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4 years ago

Zionjasean17 zionjasean17

Engineering

2 answers:

yuradex [85]4 years ago

8 0

Me ksidhejdndhshdjdnhdhdnnd

Julli [10]4 years ago

4 0

Answer:

Zion

Explanation:

Zionjasean17 is your answer because of the capitalization. Good article btw.

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When using bits to represent fractions of a number, can you create all possible fractions? Why or why not?

Strike441 [17]

Answer:

Explanation:

There are an infinite number of fractions. You cannot create all possible fractions with a finite number of bits.

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4 years ago

When lining up the song on the tempo grid it is important to allow

Reika [66]

Tempo decides the speed at which the music is played.

<u>Explanation:</u>

The Tempo of a bit of music decides the speed at which it is played, and is estimated in beats per minute (BPM). The 'beat' is dictated when mark of the piece, so 100 BPM in 4/4 compares to 100 quarter notes in a single moment.

A quick tempo, prestissimo, has somewhere in the range of 200 and 208 beats for each moment, presto has 168 to 200 beats for every moment, allegro has somewhere in the range of 120 and 168 beats for every moment, moderato has 108 to 120 beats for every moment, moderately slow and even has 76 to 108, adagio has 66 to 76, larghetto has 60 to 66, and largo, the slowest rhythm, has 40 to 60.

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3 years ago

Rewrite the following loops, using the enhanced for an array of floating-point numbers. Write your code within comments

dmitriy555 [2]

Answer:

Explanation:poop

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3 years ago

Java Programming: Loops and Conditionals

velikii [3]

Answer:

import java.util.Scanner;
import java.util.Random;
public class Main {
public static void main(String[] args) {
Scanner input = new Scanner(System.in);
Random rand = new Random();
System.out.print("Enter number of questions you wish to practice: ");
int n = input.nextInt();
int i = 1;
while(i <= n){
int num1 = 1 + rand.nextInt(10);
int num2 = 1 + rand.nextInt(10);
int answer = num1 * num2;
System.out.print("Question " + i + ": " + num1 + " x " + num2 + " = ");
int response = input.nextInt();
if(response == answer){
System.out.println("Correct answer!");
}
else{
System.out.println("Wrong answer. It should be " + answer);
}
i++;
}
}
}

Explanation:

Firstly, we import Scanner and Random classes (Line 1-2) as we need to get user input for number of questions and generate random number in each question.

Next, create a Scanner and Random object (Line 7-8). Prompt user to input number of question and get the input using the Scanner object (Line 10-11).

Next create a while loop and set the condition to loop for n number of times (based on number of questions) (Line 14). In the loop, use Random object nextInt method to generate two integers between 1 -10 (Line 15-16). Prompt user to input an answer using the Scanner object (Line 18-19). If the response is matched with answer print a correct message (Line 21-22) other wise inform user they write a wrong answer and show the correct answer (Line 24-25).

8 0

4 years ago

Five Kilograms of continuous boron fibers are introduced in a unidirectional orientation into of an 8kg aluminum matrix. Calcula

Lunna [17]

Answer:

Explanation:

Given that,

Mass of boron fiber in unidirectional orientation

Mb = 5kg = 5000g

Mass of aluminum fiber in unidirectional orientation

Ma = 8kg = 8000g

A. Density of the composite

Applying rule of mixing

ρc = 1•ρ1 + 2•ρ2

Where

ρc = density of composite

1 = Volume fraction of Boron

ρ1 = density composite of Boron

2 = Volume fraction of Aluminum

ρ2 = density composite of Aluminum

ρ1 = 2.36 g/cm³ constant

ρ2 = 2.7 g/cm³ constant

To Calculate fractional volume of Boron

1 = Vb / ( Vb + Va)

Vb = Volume of boron

Va = Volume of aluminium

Also

To Calculate fraction volume of aluminum

2= Va / ( Vb + Va)

So, we need to get Va and Vb

From density formula

density = mass / Volume

ρ1 = Mb / Vb

Vb = Mb / ρ1

Vb = 5000 / 2.36

Vb = 2118.64 cm³

Also ρ2 = Ma / Va

Va = Ma / ρ2

Va = 8000 / 2.7

Va = 2962.96 cm³

So,

1 = Vb / ( Vb + Va)

1 = 2118.64 / ( 2118.64 + 2962.96)

1 = 0.417

Also,

2= Va / ( Vb + Va)

2 = 2962.96 / ( 2118.64 + 2962.96)

2 = 0.583

Then, we have all the data needed

ρc = 1•ρ1 + 2•ρ2

ρc = 0.417 × 2.36 + 0.583 × 2.7

ρc = 2.56 g/cm³

The density of the composite is 2.56g/cm³

B. Modulus of elasticity parallel to the fibers

Modulus of elasticity is defined at the ratio of shear stress to shear strain

The relation for modulus of elasticity is given as

Ec = = 1•Eb+ 2•Ea

Ea = Elasticity of aluminium

Eb = Elasticity of Boron

Ec = Modulus of elasticity parallel to the fiber

Where modulus of elastic of aluminum is

Ea = 69 × 10³ MPa

Modulus of elastic of boron is

Eb = 450 × 10³ Mpa

Then,

Ec = = 1•Eb+ 2•Ea

Ec = 0.417 × 450 × 10³ + 0.583 × 69 × 10³

Ec = 227.877 × 10³ MPa

Ec ≈ 228 × 10³ MPa

The Modulus of elasticity parallel to the fiber is 227.877 × 10³MPa

OR Ec = 227.877 GPa

Ec ≈ 228GPa

C. modulus of elasticity perpendicular to the fibers?

The relation of modulus of elasticity perpendicular to the fibers is

1 / Ec = 1 / Eb+ 2 / Ea

1 / Ec = 0.417 / 450 × 10³ + 0.583 / 69 × 10³

1 / Ec = 9.267 × 10^-7 + 8.449 ×10^-6

1 / Ec = 9.376 × 10^-6

Taking reciprocal

Ec = 106.66 × 10^3 Mpa

Ec ≈ 107 × 10^3 MPa

Note that the unit of Modulus has been in MPa,

7 0

3 years ago