI = L / RWL ............. Eq (A)
NIOSH Recommended Weight Limit equation is following,
RWL = LC * HM * VM * DM * AM * FM * CM ........... Eq (B)
Where,
LC = Load constant
HM = Horizontal multiplier
VM = Vertical multiplier
DM = Distance multiplier
AM = Asymmetric multiplier
FM = Frequency multiplier
CM = Coupling multiplier
Given data
V = 30 + (8/2) = 34 in
H = 7 in
D = 55 - 30 = 25 in
A = 0
F = 10 boxes/min
C = 1 = Good coupling
According to NIOSH lifting guide
LC = 51 lb
HM = 10/H
VM = 1 - {0.0075*(v-30)}
DM = 0.82 + (1.8/D)
AM = 1 - (0.0032*A)
FM = 0.45 (Table 5 from NIOSH lifting guide)
CM = 1 (Table 7 from NIOSH lifting guide)
Solution:
RWL = 51 * (10/7) * [1-{0.0075(34-30)}] * (0.82+ 1.8/25) * (1-0.0032*0) * 0.45 * 1
RWL = 51 * (10/7) * 0.97 * 0.892 * 1 * 0.45 * 1
RWL = 28.37 lb
Using equation A
LI = L / RWL
LI = 20 / 28.37
LI = 0.71
According to NIOSH lifting guide LI <= 1
So No lifting risk is involved
) You are using a load cell to measure the applied load to a test in the Civil Engineering Structures Lab. What should you do if the measured load does not return to zero? How should you troubleshoot this to determine if this is a load cell or a mechanical problem?
Answer:
Insulation Resistance Tests
Explanation:
An insulation resistance test is carried out when there are unstable readings and random changes in the zero balance point of the load cell. It is done by measuring the resistance between the load cell body and all its connected wires, as follows:
First, disconnect the load cell from the summing box and indicator panel.
Connect all the input, output and sense (if equipped) wires together.
Measure the insulator resistance between the connected wires and the load cell body with a mega-ohmmeter.
Measure the insulation resistance between the connected wires and the cable shield.
Measure the insulation resistance between the load cell body and the cable shield.
The insulation resistance should match the value in the product’s load cell datasheet. A lower value shows an electrical leakage caused by moisture; this causes short circuits, giving unstable load cell outputs.
Block A weighs 10 lb and block B weighs 3 lb. If B is moving downward with a velocity (vB)1 = 3 ft>s at t = 0, determine the velocity of A when t = 1 s. Assume that the horizontal plane is smooth. Neglect the mass of the pulleys and cords
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
P4.36. Real inductors have series resistance associated with the wire used to wind the coil. Suppose that we want to store energy in a 10-H inductor. Determine the limit on the series resistance so the energy remaining after one hour is at least 75 percent of the initial energy.
Answer:
The limit on the series resistance is R ≤ 400μΩ
Explanation:
Considering the circuit has a series of inductance and resistance. The current current in the current in the circuit in time is
[tex]i(t) = Iie^{\frac{R}{L} t}[/tex] (li = initial current)
So, the initial energy stored in the inductor is
[tex]Wi = \frac{1}{2} Li^{2}_{i}[/tex]
After 1 hour
[tex]w(3600) = \frac{1}{2} Li_{i}e^{-\frac{R}{L} 3600 }[/tex]
Knowing it is equal to 75
[tex]w(3600) = 0.75Wi = 0.75 \frac{1}{2} Li^{2}_{i} = \frac{1}{2} Li_{i}e^{-\frac{R}{L} 3600 }\\[/tex]
This way we have,
R = [tex]-10 \frac{ln 0.75}{2 * 3600} = 400[/tex] μΩ
Than, the resistance is R ≤ 400μΩ
Pin, Password, Passphrases, Tokens, smart cards, and biometric devices are all items that can be
used for Authentication. When one of these item listed above in conjunction with a second factor to validate authentication, it provides robust authentication of the individual by practicing which of the following?
