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sveticcg [70]
3 years ago
6

Consider the two wood pieces that are connected by a velcro as indicated below. The block is subjected to a tension force P and

has dimensions b 2 in and h = 4 in. The velcro can sustain the maximum normal stress Om-1 psi and the maximum shear stress Tn-7 psi. Use α = 20°. What is the maximum value of the force P that can be applied to the block before the velcro breaks connection?

Engineering
2 answers:
sweet [91]3 years ago
7 0

Answer:

Answer is 68.42.

Refer below for the explanation.

Explanation:

Refer to the pictures please.

Amanda [17]3 years ago
3 0

Answer:

P= 68.42 lb

Explanation:

The solution and complete explanation for the above question and mentioned conditions is given below in the attached document.I hope my explanation will help you in understanding this particular question.

You might be interested in
Your application must generate: - an array of thirteen random integers from 1-99, - then prompt the user to select a sorting opt
Fynjy0 [20]

Answer:

The code is given which can be pasted in the Javascript file

Explanation:

The code is given as below

package Sorters;

public class javasort

{

private int[] arr=new int[13];

public void bubbleSort(int[] a){

int c,d,temp;

for (c = 0; c < ( 13 - 1 ); c++) {

for (d = 0; d < 13- c - 1; d++) {

if (a[d] > a[d+1]) /* For descending order use < */

{

temp = a[d];

a[d] = a[d+1];

a[d+1] = temp;

}

}

 

System.out.print("\n[");

for (int i = 0; i < 13; i++){

  System.out.print(a[i]+" ");

}

System.out.print("]");

}

System.out.println("\nSorted list of numbers");

System.out.print("[");

for (int i = 0; i < 13; i++){

  System.out.print(a[i]+" ");

}

System.out.print("]");

}

public void insertionSort(int[] a){

int temp;

for (int i = 1; i < 13; i++) {

for(int j = i ; j > 0 ; j--){

if(a[j] < a[j-1]){

temp = a[j];

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

a[j-1] = temp;

}

}

System.out.print("\n[");

for (int c = 0; c < 13; c++){

  System.out.print(a[c]+" ");

}

System.out.print("]");

}

System.out.println("\nSorted list of numbers");

System.out.print("[");

for (int i = 0; i < 13; i++){

  System.out.print(a[i]+" ");

}

System.out.print("]");

}

public void shellSort(int[] a){

int increment = a.length / 2;

while (increment > 0)

{

for (int i = increment; i < a.length; i++)

{

int j = i;

int temp = a[i];

while (j >= increment && a[j - increment] > temp)

{

a[j] = a[j - increment];

j = j - increment;

}

a[j] = temp;

}

if (increment == 2)

increment = 1;

else

increment *= (5.0 / 11);

System.out.print("\n[");

for (int c = 0; c < 13; c++){

  System.out.print(a[c]+" ");

}

System.out.print("]");

}

System.out.println("\nSorted list of numbers");

System.out.print("[");

for (int i = 0; i < 13; i++){

  System.out.print(a[i]+" ");

}

System.out.print("]");

}

public void MergeSort(int[] a, int low, int high){

int N = high - low;  

if (N <= 1)

return;

int mid = low + N/2;

// recursively sort

MergeSort(a, low, mid);

MergeSort(a, mid, high);

// merge two sorted subarrays

int[] temp = new int[N];

int i = low, j = mid;

for (int k = 0; k < N; k++)

{

if (i == mid)

temp[k] = a[j++];

else if (j == high)

temp[k] = a[i++];

else if (a[j]<a[i])

temp[k] = a[j++];

else

temp[k] = a[i++];

}

for (int k = 0; k < N; k++)

a[low + k] = temp[k];

System.out.print("\n[");

for (int c = 0; c < 13; c++){

  System.out.print(a[c]+" ");

}

System.out.print("]");

printM(a);

}

public void quickSort(int[] a,int low,int high){

  System.out.print("\n[");

  for (int c = 0; c < 13; c++){

      System.out.print(a[c]+" ");

  }

  System.out.print("]");

int i =low, j = high;

int temp;

int pivot = a[(low + high) / 2];

/** partition **/

while (i <= j)

{

while (a[i] < pivot)

i++;

while (a[j] > pivot)

j--;

if (i <= j)

{

/** swap **/

temp = a[i];

a[i] = a[j];

a[j] = temp;

i++;

j--;

}

}

/** recursively sort lower half **/

if (low < j)

quickSort(a, low, j);

/** recursively sort upper half **/

if (i < high)

quickSort(a, i, high);

printM(a);

}

public void printM(int[] a){

arr=a;

}

public void fPrint(){

  System.out.println("\nSorted list:");

  System.out.print("\n[");

  for (int c = 0; c < 13; c++){

      System.out.print(arr[c]+" ");

  }

  System.out.print("]");

}

}

package mani;

import java.util.Random;

import java.util.Scanner;

public class javasorttest

{

 

public static void main(String[] args){

int[] a=new int[13];

Random r=new Random();

for(int i=0;i<13;i++){

a[i]=r.nextInt(99)+1;

}

System.out.print("[");

for (int c = 0; c < 13; c++){

  System.out.print(a[c]+" ");

}

System.out.print("]");

javasort j=new javasort();

