The basic functions of Brake Control Valves is?

Computers and Technology

1 answer:

GrogVix [38]2 years ago

3 0

Answer:

This valve is necessary for optimizing front-to-rear bias, also referred to as brake balance. It is a spring-loaded component that activates when fluid pressure builds when you step on the brake pedal. Then, the valve's plunger unseats and fluid rushes into the calibrated range.

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Which of the following is NOT an advantage to using inheritance?

IrinaK [193]

Answer:

D)One big superclass can be used instead of many little classes.

Explanation:

Inheritance could be described as getting new class on another existing class with almost same execution,

inheritance allows codes to be re-used. It should be noted that inheritance bring about the code sharing among class.

The advantages of using inheritance includes but not limited to

✓ Code that is shared between classes needs to be written only once.

✓Similar classes can be made to behave consistently.

✓ Enhancements to a base class will automatically be applied to derived classes.

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3 years ago

What do you find when you first open a word processor and how do you adjust the document Indentitions

Ilia_Sergeevich [38]

When you first open a word processor document you will find a document with the default indents and margins.

In order to adjust the document indentations the easiest way to do this is to place the cursor where you want the indent to be and then adjust the indent arrow on the ruler at the top of the document to the appropriate value.

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2 years ago

. (a) Prove or disprove carefully and in detail: (i) Θ is transitive and (ii) ω is transitive. (b) Assume n is a positive intege

Sergio [31]

Answer:

The Following are the solution to this question:

Explanation:

In Option a:

In the point (i) $\Omega$ is transitive, which means it converts one action to others object because if $\Omega(f(n))=g(n)$ indicates $c.g(n)$ . It's true by definition, that becomes valid. But if $\Omega(g(n))=h(n)$ , which implies $c.h(n)$ . it's a very essential component. If $c.h(n) < = g(n) = f(n) \$ . They $\Omega(f(n))$ will also be $h(n)$ .

In point (ii), The value of $\Theta$ is convergent since the $\Theta(g(n))=f(n)$ . It means they should be dual a and b constant variable, therefore $a.g(n)$ could only be valid for the constant variable, that is $\frac{1}{a}\ \ and\ \ \frac{1}{b}$ .

In Option b:

In this algorithm, the input size value is equal to 1 object, and the value of A is a polynomial-time complexity, which is similar to its outcome that is $O(n^{2})$ . It is the outside there will be a loop(i) for n iterations, that is also encoded inside it, the for loop(j), which would be a loop $(n^{2})$ . All internal loops operate on a total number of $N^{2}$ generations and therefore the final time complexity is $O(n^{2})$ .