Answer:
The Java Program for the given problem is as below. Directly copy the code and run it on your machine.
Explanation:
Refer the Screenshots attached for the output.
import java.io.BufferedReader;
import java.io.InputStreamReader;
public class TheSpeedOfSound {
public static void main(String[] s)
{
String medium;
double distance;
double time;
try{
BufferedReader choice = new BufferedReader(new InputStreamReader(System.in));
System.out.println("Enter one of the following: air, water, or steel: ");
medium = choice.readLine(); // reading input i.e. air, water or steel
//check for air water and steel
if (medium.equalsIgnoreCase("air") || medium.equalsIgnoreCase("water") || medium.equalsIgnoreCase("steel")){
System.out.println("Enter the distance the sound wave will travel: ");
distance = Double.parseDouble(choice.readLine()); // read distance value if it is air water or steel
switch (medium)
{
//if medium is air
case "air":
time = distance/1100;
System.out.print("It will take " + time + " seconds.");
break;
//if medium is water
case "water":
time = distance/4900;
System.out.print("It will take " + time + " seconds.");
break;
//if medium is steel
case "steel":
time = distance/16400;
System.out.print("It will take " + time + " seconds.");
break;
}
}
else{
System.out.print("Sorry, you must enter air, water, or steel.");
}
}
catch(Exception e){
e.printStackTrace();
}
}
}
ATM (Automatic Teller Machine) is a banking terminal that accepts deposits and dispenses cash. ATMs are activated by inserting cash (in cases of ATM Depositing) or debit /credit card that contain the user's account number and PIN on a magnetic stripe (for cash withdrawals)
<h2>A <u>Rogue attack</u> utilizes software just to attempt hundreds of frequent phrases in a row.</h2>
- Rogue access equipment would be any WLAN radio channel that isn't even authorized to communicate or join the corporate connection.
- Whenever they have been misconfigured as well as set up without authentication, it creates a new security vulnerability potentially gaining simple access to a private network.
Thus the response above is correct.
Learn more about software attacks here:
brainly.com/question/25407509
Code:
def myAppend( str, ch ):
# Return a new string that is like str but with
# character ch added at the end
return str + ch
def myCount( str, ch ):
# Return the number of times character ch appears
# in str.
# initiaalizing count with 0
count = 0
# iterating over every characters present in str
for character in str:
# incrementing count by 1 if character == ch
if character == ch:
count += 1
# returning count
return count
def myExtend( str1, str2 ):
# Return a new string that contains the elements of
# str1 followed by the elements of str2, in the same
# order they appear in str2.
# concatenating both strings and returning its result
return str1 + str2
def myMin( str ):
# Return the character in str with the lowest ASCII code.
# If str is empty, print "Empty string: no min value"
# and return None.
if str == "":
print("Empty string: no min value")
return None
# storing first character from str in char
char = str[0]
# iterating over every characters present in str
for character in str:
# if current character is lower than char then
# assigning char with current character
if character < char:
char = character
# returning char
return char
def myInsert( str, i, ch ):
# Return a new string like str except that ch has been
# inserted at the ith position. I.e., the string is now
# one character longer than before.
# Print "Invalid index" if
# i is greater than the length of str and return None.
if i > len(str):
print("Invalid index")
return None
# str[:i] gives substring starting from 0 and upto ith position
# str[i:] gives substring starting from i and till last position
# returning the concatenated result of all three
return str[:i]+ch+str[i:]
def myPop( str, i ):
# Return two results:
# 1. a new string that is like str but with the ith
# element removed;
# 2. the value that was removed.
# Print "Invalid index" if i is greater than or
# equal to len(str), and return str unchanged and None
if i >= len(str):
print("Invalid index")
return str, None
# finding new string without ith character
new_str = str[:i] + str[i+1:]
# returning new_str and popped character
return new_str, str[i]
def myFind( str, ch ):
# Return the index of the first (leftmost) occurrence of
# ch in str, if any. Return -1 if ch does not occur in str.
# finding length of the string
length = len(str)
# iterating over every characters present in str
for i in range(length):
# returning position i at which character was found
if str[i]==ch:
return i
# returning -1 otherwise
return -1
def myRFind( str, ch ):
# Return the index of the last (rightmost) occurrence of
# ch in str, if any. Return -1 if ch does not occur in str.
# finding length of the string
length = len(str)
# iterating over every characters present in str from right side
for i in range(length-1, 0, -1):
# returning position i at which character was found
if str[i]==ch:
return i
# returning -1 otherwise
return -1
def myRemove( str, ch ):
# Return a new string with the first occurrence of ch
# removed. If there is none, return str.
# returning str if ch is not present in str
if ch not in str:
return str
# finding position of first occurence of ch in str
pos = 0
for char in str:
# stopping loop if both character matches
if char == ch:
break
# incrementing pos by 1
pos += 1
# returning strig excluding first occurence of ch
return str[:pos] + str[pos+1:]
def myRemoveAll( str, ch ):
# Return a new string with all occurrences of ch.
# removed. If there are none, return str.
# creating an empty string
string = ""
# iterating over each and every character of str
for char in str:
# if char is not matching with ch then adding it to string
if char!=ch:
string += char
# returning string
return string
def myReverse( str ):
# Return a new string like str but with the characters
# in the reverse order.
return str[::-1]