A. Multi-party authentication
B. Two-factor authentication
C. Mandatory authentication
D. Discretionary authentication
Answer:
B. Two-factor authentication
Explanation:
As far as identity as been established, authentication must be carried out. There are various technologies and ways of implementing authentication, although most method falls under the same category.
The three major types of authentication.
a. Authentication through knowledge, what someone knows.
b. Authentication through possessions, what someone has.
c. Authentication by characteristic features, who a person is.
Logical controls that are in relations to these types of authentication are known as factors.
Two factor authentication has to do with the combination of 2 out of the 3 factors of authentication.
The general term used when more than one factor is adopted is Multi-party authentication
Determine the average and rms values for the function, y(t)=25+10sino it over the time periods (a) 0 to 0.1 sec and (b) 0 to 1/3 sec. Discuss which case represents the long-term behavior of the signal (Hint, consider the period of the signal).
Answer:
Explanation:
AVERAGE: the average value is given as
[tex]\frac{1}{0.1} \int\limits^\frac{1}{10} _0 {25+10sint} \, dt = \frac{1}{0.1} [ 25t- 10cos\ t]_0^{0.1}[/tex]
=[tex]\frac{1}{0.1} ([2.5-10]-10)=-175[/tex]
RMS= [tex]\sqrt{\int\limits^\frac{1}{3} _0 {y(t)^2} \, dt }[/tex]
[tex]y(t)^2 = (25 + 10sin \ t)^2 = 625 +500sin \ t + 10000sin^2 \ t[/tex]
[tex]\frac{1}{\frac{1}{3} } \int\limits^\frac{1}{3} _0 {y(t)^2} \, dt =3[ 625t -500cos \ t + 10000(\frac{t}{2} - \frac{sin2t}{4} )]_0^{\frac{1}{3} }[/tex]
=[tex]3[[\frac{625}{3} - 500 + 10000(\frac{1}{6} - 0.002908)] + 500] = 2845.92\\[/tex]
therefore, RMS = [tex]\sqrt{2845.92} = 53.3[/tex]
If the wire has a diameter of 0.5 inin., determine how much it stretches when a distributed load of 140 lb/ftlb/ft acts on the beam. The material remains elastic. Express your answer to three significant figures and include appropriate units.
Answer:
δ_AB = 0.0333 in
Explanation:
Given:
- The complete question is as follows:
" The rigid beam is supported by a pin at C and an A−36
steel guy wire AB. If the wire has a diameter of 0.5 in.
determine how much it stretches when a distributed load of
w=140 lb / ft acts on the beam. The material remains elastic."
- Properties for A-36 steel guy wire:
Young's Modulus E = 29,000 ksi
Yield strength σ_y = 250 MPa
- The diameter of the wire d = 0.5 in
- The distributed load w = 140 lb/ft
Find:
Determine how much it stretches under distributed load
Solution:
- Compute the surface cross section area A of wire:
A = π*d^2 / 4
A = π*0.5^2 / 4
A = π / 16 in^2
- Apply equilibrium conditions on the rigid beam ( See Attachment ). Calculate the axial force in the steel guy wire F_AB
Sum of moments about point C = 0
-w*L*L/2 + F_AB*10*sin ( 30 ) = 0
F_AB = w*L*L/10*2*sin(30)
F_AB = 140*10*10/10*2*sin(30)
F_AB = 1400 lb
- The normal stress in wire σ_AB is given by:
σ_AB = F_AB / A
σ_AB = 1400*16 / 1000*π
σ_AB = 7.13014 ksi
- Assuming only elastic deformations the strain in wire ε_AB would be:
ε_AB = σ_AB / E
ε_AB = 7.13014 / (29*10^3)
ε_AB = 0.00024
- The change in length of the wire δ_AB can be determined from extension formula:
δ_AB = ε_AB*L_AB
δ_AB = 0.00024*120 / cos(30)
δ_AB = 0.0333 in
A freshwater jet boat takes in water through side vents and ejects it through a nozzle of diameter D = 75 mm; the jet speed is Vj. The drag on the boat is given by Fdrag =kV2, where V is the boat speed. Find an expression for the steady speed, V, in terms of water density rho, flow rate through the system of Q, constant k, and jet speed Vj. A jet speed Vj = 15 m=s produces a boat speed of V = 10 m=s. (a) Under these conditions, what is the new flow rate Q? (b) Find the value of the constant k. (c) What speed V will be produced if the jet speed is increased to Vj = 25 m=s? (d) What will be the new flow rate?
