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Inga [223]
3 years ago
13

- WHEN YOU ARE TOWING A TRAILER:

Engineering
1 answer:
zheka24 [161]3 years ago
7 0

Answer:

And Im still going with B..

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Differences between acidic and basic Bessemer process​
dybincka [34]

Answer:

In the acid processes, deoxidation can take place in the furnaces, leaving a reasonable time for the inclusions to rise into the sla*g and so be removed before casting. Whereas in the basic furnaces, deoxidation is rarely carried out in the presence of the sla*g, otherwise phosphorus would return to the metal.

5 0
2 years ago
In order to impress your neighbors and improve your vision in traffic jams, you decide to mount a cylindrical periscope 2.0 m hi
kondaur [170]
Follow @richard.gbe on Instagram for the answer
5 0
3 years ago
Which answer best describes the relationship between the left-hand and right-hand panels in both File Explorer and Finder?
Art [367]

Answer: I think The left-hand panel shows like a high level list of folders. And Right-hand panel shows the contents of selected folder.

4 0
2 years ago
Velocity components in an incompressible flow are: v = 3xy + x^2 y: w = 0. Determine the velocity component in the x-direction.
cupoosta [38]

Answer:

Velocity component in x-direction u=-\frac{3}{2}x^2-\frac{1}{3}x^3.

Explanation:

   v=3xy+x^{2}y

We know that for incompressible flow

   \frac{\partial u}{\partial x}+\frac{\partial v}{\partial y}=0

\frac{\partial v}{\partial y}=3x+x^{2}

So   \frac{\partial u}{\partial x}+3x+x^{2}=0

\frac{\partial u}{\partial x}= -3x-x^{2}

By integrate with respect to x,we will find

u=-\frac{3}{2}x^2-\frac{1}{3}x^3+C

So the velocity component in x-direction u=-\frac{3}{2}x^2-\frac{1}{3}x^3.

3 0
3 years ago
Number pattern Write a recursive method called print Pattern() to output the following number pattern. Given a positive integer
lilavasa [31]

Answer:

See explaination

Explanation:

Code;

import java.util.Scanner;

public class NumberPattern {

public static int x, count;

public static void printNumPattern(int num1, int num2) {

if (num1 > 0 && x == 0) {

System.out.print(num1 + " ");

count++;

printNumPattern(num1 - num2, num2);

} else {

x = 1;

if (count >= 0) {

System.out.print(num1 + " ");

count--;

if (count < 0) {

System.exit(0);

}

printNumPattern(num1 + num2, num2);

}

}

}

public static void main(String[] args) {

Scanner scnr = new Scanner(System.in);

int num1;

int num2;

num1 = scnr.nextInt();

num2 = scnr.nextInt();

printNumPattern(num1, num2);

}

}

See attachment for sample output

3 0
3 years ago
Read 2 more answers
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