Answer:
Explanation:
g = Acceleration due to gravity at sea level = 
R = Radius of Earth = 6371000 m
h = Altitude of observatory = 4205 m
Change in acceleration due to gravity due to change in altitude is given by
Weight at sea level
Weight at the given height
Change in weight 
Her weight reduces by .
R=10+15+30
55 is the answer to the question
When an object has the same number of positive and negative charges, its electrical charge will become neutral.
What is an electric charge?
When the matter is put in an electromagnetic field, it has an electric charge, which causes it to experience a force. A positive or negative electric charge can exist.
Now when the two equal magnitude charges with opposite natures come together they become neutral.
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EPA Regulations provides a certified course for the technicians involved in the Air-conditioning system.
Answer: Option (b)
<u>Explanation:</u>
The EPA regulation has implemented an act called the "Clean Air Act" under the "section of 609".
This act provides some basic requirements for EPA Regulation such as follows;
- Refrigerant: This unit must be approved by EPA Regulations before being implemented into the atmosphere.
- Servicing: This system provides a certified course for technicians in service and also approve them with proper refrigerant equipment.
- Reuse Refrigerants: The use of recycled refrigerants must be properly monitored before it comes in to serve.
Answer:
- public class Main {
- public static void main(String[] args) {
- String testString = "abscacd";
-
- String evenStr = "";
- String oddStr = "";
-
- for(int i=testString.length() - 1; i >= 0; i--){
-
- if(i % 2 == 0){
- evenStr += testString.charAt(i);
- }
- else{
- oddStr += testString.charAt(i);
- }
- }
-
- System.out.println(evenStr + oddStr);
- }
- }
Explanation:
Firstly, let declare a variable testString to hold an input string "abscacd" (Line 1).
Next create another two String variable, evenStr and oddStr and initialize them with empty string (Line 5-6). These two variables will be used to hold the string at even index and odd index, respectively.
Next, we create a for loop that traverse the characters of the input string from the back by setting initial position index i to testString.length() - 1 (Line 8). Within the for-loop, create if and else block to check if the current index, i is divisible by 2, (i % 2 == 0), use the current i to get the character of the testString and join it with evenStr. Otherwise, join it with oddStr (Line 10 -14).
At last, we print the concatenated evenStr and oddStr (Line 18).