To find an expression for the jet boat's steady speed, equate the water jet force (ρQVj) with the drag force (kV^2). Solve for V to get the equation for speed in terms of ρ, Q, k, and Vj. Then use known values for Vj and V to find Q and k and repeat the process for a different Vj to get new V and Q values.
Explanation:To find an expression for the steady speed V of a freshwater jet boat in terms of the water density ρ, flow rate Q, constant k, and jet speed Vj, we can apply Newton's second law, assuming that the force provided by the water jet equals the drag force on the boat when at steady speed. The water jet force can be expressed as the rate of change of momentum of the water, which is ρQVj (since Q is the mass flow rate ρQ is the momentum flow rate), and the drag force is given as kV^2. Equating these two forces gives us:
ρQVj = kV^2
Solving for V will give us the steady speed expression we are looking for:
V = √(ρQVj / k)
To calculate the new flow rate Q given Vj = 15 m/s and V = 10 m/s, we plug these values into the expression obtained from above:
Q = kV^2 / (ρVj)
With given values, we would have:
Q = k * 10^2 / (ρ * 15)
The constant k can be determined using the known conditions and solving for k.
k = ρQVj / V^2
For (c) and (d), the same equations can be applied with the jet speed Vj changed to 25 m/s to find the new boat speed and flow rate.
The velocity field of a flow is given by where x and y are in feet. Determine the fluid speed at points along the x axis; along the y axis. What is the angle between the velocity vector and the x axis at points 15, 52, and 10, 52
There is a part of the question missing and it says;
The velocity field of a flow is given by V = [20y/(x² + y²)^(1/2)]î − [20x/(x² + y²)^(1/2)] ĵ ft/s, where x and y are in feet.
Answer:
A) At (1,5),angle is -11.31°
B) At (5,2),angle is -68.2°
C) At (1,0), angle is -90°
Explanation:
From the question ;
V = [20y/(x² + y²)^(1/2)]î − [20x/(x² + y²)^(1/2)] ĵ ft/s
Let us assume that u and v are the flow velocities in x and y directions respectively.
Thus we have the expression;
u = [20y/(x² + y²)^(1/2)]
and v = - [20x/(x² + y²)^(1/2)]
Thus, V = √(u² + v²)
V = √[20y/(x² + y²)^(1/2)]² + [-20x/(x² + y²)^(1/2)]²
V = √[400y²/(x²+y²)] +[400x²/(x²+y²)
V = √(400x² + 400y²)/(x²+y²)
Now for the angle;
tan θ = opposite/adjacent
And thus, in this question ;
tan θ = v/u
tan θ = [-20x/(x² + y²)^(1/2)]/ [20y/(x² + y²)^(1/2)]
Simplifying this, we have;
tan θ = - 20x/20y = - x/y
so the angle is ;
θ = tan^(-1)(-x/y)
So let's now find the angle at the various coordinates.
At, 1,5
θ = tan^(-1)(-1/5) = tan^(-1)(-0.2)
θ = -11.31°
At, 5,2;
θ = tan^(-1)(-5/2) = tan^(-1)(-2.5)
θ = -68.2°
At, 1,0;
θ = tan^(-1)(-1/0) = tan^(-1)(-∞)
θ = -90°
At,
A fatigue test was conducted in which the mean stress was 50 MPa (7250 psi) and the stress amplitude was 225 MPa (32,625 psi). (a) Compute the maximum and minimum stress levels.
