All categories

2 years ago

Technician A says that reinforcements may be made of plastic.

Engineering

1 answer:

vlada-n [284]2 years ago

6 0

Answer:

Technician A Is right

Explanation:

Reinforcements, as the name suggests, are used to enhance the mechanical properties of a plastic. Finely divided silica, carbon black, talc, mica, and calcium carbonate, as well as short fibres of a variety of materials, can be incorporated as particulate fillers.

You might be interested in

Can crushers help us recycle in a space efficient way which is good for saving the earth and for giving you more room in your ap

777dan777 [17]

Answer:

Force magnitude = 296.7 N

Explanation:

Detailed illustration is given in the attached document.

6 0

3 years ago

A rectangular steel bar, with 8" x 0.75" cross-sectional dimensions, has equal and opposite moments applied to its ends.

denpristay [2]

Answer:

Part a: The yield moment is 400 k.in.

Part b: The strain is $8.621 \times 10^{-4} in/in$

Part c: The plastic moment is 600 ksi.

Explanation:

Part a:

As per bending equation

$\frac{M}{I}=\frac{F}{y}$

Here

M is the moment which is to be calculated
I is the moment of inertia given as

$I=\frac{bd^3}{12}$

Here

b is the breath given as 0.75"
d is the depth which is given as 8"

$I=\frac{bd^3}{12}\\I=\frac{0.75\times 8^3}{12}\\I=32 in^4$

y is given as

$y=\frac{d}{2}\\y=\frac{8}{2}\\y=4"\\$

Force is 50 ksi

$\frac{M_y}{I}=\frac{F_y}{y}\\M_y=\frac{F_y}{y}{I}\\M_y=\frac{50}{4}{32}\\M_y=400 k. in$

The yield moment is 400 k.in.

Part b:

The strain is given as

$Strain=\frac{Stress}{Elastic Modulus}$

The stress at the station 2" down from the top is estimated by ratio of triangles as

$F_{2"}=\frac{F_y}{y}\times 2"\\F_{2"}=\frac{50 ksi}{4"}\times 2"\\F_{2"}=25 ksi$

Now the steel has the elastic modulus of E=29000 ksi

$Strain=\frac{Stress}{Elastic Modulus}\\Strain=\frac{F_{2"}}{E}\\Strain=\frac{25}{29000}\\Strain=8.621 \times 10^{-4} in/in$

So the strain is $8.621 \times 10^{-4} in/in$

Part c:

For a rectangular shape the shape factor is given as 1.5.

Now the plastic moment is given as

$shape\, factor=\frac{Plastic\, Moment}{Yield\, Moment}\\{Plastic\, Moment}=shape\, factor\times {Yield\, Moment}\\{Plastic\, Moment}=1.5\times400 ksi\\{Plastic\, Moment}=600 ksi$

The plastic moment is 600 ksi.

3 0

3 years ago

I don’t get this it’s hella hard

qwelly [4]

Answer:

V₂ = 20 V

Vt = 20 V

V₁ = 20 V

V₃ = 20 V

I₁ = 10 mA

I₃ = 3.33 mA

It = 18.33 mA

Rt = 1090.91 Ω

Pt = 0.367 W

P₁ = 0.2 W

P₂ = 0.1 W

P₃ = 0.067 W

Explanation:

Part of the picture is cut off. I assume there is a voltage source Vt there?

First, use Ohm's law to find V₂.

V = IR

V₂ = (0.005 A) (4000 Ω)

V₂ = 20 V

R₁ and R₃ are in parallel with R₂ and the voltage source Vt. That means V₁ = V₂ = V₃ = Vt.

V₁ = 20 V

V₃ = 20 V

Vt = 20 V

Now we can use Ohm's law again to find I₁ and I₃.

V = IR

I = V/R

I₁ = (20 V) / (2000 Ω)

I₁ = 0.01 A = 10 mA

I₃ = (20 V) / (6000 Ω)

I₃ = 0.00333 A = 3.33 mA

The current It passing through Vt is the sum of the currents in each branch.

It = I₁ + I₂ + I₃

It = 10 mA + 5 mA + 3.33 mA

It = 18.33 mA

The total resistance is the resistance of the parallel resistors:

1/Rt = 1/R₁ + 1/R₂ + 1/R₃

1/Rt = 1/2000 + 1/4000 + 1/6000

Rt = 1090.91 Ω

Finally, the power is simply each voltage times the corresponding current.

P = IV

Pt = (0.01833 A) (20 V)

Pt = 0.367 W

P₁ = (0.010 A) (20 V)

P₁ = 0.2 W

P₂ = (0.005 A) (20 V)

P₂ = 0.1 W

P₃ = (0.00333 A) (20 V)

P₃ = 0.067 W

7 0

3 years ago

Q4 a.

dedylja [7]

The Java program that accepts a matrix of M × N order and then interchanges diagonals of the matrix is given below:

<h3>Steps: </h3>

1. We can only interchange diagonals for a square matrix.
2. Therefore, we would have to create a square matrix of size [M × M].
3. We would check whether the matrix is a square matrix or not. If the matrix is square then follow step 3 else terminate the program.
4. Apply logic for interchange diagonal of the matrix some logic is given below.

// Java Program to Accept a Matrix of Order M x N &

// Interchange the Diagonals

import java.util.Scanner;

public class InterchangeDiagonals {

public static void main(String[] args)

{

// declare variable

int m, n, i, j, temp;

// create a object of scanner class

Scanner sc = new Scanner(System.in);

System.out.print("Enter number of rows ");

// take number of rows

m = sc.nextInt();

System.out.print("Enter number of columns ");

// take number of columns

n = sc.nextInt();

// declare a mxn order array

int a[][] = new int[m][n];

// if block it's execute when m is equals to n

if (m == n) {

System.out.println(

"Enter all the values of matrix ");

// take the matrix inputs

for (i = 0; i < m; i++) {

for (j = 0; j < n; j++) {

a[i][j] = sc.nextInt();

}

System.out.println("original Matrix:");

// print the original matrix

for (i = 0; i < m; i++) {

for (j = 0; j < n; j++) {

System.out.print(a[i][j] + " ");

}

System.out.println("");

}

// perform interchange

for (j = 0; j < m; j++) {

temp = a[j][j];

a[j][j] = a[j][n - 1 - j];

a[j][n - 1 - j] = temp;

}

System.out.println(

" after interchanging diagonals of matrix ");

// print interchanged matrix

for (i = 0; i < m; i++) {

for (j = 0; j < n; j++) {

System.out.print(a[i][j] + " ");

}

System.out.println("");

}

// else block it's only execute when m is not equals

// to n

else {

System.out.println("Rows not equal to columns");

}

Read more about java programming here:

brainly.com/question/18554491

#SPJ1

7 0

1 year ago

Do not discharge wastewater into storm sewers because they flow directly into local surface waters. A. true B. false

enot [183]

Answer:

True

Explanation:

Wastewater should never be discharged into storm sewers. Storm sewers are constructed to store rainwater.

Rainwater passes through storm sewers to the nearest underground water resource if the wastewater discharge into storm sewers, then they will also pollute the ground water, so usually the dirty water is not put in the storm sewers.

3 0

3 years ago