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3 years ago

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12. A structural component is fabricated from an alloy that has a plane strain fracture toughness of It has been determined that

this component fails at a stress of 300 MPa when the maximum length of a surface crack is 0.95 mm. What is the maximum allowable surface crack length (in mm) without fracture for this same component exposed to a stress of 300 MPa and made from another alloy with a plane-strain fracture toughness of 35 MPa m (31.9 ksi in. ). 40 MPa m (36.4 ksi in. ). 45 MPa m. 57.5 MPa m

1 answer:

Artyom0805 [142]3 years ago

3 0

Answer:

Q1 = 4.233 mm

Explanation:

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aksik [14]

Answer:

a) $\eta = 13.455\%$ , b) $E_{day} = 812.716\,kJ$ , c) $C_{month. total} = 19.505\, USD$ , d) $t = 40.588\,years$

Explanation:

a) The area of the solar panel is:

$A = (20\,ft^{2})\cdot (\frac{0.3048\,m}{1\,ft} )^{2}$

$A = 1.858\,m^{2}$

The energy potential is determined herein:

$\dot E_{o} = (1000\,\frac{W}{m^{2}} )\cdot (1.858\,m^{2})$

$\dot E_{o} = 1858\,W$

The efficiency of the solar panel is:

$\eta = \frac{\dot E}{\dot E_{o}}\times 100\%$

$\eta = \frac{250\,W}{1858\,W}\times 100\%$

$\eta = 13.455\%$

b) The energy generated by the solar panel is presented below:

$E_{day} = (0.135)\cdot (150\,\frac{W}{m^{2}} )\cdot (20\,ft^{2})\cdot \left(\frac{0.3048\,m}{1\,ft} \right)^{2}\cdot (6\,h)\cdot (\frac{3600\,s}{1\,h} )\cdot (\frac{1\,kJ}{1000\,J} )$

$E_{day} = 812.716\,kJ$

c) The energy generated per month and per panel is:

$E_{month} = 30\cdot E_{day}$

$E_{month} = 30 \cdot (812.716\,kJ)\cdot \left(\frac{1\,kWh}{3600\,kJ} \right)$

$E_{month} = 6.773\,kWh$

Monthly energy savings due to the use of 18 panels are:

$C_{month, total} = 18\cdot E_{month}\cdot c$

$C_{month, total} = 18\cdot (6.773\,kWh)\cdot (\frac{0.16\,USD}{1\,kWh} )$

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d) The payback of the solar energy system is:

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$t = 40.588\,years$

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3 years ago

A specimen has circular cross-sectional area with initial diameter d = 26 mm and a gauge length Lo = 200 mm. A force P = 127 kN

steposvetlana [31]

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3 years ago

A plane wall of length L, constant thermal conductivity k and no thermal energy generation undergoes one-dimensional, steady-sta

KIM [24]

Answer:

Temperature distribution is $T(x)=\dfrac{q}{k}(L-x)+T$

Heat flux=q

Heat rate=q A

Explanation:

We know that for no heat generation and at steady state

$\dfrac{\partial^2 T}{\partial x^2}=0$

$\dfrac{\partial T}{\partial x}=a$

$T(x)=ax+b$

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Given that heat flux=q

We know that heat flux given as

$q=-K\dfrac{dT}{dx}$

From above we can say that

$a= -\dfrac{q}{K}$

Alos given that when x= L temperature is T(L)=T

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$b=T+\dfrac{q}{K}L$

So temperature T(x)

$T(x)=-\dfrac{q}{K}x+T+\dfrac{q}{K}L$

$T(x)=\dfrac{q}{k}(L-x)+T$

So temperature distribution is $T(x)=\dfrac{q}{k}(L-x)+T$

Heat flux=q

Heat rate=q A (where A is the cross sectional area of wall)

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3 years ago

Milestone 1: Write code which asks for a length input until it gets an integer 10 or greater, then creates 2 arrays of this leng

jonny [76]

Answer:

import java.util.Scanner;

import java.lang.Math;

class Main {

public static void main(String[] args) {

int length = 0;

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Scanner scan = new Scanner(System.in);

while (lengthCheck == true)

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System.out.println("Enter an array length (must be 10 or greater):");

length = scan.nextInt();

if (length >= 10)

{

lengthCheck = false;

}

}

int[] firstArray = new int[length];

int[] secondArray = new int[length];

System.out.print("\nFirst Array: ");

for (int i = 0; i < length; i++)

{

firstArray[i] = (int) (Math.random() * 100) + 1;

System.out.print(firstArray[i] + " ");

}

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for (int i = 0; i < length; i++)

{

secondArray[i] = (int) (Math.random() * 100) + 1;

System.out.print(secondArray[i] + " ");

}

System.out.println("\n");

boolean[] isAdded = new boolean[100];

int[] merge = new int[(firstArray.length + secondArray.length)];

int j=0;

for (int i = 0; i < length; i++)

{

if(!isAdded[firstArray[i] - 1]) {

merge[j] = firstArray[i];

j++;

isAdded[firstArray[i] - 1] = true;

}

if(!isAdded[secondArray[i] - 1]) {

merge[j] = secondArray[i];

j++;

isAdded[secondArray[i] - 1] = true;

}

}

System.out.print("Merged Array: ");

for (int i = 0; i < 2*length && merge[i] != 0; i++)

{

System.out.print(merge[i] + " ");

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System.out.println("\n");

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}

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3 years ago

Batteries don’t run forever. Why do you think batteries run down?

Marina86 [1]

Because there is no such thing as immortal batteries.

When a battery is connected to a circuit, the charge moves through the circuit, and a chemical reaction occurs inside that separate the charges. The strength of this reaction diminishes over time and the battery eventually dies.

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3 years ago

Read 2 more answers