Answer:
[tex]\sigma_{max} = 275\,MPa[/tex], [tex]\sigma_{min} = - 175\,MPa[/tex]
Explanation:
Maximum stress:
[tex]\sigma_{max}=\overline \sigma + \sigma_{a}\\\sigma_{max}= 50\,MPa + 225\,MPa\\\sigma_{max} = 275\,MPa[/tex]
Minimum stress:
[tex]\sigma_{min}=\overline \sigma - \sigma_{a}\\\sigma_{min}= 50\,MPa - 225\,MPa\\\sigma_{min} = - 175\,MPa[/tex]
A fatigue test was conducted in which the mean stress was 50 MPa (7,250 psi) and the stress amplitude was 225 MPa (32,625 psi).
(a) Compute the maximum and minimum stress levels.
(b) Compute the stress ratio.
(c) Compute the magnitude of the stress range.
Answer:(a) The maximum and minimum stress levels are 275MPa and -175MPa respectively.
(b) The stress ratio is 0.6
(c) The magnitude of the stress range is 450MPa
Explanation:(a )In fatigue, the mean stress ([tex]S_{m}[/tex]) is found by finding half of the sum of the maximum stress ([tex]S_{max}[/tex]) and minimum stress ([tex]S_{min}[/tex]) levels. i.e
[tex]S_{m}[/tex] = [tex]\frac{S_{max} + S_{min}}{2}[/tex] ------------------------(i)
Also, the stress amplitude (also called the alternating stress), [tex]S_{a}[/tex], is found by finding half of the difference between the maximum stress ([tex]S_{max}[/tex]) and minimum stress ([tex]S_{min}[/tex]) levels. i.e
[tex]S_{a}[/tex] = [tex]\frac{S_{max} - S_{min}}{2}[/tex] ------------------------(ii)
From the question,
[tex]S_{m}[/tex] = 50 MPa (7250 psi)
[tex]S_{a}[/tex] = 225 MPa (32,625 psi)
Substitute these values into equations(i) and (ii) as follows;
50 = [tex]\frac{S_{max} + S_{min}}{2}[/tex]
=> 100 = [tex]S_{max}[/tex] + [tex]S_{min}[/tex] -------------------(iii)
225 = [tex]\frac{S_{max} - S_{min}}{2}[/tex]
=> 450 = [tex]S_{max}[/tex] - [tex]S_{min}[/tex] -------------------(iv)
Now, solve equations (iii) and (iv) simultaneously as follows;
(1) add the two equations;
100 = [tex]S_{max}[/tex] + [tex]S_{min}[/tex]
450 = [tex]S_{max}[/tex] - [tex]S_{min}[/tex]
________________
550 = 2[tex]S_{max}[/tex] --------------------------------(v)
_________________
(2) Divide both sides of equation (v) by 2 as follows;
[tex]\frac{550}{2}[/tex] = [tex]\frac{2S_{max} }{2}[/tex]
275 = [tex]S_{max}[/tex]
Therefore, the maximum stress level is 275MPa
(3) Substitute [tex]S_{max}[/tex] = 275 into equation (iv) as follows;
450 = 275 - [tex]S_{min}[/tex]
[tex]S_{min}[/tex] = 275 - 450
[tex]S_{min}[/tex] = -175
Therefore, the minimum stress level is -175MPa
In conclusion, the maximum and minimum stress levels are 275MPa and -175MPa respectively.
