Solve the inequality below.Use the drop-down menus to describe the solution and its graph.

Calculate the wire pressure for a round copper bar with an original cross-sectional area of 12.56 mm2 to a 30% reduction of area

dybincka [34]

Answer:153.76 MPa

Explanation:

$Initial Area\left ( A_0\right )=12.56 mm^2$

$Final Area\left ( A_f\right )=0.7\times 12.56 mm^2=8.792 mm^2$

$Die angle=30^{\circ}$

$\alpha =\frac{30}{2}=15^{\circ}$

$\mu =0.08$

$Yield stress\left ( \sigma _y \right )=350 MPa$

$B=\mu cot\left ( \aplha\right )=0.2985$

$\sigma _{pressure}=\sigma _y\left [\frac{1+B}{B}\right ]\left [ 1-\frac{A_f}{A_0}\right ]^B$

$\sigma _{pressure}=350\left [\frac{1+0.2985}{0.2985}\right ]\left [ 1-\frac{8.792}{12.56}\right ]^{0.2985}$

$\sigma _{pressure}=153.76 MPa$

8 0

3 years ago

Four of the minterms of the completely specified function f(a, b, c, d) are m0, m1, m4, and m5.

Sveta_85 [38]

Complete Question

The complete question is shown on the first uploaded image

Answer:

a) The required additional minterms for f so that f has eight primary implicants with two literals and no other prime implicant are $m_{2},m_{3},m_{7},m_{8},m_{11},m_{12},m_{13},m_{14}$ and $m_{15}$

b) The essential prime implicant are $c' d',a'b',ab$ and $cd$

c) The minimum sum-of-product expression for f are

$a'b' +ab +c'd'+cd+a'c',\\ a'b'+ab+c'd'+cd+a'd,\\a'b'+ab+c'd'+cd+bc' and \\ a'b'+ab+c'd' +cd+bd$

Explanation:

The explanation is shown on the second third and fourth image

8 0

3 years ago

Which of the following conditions is a good sign of minor

Kryger [21]

Answer:

Explanation:

d

7 0

3 years ago

In javaWrite a program that simulates flipping a coin to make decisions. The input is how many decisions are needed, and the out

Pavel [41]

Answer:

// Program is written in Java Programming Language

// Comments are used for explanatory purpose

import java.util.*;

public class FlipCoin

{

public static void main(String[] args)

{

// Declare Scanner

Scanner input = new Scanner (System.in);

int flips;

// Prompt to enter number of toss or flips

System.out.print("Number of Flips: ");

flips = input.nextInt();

if (flips > 0)

{

HeadsOrTails();

}

public static String HeadsOrTails(Random rand)

{

// Simulate the coin tosses.

for (int count = 0; count < flips; count++)

{

rand = new Random();

if (rand.nextInt(2) == 0) {

System.out.println("Tails"); }

else {

System.out.println("Heads"); }

rand = 0;

}

7 0

3 years ago

An air-standard Diesel cycle engine operates as follows: The temperatures at the beginning and end of the compression stroke are

Vika [28.1K]

This question is incomplete, the complete question is;

An air-standard Diesel cycle engine operates as follows: The temperatures at the beginning and end of the compression stroke are 30 °C and 700 °C, respectively. The net work per cycle is 590.1 kJ/kg, and the heat transfer input per cycle is 925 kJ/kg. Determine the a) compression ratio, b) maximum temperature of the cycle, and c) the cutoff ratio, v3/v2.

Use the cold air standard assumptions.

Answer:

a) The compression ratio is 18.48

b) The maximum temperature of the cycle is 1893.4 K

c) The cutoff ratio, v₃/v₂ is 1.946

Explanation:

Given the data in the question;

Temperature at the start of a compression T₁ = 30°C = (30 + 273) = 303 K

Temperature at the end of a compression T₂ = 700°C = (700 + 273) = 973 K

Net work per cycle $W_{net$ = 590.1 kJ/kg

Heat transfer input per cycle Qs = 925 kJ/kg

a) compression ratio;

As illustrated in the diagram below, 1 - 2 is adiabatic compression;

so,

Tγ $^{Y-1$ = constant { For Air, γ = 1.4 }

hence;

⇒ V₁ / V₂ = $($ T₂ / T₁ $)^{\frac{1}{Y-1}$

so we substitute

⇒ V₁ / V₂ = $($ 973 K / 303 K $)^{\frac{1}{1.4-1}$

= $($ 3.21122 $)^{\frac{1}{0.4}$

= 18.4788 ≈ 18.48

Therefore, The compression ratio is 18.48

b) maximum temperature of the cycle

We know that for Air, Cp = 1.005 kJ/kgK

Now,

Heat transfer input per cycle Qs = Cp( T₃ - T₂ )

we substitute

925 = 1.005( T₃ - 700 )

( T₃ - 700 ) = 925 / 1.005

( T₃ - 700 ) = 920.398

T₃ = 920.398 + 700

T₃ = 1620.398 °C

T₃ = ( 1620.398 + 273 ) K

T₃ = 1893.396 K ≈ 1893.4 K

Therefore, The maximum temperature of the cycle is 1893.4 K

c) the cutoff ratio, v₃/v₂;

Since pressure is constant, V ∝ T

So,

cutoff ratio S = v₃ / v₂ = T₃ / T₂

we substitute

cutoff ratio S = 1893.396 K / 973 K

cutoff ratio S = 1.9459 ≈ 1.946

Therefore, the cutoff ratio, v₃/v₂ is 1.946

8 0

3 years ago