When machining rotors, what is the reason for setting the indexing collars to zero?

A. To know how much metal was taken off of each side.

B. To make sure that no cut exceeds .010 inch

C. To determine the proper cut speed of the Brake Lathe.

D. To measure the depth of the grooves on the rotor surface.

Answers

Answer 1

Answer:

To determine the proper cut speed of the brake lathe

Answer 2

To determine the proper cut speed of the brake lathe reason for setting the indexing collars to zero.

What is indexing collars?

Thus, they can have a maximum of 256 colors, which is the amount of colors that 8 bits of information can specify (28 = 256). Fewer colors are produced by lower bit depths, which also result in smaller files. At the conclusion of this chapter, in Section 19.8, this is covered.

"Indexed color" refers to the fact that a color table contains the palette—the collection of colors—of the image. A reference (or "index") to a table cell containing the pixel's color is present in each pixel of the image.

By choosing Image Mode Color Table in Photoshop, you can view the table for an indexed color image.

Therefore, To determine the proper cut speed of the brake lathe reason for setting the indexing collars to zero.

To learn more about indexing collars, refer to the link:

https://brainly.com/question/12608731

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Related Questions

Why would an organization need to be aware of the responsibilities of hosting

personal data in its platforms?

Answers

Answer:

Providing transparent services for platform customers. Following data protection regulations to avoid disruptions from lack of compliance.

User ideas for ro blox? I prefer no numbers or underscores.

Answers

do what the other person said

For this exercise, you are going to complete the printScope() method in the Scope class. Then you will create a Scope object in the ScopeTester and call the printScope.
The method will print the name of each variable in the Scope class, as well as its corresponding value. There are 5 total variables in the Scope class, some of which can be accessed directly as instance variable, others of which need to be accessed via their getter methods.
For any variable that can be accessed directly, use the variable name. Otherwise, use the getter method.
Sample Output:
The output of the printScope method should look like this:
a = 5
b = 10
c = 15
d = 20
e = 25
public class ScopeTester
{
public static void main(String[] args)
{
// Start here!
}
}
public class Scope
{
private int a;
private int b;
private int c;
public Scope(){
a = 5;
b = 10;
c = 15;
}
public void printScope(){
//Start here
}
public int getA() {
return a;
}
public int getB() {
return b;
}
public int getC() {
return c;
}
public int getD(){
int d = a + c;
return d;
}
public int getE() {
int e = b + c;
return e;
}
}

Answers

Answer:

Explanation:

The following is the entire running Java code for the requested program with the requested changes. This code runs perfectly without errors and outputs the exact Sample Output that is in the question...

public class ScopeTester

{

public static void main(String[] args)

{

       Scope scope = new Scope();

       scope.printScope();

}

}

public class Scope

{

private int a;

private int b;

private int c;

public Scope(){

a = 5;

b = 10;

c = 15;

}

public void printScope(){

       System.out.println("a = " + a);

       System.out.println("b = " + b);

       System.out.println("c = " + c);

       System.out.println("d = " + getD());

       System.out.println("e = " + getE());

}

public int getA() {

return a;

}

public int getB() {

return b;

}

public int getC() {

return c;

}

public int getD(){

int d = a + c;

return d;

}

public int getE() {

int e = b + c;

return e;

}

}

Assume in the for loop header, the range function has the three arguments: range (1, 10, 3), if you were to print out the value of the variable
in the for loop header, what will be printed out? List the values and separate them with a comma.

Answers

Answer:

1, 4, 7

Explanation:

The instruction in the question can be represented as:

for i in range(1,10,3):

   print i

What the above code does is that:

It starts printing the value of i from 1

Increment by 3

Then stop printing at 9 (i.e.. 10 - 1)

So: The sequence is as follows

Print 1

Add 3, to give 4

Print 4

Add 3, to give 7

Print 7

Add 3, to give 10 (10 > 10 - 1).

So, it stops execution.

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