Answer:
first is the parentheses, (3+2)=5 next is the exponent 5^2=25, next is the division 5 / 5 = 1, then the multiplication 4*1=4 and then you add 4+25=29. so the answer is 29.
Answer:
4 times around
Explanation:
The total number of teeth involved will be the same for each gear. If the front gear is connected to the pedal and it goes around twice, then 2·24 = 48 teeth will have passed the reference point.
If the rear gear is attached to the wheel, and 48 teeth pass the reference point, then it will have made ...
(48 teeth)/(12 teeth/turn) = 4 turns
(a) The number of vacancies per cubic centimeter is 1.157 X 10²⁰
(b) ρ = n X (AM) / v X Nₐ
<u>Explanation:</u>
<u />
Given-
Lattice parameter of Li = 3.5089 X 10⁻⁸ cm
1 vacancy per 200 unit cells
Vacancy per cell = 1/200
(a)
Number of vacancies per cubic cm = ?
Vacancies/cm³ = vacancy per cell / (lattice parameter)³
Vacancies/cm³ = 1 / 200 X (3.5089 X 10⁻⁸cm)³
Vacancies/cm³ = 1.157 X 10²⁰
Therefore, the number of vacancies per cubic centimeter is 1.157 X 10²⁰
(b)
Density is represented by ρ
ρ = n X (AM) / v X Nₐ
where,
Nₐ = Avogadro number
AM = atomic mass
n = number of atoms
v = volume of unit cell
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).