===============================================================
(b) The stress ratio ([tex]S_{r}[/tex]) is given by;
[tex]S_{r}[/tex] = [tex]\frac{S_{min} }{S_{max} }[/tex] ----------------------------(vi)
Insert the values of [tex]S_{max}[/tex] and [tex]S_{min}[/tex] into equation (vi)
[tex]S_{r}[/tex] = [tex]\frac{-175}{275}[/tex]
[tex]S_{r}[/tex] = 0.6
Therefore, the stress ratio is 0.6
===============================================================
(c) The magnitude of the stress range ([tex]S_{R}[/tex]) is given by
[tex]S_{R}[/tex] = | [tex]S_{max}[/tex] - [tex]S_{min}[/tex] | ------------------------------(vii)
Insert the values of [tex]S_{max}[/tex] and [tex]S_{min}[/tex] into equation (vii)
[tex]S_{R}[/tex] = | 275 - (-175) |
[tex]S_{R}[/tex] = 450MPa
Therefore, the magnitude of the stress range is 450MPa
===============================================================
Note:1 MPa = 145.038psi
Therefore, the values of the maximum and minimum stress levels, the stress range can all be converted from MPa to psi (pounds per inch square) by multiplying the values by 145.038 as follows;
[tex]S_{max}[/tex] = 275MPa = 275 x 145.038psi = 39885.45psi
[tex]S_{min}[/tex] = -175MPa = -175 x 145.038psi = 25381.65psi
[tex]S_{R}[/tex] = 450MPa = 450 x 145.038psi = 65267.1psi
A sewage lagoon that has a surface area of 100,000 m2 (10 ha) and a depth of 1 m is receiving 8,640 m3/d of sewage containing 100 mg/L of biodegradable contaminant. At steady state, the effluent from the lagoon must not exceed 20 mg/L of biodegradable contaminant. Assuming the lagoon is well mixed and that there are no losses or gains of water in the lagoon other than the sewage input, what biodegradation reaction rate coefficient (d-1) must be achieved?
To achieve an effluent concentration of 20 mg/L in a sewage lagoon receiving sewage with 100 mg/L of a contaminant, a precise biodegradation reaction rate coefficient, determined by the mass balance equation under steady-state conditions, must be achieved.
Explanation:Calculating the Biodegradation Reaction Rate Coefficient
The question involves determining the biodegradation reaction rate coefficient necessary to reduce the concentration of a contaminant in a sewage lagoon, illustrating principles of environmental engineering. Given a lagoon with a surface area of 100,000 m2 and a depth of 1 m, receiving 8,640 m3/d of sewage that contains 100 mg/L of biodegradable contaminant, the goal is to lower the effluent concentration to no more than 20 mg/L.
To find the required biodegradation reaction rate coefficient (d-1), we must apply the mass balance concept in a steady-state condition, assuming the lagoon is well mixed. The mass balance equation for a contaminant undergoing a first-order degradation reaction can be expressed as: Input = Output + Decay. By substituting the given values and solving for the decay rate, we can find the coefficient that ensures the specified effluent concentration.
The calculation involves deriving relationships between the influent and effluent concentrations, the volume of the lagoon, and the decay process characterized by the reaction rate coefficient. For the lagoon described, achieving an effluent concentration of 20 mg/L from an influent concentration of 100 mg/L through biodegradation requires precise control of the treatment process and understanding of the kinetics of contaminant degradation.
A vertical piston-cylinder device contains water and is being heated on top of a range. During the process, 10 kJ of heat is transferred to the water, and heat losses from the side walls amount to 80 J. The piston rises as a result of evaporation, and 2 J of work is done by the vapor. Determine the change in the energy of the water for this process.
Answer: 9.9KJ
Explanation: Q = U + W + losses
Q is heat transfered to the water
U is the change in energy of the system
W is work done by the system = 2J
Losses = 80J
Heat into system is 10kJ = 10000KJ
Therefore
U = Q - W - losses
U = 10000 - 2 - 80 = 9990J
= 9.9kJ
The radiator of a steam heating system has a volume of 20 L and is filled with superheated water vapor at 200 kPa and 200°C. At this moment both the inlet and the exit valves to the radiator are closed. After a while it is observed that the temperature of the steam drops to 80°C as a result of heat transfer to the room air, which is at 21°C. Assuming the surroundings to be at 0°C, determine (a) the amount of heat transfer to the room and (b) the maximum amount of heat that can be 462 EXERGY supplied to the room if this heat from the radiator is supplied to a heat engine that is driving a heat pump. Assume the heat engine operates between the radiator and the surroundings.
Answer:
a = 30.1 kj
b = 115 kj
Explanation:
To determine the mass we use the formula m = V/v1
v1 =1.08m3/kg, and V = 20L
m = 20/1000 × 1.08 = 0.0185kg
Next we determine the initial specific internal energy, u1.
Using softwares and appropriate values of T1 and p1, we get
u1 = 2650kj/kg.
