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34kurt
3 years ago
9

Which key materials are less apt to retain thermal signatures and why?

Computers and Technology
1 answer:
Iteru [2.4K]3 years ago
3 0
<span>The difference between materials in relation to the time they retain a thermal signature relates to their ability to conduct and disperse heat. Since metals are generally highly conductive, metals are likely to quickly lose heat once warmed up, and therefore quickly lose their thermal signature. Plastics and rubber are however insulators and do not conduct heat very well at all. They are likely to retain their thermal signatures for longer once exposed to heat.
</span>
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What's your opinion on Pokemon​ and pixeldip
faust18 [17]

Answer:

Its a good game but an amazing show. 10/10

Explanation:

7 0
3 years ago
Consider the following program segment. ifstream inFile; //Line 1 int x, y; //Line 2 ... //Line 3 inFile &gt;&gt; x &gt;&gt; y;
MrMuchimi

Answer:

inFile.open("progdata.dat");

Explanation:

This command opens the file which is passed to it as an argument. We are quite used to the C++'s cin and cout functions contained in iostream library for reading and writing data, however, when working will real-life stuff, we find ourselves working with huge data saved as separate files and require our programs to read them, manipulate them and possibly write back to the files... The first operation usually carried out before a file can be read or written to is open

6 0
3 years ago
Function of an actuator
RSB [31]

Answer:

an actuator is a device that uses a form of power to convert a control signal into mechanical motion

6 0
3 years ago
Read 2 more answers
Write a class called Course that represents a course taken at a school. Represent each student using the modified Student class
Vinvika [58]

Answer:

The code is given below with its appropriate output

Explanation:

//Student.java

import java.io.BufferedReader;

import java.io.FileReader;

import java.io.IOException;

import java.util.Scanner;

public class Student {

public String firstName;

public String lastName;

public float[] marks;

Scanner scanner = new Scanner(System.in);

public Student(String firstName, String lastName) {

this.firstName = firstName;

this.lastName = lastName;

}

public void getdetail() {

System.out.println("enter only five marks");

marks = new float[5];

int count = 0;

for (int i = 0; i < 5; i++) {

float mark = 0;

++count;

if (count <= 5) {

System.out.println("enter score " + count);

mark = scanner.nextFloat();

if (mark >= 0.0 && mark <= 100.0)

marks[i] = mark;

else {

System.out.println("you entered mark is invalid");

marks[i] = 0;

}

} else

System.out.println(" maximum is five only");

}

}

public void tostring() {

System.out.println("firstname :" + "\t" + firstName + "\t" + "lastname :"

+ "\t" + lastName + "\n" + "the marks are");

int count = 0;

for (float mark : marks) {

count++;

System.out.println("the mark of " + count + " is " + mark);

}

}

public void Averagecal() {

float sum = 0.0f;

for (float mark : marks) {

sum = sum + mark;

}

float avg = sum / 5;

System.out.println("average of a student is " + avg);

}

}

//Course.java

import java.io.*;

import java.io.BufferedReader;

import java.io.FileReader;

import java.io.IOException;

import java.util.Scanner;

public class Course {

public String courseName;

public int number;

Student[] students;

Scanner scanner = new Scanner(System.in);

public Course(String courseName) {

this.courseName = courseName;

}

public void addStudents() {

System.out

.println("how many no of students you want to add for this particular course");

int count = scanner.nextInt();

students = new Student[count];

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

System.out.println("enter the first name");

String fname = scanner.next();

System.out.println("enter the last name");

String lname = scanner.next();

Student s = new Student(fname, lname);

s.getdetail();

s.tostring();

s.Averagecal();

students[i] = s;

}

}

public void roll() {

int count = 0;

for (Student stud : students) {

count++;

}

System.out.println("the no of students in this " + courseName

+ " course is " + count);

}

}

//Driver.java

import java.io.BufferedReader;

import java.io.FileReader;

import java.io.IOException;

import java.util.Scanner;

public class Driver {

public static void main(String[] args) {

Course course = new Course("java");

course.addStudents();

course.roll();

}

}

<u>output: </u>

C:\Users\Daniel\Desktop\New folder>javac Student.java

C:\Users\Danie;\Desktop\New folder>javac Course.java

C:\Users\Daniel\Desktop\New folder>javac Driver.java

C:\Users\Daniel\Desktop\New folder>java Driver

how many no of students you want to add for this particular course ?

1

enter the first name

john

enter the last name

rich

enter only five marks

enter score 1

89

enter score 2

91

enter score 3

85

enter score 4

78

enter score 5

96

firstname : john lastname : rich

the marks are

the mark of 1 is 89.0

the mark of 2 is 91.0

the mark of 3 is 85.0

the mark of 4 is 78.0

the mark of 5 is 96.0

average of a student is 87.8

the no of students in this java course is 1

C:\Users\Ravi\Desktop\New folder>

3 0
3 years ago
Please add thenodes given below to construct the AVL tree show all the necessarysteps,
Paha777 [63]

Answer:

30,33,37,18,23,34,15,38,40,17

To construct the AVL tree, follow these steps and diagrams are shown in the image:

• Add 30 to the tree as the root node. Then add 33 as the right child because of 33 is greater than 30 and AVL tree is a binary search tree.

• Then add 37 as the right child of 33. Here the balance factor of node 30 becomes 0-2 = -2, unbalanced.

Use RR rotation, make node 33 the root node, 30 as the left child of 33 and 37 as the right child of 33.

• Now add 18 as the left child of 30. And 23 as the right child of 18. Here the balance factor of 30 becomes 2-0 = 2. It’s unbalanced.

Use LR rotation, make 23 the parent of 18 and 30.  

• Now add 34 as the left child of 37 and 15 as the left child of 18. Add 38 as the right child of 37. And then add 40 as the right child of 38.

• Now adding 17 as the right child of 15 makes the tree unbalanced at 18.

Use LR rotation, make 17 as the parent of 15 and 18.

Explanation:

The balance factor of a node can be either 0,1 or -1. Else the tree is called unbalanced at the node.

If the inserted node is in the left subtree of the left subtree of the unbalance node, then perform LL rotation.

If the inserted node is in the right subtree of the right subtree of the unbalance node, then perform RR rotation.

If the inserted node is in the left subtree of the right subtree of the unbalance node, then perform RL rotation.

If the inserted node is in the right subtree of the left subtree of the unbalance node, then perform LR rotation.

7 0
3 years ago
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