System.out.println("\nSelect the sorting algo.\n1.bubbleSort\n2.insertionSort\n3.shellSort\n4.MergeSort\n5.QuickSort.");

Scanner s=new Scanner(System.in);

int opt=s.nextInt();

switch(opt){

case 1:

j.bubbleSort(a);

break;

case 2:

j.insertionSort(a);

 

break;

case 3:

j.shellSort(a);

break;

case 4:

j.MergeSort(a, 0, 13);

j.fPrint();

break;

case 5:

j.quickSort(a ,0, 12);

j.fPrint();

break;

}

}

}

5 0
3 years ago
2.7. The term route of entry on an SDS refers to the way a
VARVARA [1.3K]

Answer:

(B) O Chemical

Explanation:

The information contained in the SDS is largely the same as the MSDS, except now the SDSs are required to be presented in a consistent user-friendly, 16-section format. This brief provides guidance to help workers who handle hazardous chemicals to become familiar with the format and understand the contents of the SDSs.

The SDS includes information such as the properties of each chemical; the physical, health, and environmental health hazards; protective measures; and safety precautions for handling, storing, and transporting the chemical.

8 0
3 years ago
The sliders A and B are connected by a light rigid bar of length l = 20 in. and move with negligible friction in the slots, both
DedPeter [7]

Answer:

Explanation:

Given:

- The Length of the rigid bar L = 20 in

- The position of slider a, x_a = 16 in

- The position of slider b, y_b

- The velocity of slider a, v_a = 3 ft /s

- The velocity of slider b, v_b

- The acceleration of slider a, a_a

- The acceleration of slider b, a_b

Find:

-Determine the acceleration of each slider and the force in the bar at this instant.

Solution:

- The relationship between the length L of the rod and the positions x_a and x_b of sliders A & B is as follows:

                               L^2 = x_a^2 + y_b^2   ....... 1

                               y_b = sqrt( 20^2 - 16^2 )

                               y_b = 12

- The velocity expression can derived by taking a derivation of Eq 1 with respect to time t:

                               0 = 2*x_a*v_a + 2*y_b*v_b

                               0 = x_a*v_a + y_b*v_b   ..... 2

                               0 = 16*36 + 12*v_b

                               v_b = - 48 in /s = -4 ft/s

- Similarly, the acceleration expression can be derived by taking a derivative of Eq 2 with respect to time t:

                               0 = v_a^2 + x_a*a_a + v_b^2 + y_b*a_b

                               0 = 9 + 4*a_a/3 + 16 + a_b

                               4*a_a/3 + a_b = -25

                               4*a_a + 3*a_b = -75  .... 3

- Use dynamics on each slider. For Slider A, Apply Newton's second law of motion in x direction:

                               F_x = m_a*a_a

                               P - R_r*16/20 = m_a*a_a

                               

- For Slider B, Apply Newton's second law of motion in y direction:

                               F_y = m_b*a_b

                               - R_r*12/20 = m_b*a_b

- Combine the two dynamic equations:

                               P - 4*m_b*a_b / 3 = m_a*a_a

                               3P = 3*m_a*a_a + 4*m_b*a_b  ... 4

- Where,                  P = Is the force acting on slider A

                               P , m_a and m_b are known quantities but not given in question. We are to solve Eq 3 and Eq 4 simultaneously for a_a and a_b.                    

5 0
3 years ago
What is the weight of a glider with a mass of 4.9 grams?
timama [110]

Answer:

0.0481 N

Explanation:

The mass of a body is the quantity of matter stored in that body. It is a property that measures resistance to acceleration when a force is applied. The S.I unit of mass is the kilogram (kg).

The weight of a body is the force exerted on that body by gravity. The S.I unit of weight is the Newton (N). The weight of an object can also be defined as the force acting on the object. The formula for weight is:

Weight (W) = Mass (m) × acceleration due to gravity(g).

Giving that: mass (m) = 4.9 g = 0.0049 g and acceleration due to gravity (g) = 9.81 m/s².

Therefore, Weight (W) = 0.0049 × 9.81 = 0.0481 N

4 0
3 years ago
A mixture of two gases has a total mass of 80 kg. One gas has a specific volume of 0.8 m^3/kg and occupies a volume of 40 m^3. T
Nady [450]

Answer:

specific volume = 1.025 m³/kg

Explanation:

given data

total mass m1 +m2 = 80 kg

specific volume = 0.8 m³/kg

occupies volume v1 = 40 m³

other gas specific volume  = 1.4 m³

to find out

How much volume is occupied by the second gas and what is the specific volume of the mixture

solution

we know that density is reciprocal of specific volume and here gas is not interacting

so total specific volume is assume so ratio is total volume to total mass

and

specific volume = \frac{1}{\rho}

here ρ is density

so ρ1  =   \frac{1}{0.8} = 1.25 kg/m³

and ρ2  =   \frac{1}{1.4} = 0.714 kg/m³

and

so m1 = ρ1v1 = 1.25 × 40 = 50 kg

and m2 = 80 - 50 = 30 kg

so

v2 = \frac{m2}{\rho 2}

v2 =  \frac{30}{0.714} = 42 m³

so volume occupied by second das = 42 m³

and

specific volume of mixture will be

specific volume of mixture = \frac{v1+v2}{m1+m2}

specific volume = \frac{42+40}{50+30}

specific volume = 1.025 m³/kg

3 0
3 years ago
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