After this we determine the final specific internal energy, u2 using the formula u2 = uf + x2 × ufg
Therefore we need to find x2 first.
x2 = u2 - uf/ug - uf
x2 = 1.08 - 0.001029/3.4053 -0.001029
x2 = 0.3180
But u2 = uf + x2× uf=334.97 + 0.3180×2146.6 = 1017.59 kj/kg
Now heat transfer Q= DU
Q = m x (u1 - u2)
Q = 0.0185(2650-1017.59
Q = 30.1 kj
Calculating the b part of the question we use the formula
W = m( u1-u2) - m. To. (s1 - s2)
Where s1 = 7.510kj/kgk
And s2 = 3.150 kj/kgk
We need to convert To and Ta to k values by adding 273 to 0 and 21 respectively.
Putting the values into the formula, we get W = 30.1 - 0.0185 × 273 (7.510-3.150)
W = 8.179kj
Finally maximum heat transfer
Qm = W/1 - to/ta
Qm = 8.179/1 - 273/294
Qm = 115kj
An air conditioner using refrigerant-134a as the working fluid and operating on the ideal vapor-compression refrigeration cycle is to maintain a space at 22°C while operating its condenser at 1000 kPa. Determine the COP of the system when a temperature difference of 2°C is allowed for the transfer of heat in the evaporator.
Answer:
note:
solution is attached due to error in mathematical equation. please find the attachment
Answer:
COP = 13.31
Explanation:
We have an allowed temperature difference of 2°C, thus, let's make use of temperature of 20°C in the evaporator.
Now, looking at table A-11 i have attached and looking at temperature of 20°C, we will see that the enthalpy(h1) = 261.59 Kj/Kg
While the enthropy(s1) = 0.92234 Kj/KgK
Now, the enthalpy at the second state will be gotten from the given condenser pressure under the condition s2 = s1.
Thus, looking at table A-13 which i have attached, direct 20°C is not there, so when we interpolate between the enthalpy values at 15.71°C and 21.55°C, we get an enthalpy of 273.18 Kj/Kg.
Now, the enthalpy at the third and fourth states is again obtained from interpolation between values at temperatures of 18.73 and 21.55 of the saturated liquid value in table A-12 i have attached.
Thus, h3=h4 = 107.34 Kj/kg
Formula for COP = QL/w = (h1- h4) / (h2 - h1)
COP = (261.59 - 107.34)/( 273.18 - 261.59) = 13.31
Use the writeln() method of the document object to display the user agent in a
tag in the webpage. Hint: The userAgent property of the window.navigator object contains the user agent.
Answer:
Note that writeln() add a new line after each statement
var txt = "<p>User-agent header: " + navigator.userAgent + "</p>";
$("#agent").writeln(txt);
Then Anywhere in the body tag of the html file
create a div tag and include an id="agent"
Consider a computer system with a 32-bit logical address and 4-KB page size. The system
supports up to 512 MB of physical memory. How many entries are there in each of the following?
a. A conventional single-level page table?
b. An inverted page table?
Answer:
Conventional single-level page table [tex]2^{20}[/tex] pages
Inverted page table are [tex]2^{17}[/tex] frame
Explanation:
given data
logical address = 32-bit = [tex]2^{32}[/tex] Bytes
page size = 4-KB = [tex]2^{12}[/tex] Bytes
physical memory = 512 MB = [tex]2^{29}[/tex] bytes
solution
we get here number of pages that will be
number of pages = [tex]\frac{logical\ address}{page\ size}[/tex] ..............1
put here value
number of pages = [tex]\frac{2^{32}}{2^{12}}[/tex]
number of pages = [tex]2^{20}[/tex] pages
and
now we get number of frames that is
number of frames = [tex]\frac{physical\ memory}{page\ size}}[/tex] ............2
number of frames = [tex]\frac{2^{29}}{2^{12}}[/tex]
number of frames = [tex]2^{17}[/tex] frame
so
Conventional single-level page table [tex]2^{20}[/tex] pages
and
Inverted page table are [tex]2^{17}[/tex] frame