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2 years ago

15

Is EPA is the organization that was formed to ensure safe and health working conditions for workers by setting and enforcing sta

ndards

1 answer:

mr_godi [17]2 years ago

3 0

Let me search this up

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A freshly annealed glass containing flaws of maximum length of 0.1 microns breaks under a tensile stress of 120 MPa. If a sample

almond37 [142]

Answer:

0.16 micron per day

Explanation:

Given:

The initial crack length, a₁ = 0.1 micron = 0.1 × 10⁻⁶ m

Initial tensile stress, σ₁ = 120 MPa

Final stress = 30 MPa

now from Griffith's equation, we have

$\sigma=[\frac{G_cE}{\pi\ a}]^\frac{1}{2}$

where,

Gc and E are the material constants

now,

for the initial stage

$120=[\frac{G_cE}{\pi\ (0.1\times10^{-6}}]^\frac{1}{2}$ ........{1}

and for the final case

$30=[\frac{G_cE}{\pi\ a_2}]^\frac{1}{2}$ ............{2}

on dividing 1 by 2, we get

$\frac{120}{30}=[\frac{a_2}{0.1\times10^{-6}}]^\frac{1}{2}$

or

a₂ = 4² × 0.1 × 10⁻⁶ m

or

a₂ = 1.6 micron

Now,

the change from 0.1 micron to 1.6 micron took place in 10 days

therefore, the rate at which the crack is growing = $\frac{1.6-0.1}{10}$

or

average rate of change of crack = 0.16 micron per day

6 0

3 years ago

Consider laminar, fully developed flow in a channel of constant surface temperature Ts. For a given mass flow rate and channel l

Lisa [10]

Answer:

Please see attachment.

Explanation:

6 0

3 years ago

Create a C language program that can be used to construct any arbitrary Deterministic Finite Automaton corresponding to the FDA

otez555 [7]

Answer:

see the explanation

Explanation:

/* C Program to construct Deterministic Finite Automaton */

#include <stdio.h>

#include <DFA.h>

#include <stdlib.h>

#include <math.h>

#include <string.h>

#include <stdbool.h>

struct node{

struct node *initialStateID0;

struct node *presentStateID1;

};

printf("Please enter the total number of states:");

scanf("%d",&count);

//To create the Deterministic Finite Automata

DFA* create_dfa DFA(){

q=(struct node *)malloc(sizeof(struct node)*count);

dfa->initialStateID = -1;

dfa->presentStateID = -1;

dfa->totalNumOfStates = 0;

return dfa;

}

//To make the next transition

void NextTransition(DFA* dfa, char c)

{

int tID;

for (tID = 0; tID < pPresentState->numOfTransitions; tID++){

if (pPresentState->transitions[tID].condition(c))

{

dfa->presentStateID = pPresentState->transitions[tID].toStateID;

return;

}

}

dfa->presentStateID = pPresentState->defaultToStateID;

}

//To Add the state to DFA by using number of states

void State_add (DFA* pDFA, DFAState* newState)

{

newState->ID = pDFA->numOfStates;

pDFA->states[pDFA->numOfStates] = newState;

pDFA->numOfStates++;

}

void transition_Add (DFA* dfa, int fromStateID, int(*condition)(char), int toStateID)

{

DFAState* state = dfa->states[fromStateID];

state->transitions[state->numOfTransitions].toStateID = toStateID;

state->numOfTransitions++;

}

void reset(DFA* dfa)

{

dfa->presentStateID = dfa->initialStateID;

}

5 0

3 years ago

g The parameters of a certain transmission line operating at 휔휔=6 ×108 [rad/s] are 퐿퐿=0.35 [휇휇H/m], 퐶퐶=75 [pF/m], 퐺퐺=75 [휇휇S/m],

yKpoI14uk [10]

Explanation:

$\begin{aligned}\gamma &=\sqrt{Z Y}=\sqrt{(R+j \omega L)(G+j \omega C)} \\&-\sqrt{|17|} j\left(6 \times 10^{8}\right)\left(0.35 \times 10^{-6}\right)|| 75 \times 10^{-6}\left|j\left(6 \times 10^{8}\right)\left(40 \times 10^{-12}\right)\right| \\&=0.094+j 2.25 \mathrm{m}^{-1}-\alpha+j \beta\end{aligned}$

Therefore,

$-\alpha-0.094 \mathrm{Np} / \mathrm{m} . \quad 3-2.25 \mathrm{rad} / \mathrm{m}, \text { and } \lambda-2 \pi / \beta-\underline{2.79} \mathrm{m}$

$Z_{0}-\sqrt{\frac{Z}{Y}}-\sqrt{\frac{R+j \omega L}{G+j \omega C}}-\sqrt{\frac{17+j 2.1 \times 10^{2}}{75 \times 10^{-6}+j 2.4 \times 10^{-2}}}-\frac{93.6-j 3.64 \Omega}{4}$

5 0

3 years ago

If the density of states function in the conduction band of a particular semiconductor is a constant equal to K, derive the expr

s2008m [1.1K]

Answer:

full details of the answer is attached

5 0

3 years ago