About ceramics: Generally are hard and brittle. a) True b)-False

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mash [69]

Huh

hmmm

Mmmmmmmmmmmmmm

5 0

3 years ago

Read 2 more answers

At the grocery store you place a pumpkin with a mass of 12.5 lb on the produce spring scale. The spring in the scale operates su

UkoKoshka [18]

Answer:

x=2.19in

Explanation:

This is the equation that relates the force and displacement of a spring

F=Kx

m=mass=12.5lbx1slug/32.14lb=0.39slug

F=mg=0.39*32.2=12.52Lbf

then we calculate the spring count in lbf / ft

K=F/x

K=5.7lbf/1in=5.7lbf/in=68.4lbf/ft

Finally we calculate the displacement with the initial equation

X=F/k

x=12.52/68.4=0.18ft=2.19in

8 0

4 years ago

Q10. Select the correct option for the following questions – (10 points, 2 each) a. After an edge dislocation has passed through

Gnom [1K]

Answer:

a. True - Because the atomic arrangements of that region is disorderer because of the extra half plane atoms in between the line

b.Slip

C. Strength theoretical is greater than strength experimental

d. Shear stress

e. Highest linear density

4 0

3 years ago

Print "userNum1 is negative." if userNum1 is less than 0. End with newline. Assign userNum2 with 2 if userNum2 is greater than 1

miv72 [106K]

Answer:

From the question, we have two variables

1. userNum1

2. userNum2

And we are to print "userNum1 is negative" if userNum1 is less than 0.

Then Assign userNum2 with 2 if userNum2 is greater than 10.

Otherwise, print "userNum2 is less or equal 10.".

// Program is written in C++ Programming Language

// Comments are used for explanatory purpose

// Program starts here

#include<iostream>

using namespace std;

int main ()

{

// Declare variables

int userNum1, userNum2;

// Accept input for these variables

cin>>userNum1, userNum2;

// Condition 1

if(userNum1 < 0)

{

cout<<"userNum1 is negative"<<'\n';

}

// Condition 2

if(userNum2 > 10)

{

userNum2 = 2;

}

// If condition is less than 10

else

{

cout<<"userNum2 is less or equal to 10"<<\n;

}

return 0;

}

// End of Program.

3 0

3 years ago

2.24 Carbon dioxide (CO2) gas in a piston-cylinder assembly undergoes three processes in series that begin and end at the same s

Gekata [30.6K]

Answer:

Explanation:

Given that:

From process 1 → 2

$P_1 = 10 bar \\ \\ V_1 = 1 m^3 \\ \\ V_2 = 4 m^3$

$PV^{1.5} = \ constant$

$\gamma = 1.5$

Process 2 → 3

The volume is constant i.e $V_2 =V_3 = 4m^3$

$P_3 = 10 \ bar$

Process 3 → 1

P = constant i.e the compression from state 1

Now, to start with 1 → 2

$P_1V_1^{1.5} = P_2V_2^{1.5}$

$P_2 = P_1 (\dfrac{V_1}{V_2})^{1.5}$

$P_2 = 10 \times (\dfrac{1}{4})^{1.5}$

$P_2 =1.25$

The work-done for the process 1 → 2 through adiabatic expansion is:

$W = \dfrac{1}{1-\gamma}[P_2V_2-P_1V_1]$

We know that 1 bar = $10^5 \ N/m^2$

∴

$W = \dfrac{1}{1-1.5}[1.25 \times 10^5 \times 4- 10 \times 10^5 \times 1]$

$W =1000000 \ J$

$W_{1 \to 2} = 1000 kJ$

For process 2 → 3

Since V is constant

Thus:

W = PΔV = 0

$W_{2 \to 3} = 0$

For process 3 → 1

W = PΔV

$W _{3 \to 1} = P_3(V_1-V_3)$

$W _{3 \to 1} = 10 \times 10^5 (1-4)$

$W _{3 \to 1} = 10 \times 10^5 (-3)$

$W _{3 \to 1} = -3 \times 10^6 \ J$

$W _{3 \to 1} = -3000 \ kJ$

The net work-done now for the entire system is :

$W_{net} = W_{1 \to 2} + W_{2 \to 3 } + W_{ 3 \to 1 }$

$W_{net} = (1000 + 0 + (-3000)) \ kJ$

$W_{net} =-2000 \ kJ$

The sketch of the processes on p -V coordinates can be found in the image attached below.

4 0

3 years ago