To put out a class D metal fire, you must smother the fire and eliminate the oxygen element in the fire.
<h3>What is a Class D fire?</h3>
A class D fire is a type of fire that cannot be extinguished by water. This is because adding water to it reacts with other elements in the fire intensifying the fire even more.
Smothering in this context involves adding a solution like carbon dioxide (CO2) into the fire, this results in a reduction of oxygen in the atmosphere surrounding the class D fire.
By so doing, smothering the fire eliminates the oxygen element in the fire, thereby extinguishing the fire.
You can learn more about extinguishing fires here https://brainly.in/question/760550
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Answer:
See explaination
Explanation:
Let's define tuple as an immutable list of Python objects which means it can not be changed in any way once it has been created.
Take a look at the attached file for a further detailed and step by step solution of the given problem.
Answer:
Program that removes all spaces from the given input
Explanation:
// An efficient Java program to remove all spaces
// from a string
class GFG
{
// Function to remove all spaces
// from a given string
static int removeSpaces(char []str)
{
// To keep track of non-space character count
int count = 0;
// Traverse the given string.
// If current character
// is not space, then place
// it at index 'count++'
for (int i = 0; i<str.length; i++)
if (str[i] != ' ')
str[count++] = str[i]; // here count is
// incremented
return count;
}
// Driver code
public static void main(String[] args)
{
char str[] = "g eeks for ge eeks ".toCharArray();
int i = removeSpaces(str);
System.out.println(String.valueOf(str).subSequence(0, i));
}
}
Answer:
8.85 Ω
Explanation:
Resistance of a wire is:
R = ρL/A
where ρ is resistivity of the material,
L is the length of the wire,
and A is the cross sectional area.
For a round wire, A = πr² = ¼πd².
For aluminum, ρ is 2.65×10⁻⁸ Ωm, or 8.69×10⁻⁸ Ωft.
Given L = 500 ft and d = 0.03 in = 0.0025 ft:
R = (8.69×10⁻⁸ Ωft) (500 ft) / (¼π (0.0025 ft)²)
R = 8.85 Ω