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fenix001 [56]
3 years ago
8

If x is a string, then x = new String("OH"); and x = "OH"; will accomplish the same thing. Group of answer choices True False

Chemistry
2 answers:
Vinvika [58]3 years ago
5 0

Answer:

True is the correct answer to the above question.

Explanation:

  • If x is a string then it can be assigned by the help of two ways in java:
  • By the help of constructor:- When we write " x = new String("OH");", then it will create a pass a string "OH" into the constructor. It is because the String is a class in java and x is an object created by the constructor of the String class.
  • With the help of assigning: The "x= OH", which assigns the value of x which is an object of String class it can also use the constructor to initialize the "OH" string on the class.
  • The above question states the two scenarios which are defined above. Hence the question statement is true.
Akimi4 [234]3 years ago
5 0

Answer:

"True" is the correct answer to this question.

Explanation:

The program to the given question as follows:

Program:

public class data //defining class

{

  public static void main (String [] aw)//defining the main method

  {

String x="OH"; //defining string variable x and assign value

System.out.println("assign value: "+x); //print value

x = new String("OH"); //defining instance variable and assign value

System.out.println("assign value by creating instance: "+x); //print value

  }

}

Output:

assign value: OH

assign value by creating instance: OH

Explanation of the program:

In the above java program, a class data is defined, inside the class the main method is declared, In this main method a string variable "x" is defined that holds a value "OH", then we the print function to print this variable value.

In the next line, An instance of variable x is created, which holds a value "OH" in its parameter. In this question, both are correct because both hold the same value.

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Answer: The new volume at different given temperatures are as follows.

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Explanation:

Given: V_{1} = 571 mL,       T_{1} = 26^{o}C

(a) T_{2} = 5^{o}C

The new volume is calculated as follows.

\frac{V_{1}}{T_{1}} = \frac{V_{2}}{T_{2}}\\\frac{571 mL}{26^{o}C} = \frac{V_{2}}{5^{o}C}\\V_{2} = 109.81 mL

(b) T_{2} = 95^{o}F

Convert degree Fahrenheit into degree Cesius as follows.

(1^{o}F - 32) \times \frac{5}{9} = ^{o}C\\(95^{o}F - 32) \times \frac{5}{9} = 35^{o}C

The new volume is calculated as follows.

\frac{V_{1}}{T_{1}} = \frac{V_{2}}{T_{2}}\\\frac{571 mL}{26^{o}C} = \frac{V_{2}}{35^{o}C}\\V_{2} = 768.65 mL

(c) T_{2} = 1095 K = (1095 - 273)^{o}C = 822^{o}C

The new volume is calculated as follows.

\frac{V_{1}}{T_{1}} = \frac{V_{2}}{T_{2}}\\\frac{571 mL}{26^{o}C} = \frac{V_{2}}{822^{o}C}\\V_{2} = 18052.38 mL

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