Calculate the density and specific gravity of ammonia gas from the information that 8.9L of the gas weighs 6.8g

e uma. associação de profissionais que iniciou o desenvolvimento de uma metodologia de gerenciamento de projetos,para servir aos

Darina [25.2K]

Answer:

5555

Explanation:

8 0

3 years ago

2.17 LAB: Word frequencies - methods Write a program that reads a list of words. Then, the program outputs those words and their

pickupchik [31]

Using the knowledge in computational language in JAVA it is possible to write a code that reads a list of words.

<h3>Writting the code in JAVA:</h3>

import java.util.*;

//class name LabClass

public class LabClass{

//function getFrequencyOfWord with array wordList, ListSize and currWord as parameter

public static int getFrequencyOfWord(String[] wordsList, int ListSize, String currWord){

int frequency = 0;

//for loop for iteration

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

//if current word matches with words present in list

//then increase the frequency by 1

if(wordsList[i].compareTo(currWord)==0){

frequency++;

}

//return frequency at last

return frequency;

}

//main function

public static void main(String[] args) {

//creating object of Scanner class

Scanner sc = new Scanner(System.in);

int size;

//asking user to input size of list

System.out.println("Enter the size of list :");

size = sc.nextInt();

//creating array of same size as user input size

String[] wordList = new String[size];

sc.nextLine();

//asking user to input array elements

System.out.println("Enter list elements one by one :");

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

{

wordList[i] = sc.nextLine();

}

System.out.print("\n");

sc.close();

//calling function getFrequencyOfWord in every iteration by passing

//required arguments to it

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

{

String currWord = wordList[i];

int count = getFrequencyOfWord(wordList, size, currWord);

//printing result

System.err.println(currWord + " : " + count);

}

}

See more about JAVA at brainly.com/question/12978370

#SPJ1

3 0

2 years ago

In economic terms, this is a product that can be used to satisfy a want or need.

Yuki888 [10]

The term used is called “Goods”

5 0

3 years ago

Read 2 more answers

A 0. 75-kg ball is attached to a 1. 0-m rope and whirled in a vertical circle. The rope will break when the tension exceeds 450

8090 [49]

At the bottom of the circle, the ball is being pulled upward by tension in the rope and downward by its own weight, so that the net force on it is

∑ F = 450 N - (0.75 kg) g = (0.75 kg) a

where a is centripetal acceleration. At this maximum tension, the ball has a maximum centripetal acceleration of

a = (450 N - (0.75 kg) g) / (0.75 kg) = 590.2 m/s²

Then its maximum tangential speed v is such that

a = v² / (1.0 m)

⇒ v = √((1.0 m) a) ≈ 25 m/s

7 0

2 years ago

(5372 + 100y^2) + 5(10y^2 – 110) Add the two polynomials and write in the form ay^2 + b, where a and b are constants, What is th

dangina [55]

Answer:

The value of a is 150 and b is 4822.

Explanation:

First, you have to expand to remove brackets :

$(5372 + 100 {y}^{2} ) + 5(10 {y}^{2} - 110)$

$= 5372 + 100 {y}^{2} + 50 {y}^{2} - 550$

Next, you have to simplify :

$100 {y}^{2} + 50 {y}^{2} + 5372 - 550$

$= 150 {y}^{2} + 4822$

Next, you have to compare it :

$a {y}^{2} + b = 150 {y}^{2} + 4822$

$a = 150$

$b = 4822$

5 0

3 years ago