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Gelneren [198K]
3 years ago
12

The title block generally contains ________.

Engineering
1 answer:
gavmur [86]3 years ago
7 0
B the mechanical plans
You might be interested in
1. Copy the file Pay.java (see Code Listing 1.1) from the Student CD or as directed by your instructor. 2. Open the file in your
Mrac [35]

Answer:

Code Listing 1.1 (Pay.java)

import java.util.Scanner; // Needed for the Scanner class

/**

This program calculates the user's gross pay.

*/

public class Pay

{

public static void main(String[] args)

{

// Create a Scanner object to read from the keyboard. Scanner keyboard = new Scanner(System.in);

// Identifier declarations

double hours; // Number of hours worked

double rate; // Hourly pay rate double pay; // Gross pay

// Display prompts and get input. System.out.print("How many hours did you work? "); hours = keyboard.nextDouble();

System.out.print("How much are you paid per hour? ");

rate = keyboard.nextDouble();

// Perform the calculations. if(hours <= 40)

pay = hours * rate;

else

pay = (hours - 40) * (1.5 * rate) + 40 * rate;

// Display results. System.out.println("You earned $" + pay);

}

}

Code Listing 1.2 (SalesTax.java)

import java.util.Scanner; // Needed for the Scanner class

/**

This program calculates the total price which includes

sales tax.

*/

public class SalesTax

{

public static void main(String[] args)

{

// Identifier declarations final double TAX_RATE = 0.055; double price;

double tax

double total; String item;

// Create a Scanner object to read from the keyboard. Scanner keyboard = new Scanner(System.in);

// Display prompts and get input. System.out.print("Item description: "); item = keyboard.nextLine(); System.out.print("Item price: $");

price = keyboard.nextDouble();

// Perform the calculations. tax = price + TAX_RATE;

totl = price * tax;

// Display the results. System.out.print(item + "  $"); System.out.println(price); System.out.print("Tax $"); System.out.println(tax); System.out.print("Total $"); System.out.println(total);

}

}

5 0
3 years ago
The normal stress at gage H calculated in Part 1 includes four components: an axial component due to load P, σaxial, P, a bendin
Degger [83]

Answer:

hello your question has some missing information attached to the answer is the missing component

Answer : αaxial,p = -6.034 ksi ( compressive )

             αbend,p = 19.648 ksi ( tensile )

Explanation:

αaxial, p = \frac{-p}{A}   equation 1

αbend, p = \frac{(P*A)*\frac{d}{2} }{I_{z} } equation 2

P = load = 35 kips

A = area of column = 5.8 in^{2}

d = column cross section depth = 9.5 in

I_{Z} = 55.0 in^{4}

Hence equation 1 becomes

αaxial,p = -35 / 5.8 = - 6.034 ksi ( compressive )

equation 2 becomes

αbend, p = \frac{(35*6.5)(\frac{9.2}{2}) }{55} = + 19.648 ksi ( tensile )

7 0
3 years ago
A cylinder contains 480 cm3 of loose dry sand which weighs 820 g. Under a static load of 200 kPa the volume is reduced 1%, and t
goblinko [34]

Answer:

a.

b.

c.

Explanation:

a. void  ratio is provided by the formula: e = \frac{V_{p} }{V_{s}  }

   where , V_{p} = volume of voids

                V_{s} = volume of solid grains

for loose sand, the void space = \frac{480}{480}

                                                   = 1

b. void ratio after static load = 0.1/(480)/ (480)

                                               = 0.1

c. void ratio after vibration = [480- ( 0.1 * 480) ]/ 480

                                             = 0.9

5 0
3 years ago
Which of the following describes fibers? a)- Single crystals with extremely large length-to-diameter ratios. b)- Polycrystalline
ElenaW [278]

Answer:

b)Poly crystalline and amorphous materials with small diameter

Explanation:

Fibers have high length to diameter ratio and also have high strength.Generally length of fibers is very high and diameter is very low as compare to length.

Mostly fibers is used to transfer data from one place to another place with help of  fiber optical cables.Fiber optic cables is used in telecommunication.In these cables data covert in to the electric single and reach at define location and after data is decode and covert from electric single in to original data.

Fibers poly crystalline and amorphous materials with small diameter.

6 0
3 years ago
This question allows you to practice proving a language is non-regular via the Pumping Lemma. Using the Pumping Lemma (Theorem 1
Ulleksa [173]

Answer:

<em>L is not a regular language with formal proofs  </em>

Explanation:

<em>(a) To prove that L is not a regular language, we will use a proof by contradiction. the assumption entails  that L is a regular language. Then by the Pumping Lemma for Regular Languages, </em>

<em>there exists a pumping length p for L such that for any string s ∈ L where |s| ≥ p, </em>

<em>s = xyz subject to the following conditions: </em>

<em>(a) |y| > 0 </em>

<em>(b) |xy| ≤ p, and </em>

<em>(c) ∀i > 0, xyi </em>

<em>z ∈ L</em>

<em />

<em>(b) To determine that L is not a regular language, we mke use of proof by contradiction.  lets assume, that L is regular. Then by the Pumping Lemma for Regular Languages, it states also,</em>

<em>The pumping length, p for L such that for any string s ∈ L where |s| ≥ p, s = xyz subject  to the condtions as follows : </em>

<em>(a) |y| > 0 </em>

<em>(b) |xy| ≤ p, and </em>

<em>(c) ∀i > 0, xyi </em>

<em>z ∈ L. </em>

<em>Choose s = 0p10p </em>

<em>. Clearly, |s| ≥ p and s ∈ L. By condition (b) above, it follows is shown. by the first condition x and y are zeros.</em>

<em>for some  k > 0. Per (c), we can take i = 0 and the resulting string will still be in L. Thus,  xy0 </em>

<em>z should be in L. xy0 </em>

<em>z = xz = 0(p−k)10p </em>

<em>It is shown that is is  not in L. This is a  contraption with the pumping lemma.  our assumption that L is regular is  incorrect, and L is not a regular language</em>

6 0
3 years ago
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