Answer:
See attachment below
Explanation:
Consider a fully-clamped circular diaphragm poly-Si with a radius of 250 μm and a thickness of 4 μm. Assume that Young’s modulus
is 130 GPa and Poisson’s ratio is 0.3. A uniform pressure of 20 kPa is applied.
(a) What is the stress at the center of the diaphragm?
(b) What is the radial stress at the fully-clamped edge of the diaphragm?
(c) What is the transverse stress at the fully-clamped edge of the diaphragm?
(d) What is the amount of deflection at the center of the diaphragm?
(a) What is the stress at the center of the diaphragm?
(b) What is the radial stress at the fully-clamped edge of the diaphragm?
(c) What is the transverse stress at the fully-clamped edge of the diaphragm?
(d) What is the amount of deflection at the center of the diaphragm?
1 answer:
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Answer:
See explaination
Explanation:
Code;
import java.util.Scanner;
public class NumberPattern {
public static int x, count;
public static void printNumPattern(int num1, int num2) {
if (num1 > 0 && x == 0) {
System.out.print(num1 + " ");
count++;
printNumPattern(num1 - num2, num2);
} else {
x = 1;
if (count >= 0) {
System.out.print(num1 + " ");
count--;
if (count < 0) {
System.exit(0);
}
printNumPattern(num1 + num2, num2);
}
}
}
public static void main(String[] args) {
Scanner scnr = new Scanner(System.in);
int num1;
int num2;
num1 = scnr.nextInt();
num2 = scnr.nextInt();
printNumPattern(num1, num2);
}
}
See attachment for sample output
Answer:
Maximum number that can be represented by 13 bits is 8192 Instructions
Explanation:
number of instructions = 1000
number of bits = log(1000) x number of register
= 6 bits
Since the complete instruction must have 32 bits, then
remaining number of bits = 32 - 6 = 236
number of registers in instruction = 2
number of bits per register = 26/2 = 13
Maximum number that can be represented by 13 bits =
= 2¹³ = 8192