Design an algorithm for computing √n

10. True or False? A disruptive technology<br> radically changes the way people live and<br> work.

Anettt [7]

Answer:

Explanation:

According to Investopedia.com, "Disruptive technology is an innovation that significantly alters the way that consumers, industries, or businesses operate".

So yes, a disruptive does radically change the way people live and work.

5 0

3 years ago

Read 2 more answers

6. Question

valkas [14]

Answer:

Check the 2nd, 3rd and 4th statements.

Explanation:

4 0

3 years ago

Calculate KI for a rectangular bar containing an edge crack loaded in three-point bending where P=35.0 kN, W=50.8 mm, B=25 mm, a

Katyanochek1 [597]

Answer:

$K_{I}=5.21 MPa\sqrt{m}$

Explanation:

given data

Load P = 35 kN

Width of bar W = 50.8 mm

Breadth of bar B = 25 mm

Ratio of crack length to width α = a/W = 0.2

solution

we get here KI for a rectangular bar that is express as

$K_{I} = \frac{6P}{BW}Y\sqrt{\pi a}$ ................................1

here Y is the geometrical function

so

Y = $\frac{1.12+\alpha (3.43\alpha -1.89)}{1-0.55\alpha}$

Y = $\frac{1.12+0.2(3.43\times 0.2-1.89)}{1-0.55\times 0.2}$

Y = $\frac{0.8792}{0.89}$

Y = 0.9878

so put here value in equation 1

$K_{I} = \frac{6\times 35\times 10^{3}}{0.025\times 0.0508}\times 0.9878\times \sqrt{3.1415\times (0.2\times 0.0508)}$

$K_{I} = 165354.33\times 10^{3}\times 0.9878\times 0.0319$

$K_{I}$ = 5210.45 × 10³ $Pa\sqrt{m}$

$K_{I}$ = 5.21 MPa $\sqrt{m}$

5 0

3 years ago

The assembly consists of a brass shell (1) fully bonded to a ceramic core (2). The brass shell [E = 93 GPa, α= 15.1 × 10−6/°C] h

marshall27 [118]

Answer:

ΔT = 62.11°C

Explanation:

Given:

- Brass Shell:

Inner Diameter d_i = 32 mm

Outer Diameter d_o = 39 mm

E_b = 93 GPa

α_b = 15.1*10^-6 / °C

- Ceramic Core:

Outer Diameter d_o = 32 mm

E_c = 310 GPa

α_c = 3.2*10^-6 / °C

- Unstressed @ T = 8°C

- Total Length of the cylinder L = 160 mm

Find:

Determine the largest temperature increase Δ⁢t that is acceptable for the assembly if the normal stress in the longitudinal direction of the brass shell must not exceed 60 MPa.

Solution:

- Since, α_b > α_c the brass shell is in compression and ceramic core is in tension. The stress in shell is given as б_a:

б_b = - 60 MPa

- The force equilibrium can be written as:

б_b*A_b + б_c*A_c = 0

Where, б_b is the stress in core

A_b is the cross sectional area of the shell

A_c is the cross sectional area of the core

б_b*pi*( d_o^2 - d_i^2) / 4 + б_c*pi*( d_i^2) / 4 = 0

б_b*( d_o^2 - d_i^2) + б_c*( d_i^2) = 0

б_c = - б_b*( d_o^2 - d_i^2) / ( d_i^2)

Plug in the values:

б_c = 60*( 0.039^2 - 0.032^2) / ( 0.032^2)

б_c = 29.121 MPa , б_b = - 60 MPa

- The total strains in both brass shell and ceramic core is given by:

ξ_b = α_b*ΔT + б_b / E_b

ξ_c = α_c*ΔT + б_c / E_c

- The compatibility relation is:

ξ_b = ξ_c

α_b*ΔT + б_b / E_b = α_c*ΔT + б_c / E_c

ΔT*(α_b - α_c ) = б_c / E_c - б_b / E_b

ΔT = [ б_c / E_c - б_b / E_b ] / (α_b - α_c )

Plug in values and solve:

ΔT = [ 0.029121 / 310 + 0.06 / 93 ]*10^6 / (15.1 - 3.2 )

ΔT = 62.11°C

8 0

3 years ago

code a class named sllqueue that uses a double-ended singly linked list to implement a queue as described in this chapter. provi

julia-pushkina [17]

Answer:

See attached file please.

Explanation:

See attached file for detailed explanation and code.

import java.util.*;

class LinklistImplementQueue {

public static void main(String[] args)

{

Scanner scan = new Scanner(System.in);

/* Creating object of class SLLQueue */

SLLQueue lq = new SLLQueue();

char ch;

do

{

System.out.println("\nQueue Operations");

System.out.println("1. ENQUEUE");

System.out.println("2. DEQUEUE");

int choice = scan.nextInt();

.

See attached file for complete code.

Download txt

6 0

3 years ago