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andrew-mc [135]
3 years ago
12

The mechanical properties of some metals may be improved by incorporating fine particles of their oxides. If the moduli of elast

icity of a hypothetical metal and its oxide are, respectively, 55 GPa and 430 GPa, what is the upper-bound modulus of elasticity value for a composite that has a composition of 31 vol% of oxide particles

Engineering
1 answer:
mezya [45]3 years ago
3 0

Answer:

171.2 GPa

Explanation:

I explained every step in the attached file, please kindly go through it.

But the answer is 171.2 GPa

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Write a complete C++ program that is made of functions main() and rShift(). The rShift() function must be a function without ret
erma4kov [3.2K]

Answer:

Explanation:

attached is an uploaded picture to support the answer.

the program is written thus;

#include<iostream>

using namespace std;

// function declaration

void rShift(int&, int&, int&, int&, int&, double&);

int main()

{

   // declare the variables

   int a1, a2, a3, a4;

   int maximum = 0;

   double average = 0;

   // inputting the numbers

   cout << "Enter all the 4 integers seperated by space -> ";

   cin >> a1 >> a2 >> a3 >> a4;

   cout << endl << "Value before shift." << endl;

   cout << "a1 = " << a1 << ", a2 = " << a2 << ", a3 = "  

        << a3 << ", a4 = " << a4 << endl << endl;

   // calling rSift()

   // passing the actual parameters

   rShift(a1,a2,a3,a4,maximum,average);

   // printing the values

   cout << "Value after shift." << endl;

   cout << "a1 = " << a1 << ", a2 = " << a2 << ", a3 = "  

           << a3 << ", a4 = " << a4 << endl << endl;

   cout << "Maximum value is: " << maximum << endl;

   cout << "Average is: " << average << endl;

}

// function to right shift the parameters circularly

// and calculate max, average of the numbers

// and return it to main using pass by reference

void rShift(int &n1, int &n2, int &n3, int &n4, int &max, double &avg)

{

   // calculating the max

   max = n1;

   if(n2 > max)

     max = n2;

   if(n3 > max)

     max = n3;

   if(n4 > max)

     max = n4;

   // calculating the average

   avg = (double)(n1+n2+n3+n4)/4;

   // right shifting the numbers circulary

   int temp = n2;

   n2 = n1;

   n1 = n4;

   n4 = n3;

   n3 = temp;

}

8 0
4 years ago
a. Determine R for a series RC high-pass filter with a cutoff frequency (fc) of 8 kHz. Use a 100 nF capacitor. b. Draw the schem
Readme [11.4K]

Answer:

a) 199.04 ohms

b) attached in image

c) -0.696dB

Explanation:

We are given:

Fc = 8Khz = 8000hz

C = 100nF = 100*10^-^9F

a)Using the formula:

F_c = \frac{1}{2pie*Rc}

8000= \frac{1}{2*3.14*R*100*10^-^9}

R =\frac{1}{2*3.14*100*10^-^9*8000}

R = 199.04 ohms

b) diagram is attached

c) H(w) = \frac{V_out(w)}{Vin(w)} = \frac{1}{1-j\frac{wc}{w}}

H(F) = \frac{1}{1-j\frac{fc}{f}}

At F = 20KHz and Fc= 8KHz we have:

H(F)= \frac{1}{1-j\frac{8}{20}} = \frac{1}{1-j(0.4)}

|H(F)|= \frac{1}{\sqrt{1^2+(0.4)^2}}

=0.923

|H(F)| in dB = 20log |H(F)|

=20log0.923

= -0.696dB

5 0
3 years ago
on the same scale for stress, the tensile true stress-true strain curve is higher than the engineeringstress-engineering strain
Bess [88]

Answer:

The condition does not hold for a compression test

Explanation:

For a compression test the engineering stress - strain curve is higher than the actual stress-strain curve and this is because the force needed in compression is higher than the force needed during Tension.  The higher the force in compression leads to increase in the area therefore for the same scale of stress the there is more stress on the Engineering curve making it higher than the actual curve.

<em>Hence the condition of : on the same scale for stress, the tensile true stress-true strain curve is higher than the engineering stress-engineering strain curve.</em><em> </em>does not hold for compression test

5 0
3 years ago
At a point on the free surface of a stressed body, the normal stresses are 20 ksi (T) on a vertical plane and 30 ksi (C) on a ho
victus00 [196]

Answer:

The principal stresses are σp1 = 27 ksi, σp2 = -37 ksi and the shear stress is zero

Explanation:

The expression for the maximum shear stress is given:

\tau _{M} =\sqrt{(\frac{\sigma _{x}^{2}-\sigma _{y}^{2}  }{2})^{2}+\tau _{xy}^{2}    }

Where

σx = stress in vertical plane = 20 ksi

σy = stress in horizontal plane = -30 ksi

τM = 32 ksi

Replacing:

32=\sqrt{(\frac{20-(-30)}{2} )^{2} +\tau _{xy}^{2}  }

Solving for τxy:

τxy = ±19.98 ksi

The principal stress is:

\sigma _{x}+\sigma _{y} =\sigma _{p1}+\sigma _{p2}

Where

σp1 = 20 ksi

σp2 = -30 ksi

\sigma _{p1}  +\sigma _{p2}=-10 ksi (equation 1)

\tau _{M} =\frac{\sigma _{p1}-\sigma _{p2}}{2} \\\sigma _{p1}-\sigma _{p2}=2\tau _{M}\\\sigma _{p1}-\sigma _{p2}=32*2=64ksi equation 2

Solving both equations:

σp1 = 27 ksi

σp2 = -37 ksi

The shear stress on the vertical plane is zero

4 0
4 years ago
Read two numbers from user input. Then, print the sum of those numbers. Hint -- Copy/paste the following code, then just type co
Softa [21]

Answer:

I am Providing Answer in C Language Program.

Explanation:

Please find attachment regarding code of taking two numbers input and adding them.

I would like to recommend you please use software which supports C language.

#include <stdio.h>

int main () {

int a, b, sum;

printf ("\ nEnter two no:");

scanf ("% d% d", & d, & e);

sum1 = d + e;

printf ("Sum:% d", sum1);

return (0);

}

4 0
3 years ago
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