Answer:
None of these is correct.
Explanation:
The wave particle duality has to do with Louis de Broglie's proposition that matter could exist as waves or particles.
According to him, matter poseses an associated wavelength. Hence, a certain wavelength is traceable to the hydrogen atom.
This wavelength is the ratio of Plank's constant to the momentum of the hydrogen atom
Answer:
SIR IT IS D HOPE THIS HELPS (☞ﾟヮﾟ)☞☜(ﾟヮﾟ☜)
Explanation:
Answer:
See attached picture.
Explanation:
See attached picture for explanation.
Answer:
The pressure difference across hatch of the submarine is 3217.68 kpa.
Explanation:
Gauge pressure is the pressure above the atmospheric pressure. If we consider gauge pressure for finding pressure differential then no need to consider atmospheric pressure as they will cancel out. According to hydrostatic law, pressure varies in the z direction only.
Given:
Height of the hatch is 320 m
Surface gravity of the sea water is 1.025.
Density of water 1000 kg/m³.
Calculation:
Step1
Density of sea water is calculated as follows:
Here, density of sea water is
, surface gravity is S.G and density of water is 
.
Substitute all the values in the above equation as follows:
kg/m³.
Step2
Difference in pressure is calculated as follows:
pa.
Or
kpa.
Thus, the pressure difference across hatch of the submarine is 3217.68 kpa.
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).
