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3 years ago

What are the four basic parts of process plan

Engineering

1 answer:

Rus_ich [418]3 years ago

8 0

Answer:

1. Goal setting

2. Developing the planning premises

3. Reviewing Limitations

4. Deciding the planning period

5. Formulation of policies and strategies

6. Preparing operating plans

7. Integration of plans

Explanation:

I don't have an explanation but i thru in a few extra answers for like extra credit or something if your teacher does that

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For a ceramic compound, what are the two characteristics of the component ions that determine the crystal structure?

DiKsa [7]

Answer:

The two characteristics of component ions that determine the crystal structure of a ceramic compound are:

1) The magnitude of electrical charge on each ion.

2) The relative sizes of both cations and anions.

Explanation:

Most ceramics normally contain both metallic and nonmetallic elements with ionic or covalent bonds. Thus, the structure of the metallic atoms, structure of the non-metallic atoms, and also the balance of charges produced by the valence electrons must be considered.

These ionic and covalent bonds i talked about earlier are the strong primary bonds that hold the atoms together and form the ceramic material. These chemical bonds are of two types:

i) they could either be ionic in character, meaning they involve a transfer of bonding electrons from electropositive atoms (cations) to electronegative atoms (anions),

ii) or they could be covalent in character, which involves orbital sharing of electrons between the constituent atoms or ions.

Thus, Covalent bonds are generally directional in nature, often dictating the types of crystal structure possible. Whereas, Ionic bonds, on the other hand, are entirely nondirectional. This nondirectional nature allows for hard-sphere packing arrangements of the ions into a variety of crystal structures.

So, we can deduce that;

The two characteristics of component ions that determine the crystal structure of a ceramic compound are:

1) The magnitude of electrical charge on each ion.

2) The relative sizes of both cations and anions.

3 0

2 years ago

Thick fluids such as asphalt and waxes and the pipes in which they flow are often heated in order to reduce the viscosity of the

Gala2k [10]

Answer:

what id the answer

Explanation:

7 0

3 years ago

Steam enters a turbine in a Rankine cycle power plant at 200 psia and 500 °F. a) Calculate the isentropic thermal efficiency if

Aleks04 [339]

Answer:

η=0.19=19% for p=14.7psi

η=0.3=30% for p=1psi

Explanation:

enthalpy before the turbine, state: superheated steam

h1(p=200psi,t=500F)=2951.9KJ/kg

s1=6.8kJ/kgK

Entalpy after the turbine

h2(p=14.7psia, s=6.8)=2469KJ/Kg

Entalpy before the boiler

h3=(p=14.7psia,x=0)=419KJ/Kg

Learn to pronounce

the efficiency for a simple rankine cycle is

η= $\frac{h1-h2}{h1-h3}$

η=(2951.9KJ/kg-2469KJ/Kg)/(2951.9KJ/kg-419KJ/Kg)

η=0.19=19%

second part

h2(p=1psia, s=6.8)=2110

h3(p=1psia, x=0)=162.1

η=(2951.9KJ/kg-2110KJ/Kg)/(2951.9KJ/kg-162.1KJ/Kg)

η=0.3=30%

7 0

3 years ago

The time to failure for a gasket follows the Weibull distribution with ß = 2.0 and a characteristic life of 300 days. What is th

Aleks04 [339]

Answer:

64.11% for 200 days.

t=67.74 days for R=95%.

t=97.2 days for R=90%.

Explanation:

Given that

β=2

Characteristics life(scale parameter α)=300 days

We know that Reliability function for Weibull distribution is given as follows

$R(t)=e^{-\left(\dfrac{t}{\alpha}\right)^\beta}$

Given that t= 200 days

$R(200)=e^{-\left(\dfrac{200}{300}\right)^2}$

R(200)=0.6411

So the reliability at 200 days 64.11%.

When R=95 %

$0.95=e^{-\left(\dfrac{t}{300}\right)^2}$

by solving above equation t=67.74 days

When R=90 %

$0.90=e^{-\left(\dfrac{t}{300}\right)^2}$

by solving above equation t=97.2 days

7 0

3 years ago

6.32 LAB: Exact change - functions

Neko [114]

Answer:

Below is the desired C++ program for the problem. Do feel free to edit it according to your preference

Explanation:

#include <iostream>

#include <vector>

using namespace std;

void ExactChange(int userTotal, vector<int> &coinVals) {

coinVals.reserve(5);

coinVals[0] = userTotal / 100;

userTotal %= 100;

coinVals[1] = userTotal / 25;

userTotal %= 25;

coinVals[2] = userTotal / 10;

userTotal %= 10;

coinVals[3] = userTotal / 5;

userTotal %= 5;

coinVals[4] = userTotal;

}

int main() {

vector<int> coins;

int value;

cin >> value;

if (value <= 0) {

cout << "no change" << endl;

} else {

ExactChange(value, coins);

if (coins[0] != 0) cout << coins[0] << " " << (coins[0] == 1 ? "dollar" : "dollars") << endl;

if (coins[1] != 0) cout << coins[1] << " " << (coins[1] == 1 ? "quarter" : "quarters") << endl;

if (coins[2] != 0) cout << coins[2] << " " << (coins[2] == 1 ? "dime" : "dimes") << endl;

if (coins[3] != 0) cout << coins[3] << " " << (coins[3] == 1 ? "nickel" : "nickels") << endl;

if (coins[4] != 0) cout << coins[4] << " " << (coins[4] == 1 ? "penny" : "pennies") << endl;

}

return 0;

}

5 0

3 years ago