3. A competency-based training program is

A spring-mass-damper instrument is employed for acceleration measurements. The spring constant is 12000 N/m. The mass is 5 g. Th

shepuryov [24]

Answer:

a) 246.56 Hz

b) 203.313 Hz

c) Add more springs

Explanation:

Spring constant = 12000 N/m

mass = 5g = 5 * 10^-3 kg

damping ratio = 0.4

<u>a) Calculate Natural frequency </u>

Wn = √k/m = $\sqrt{12000 / 5*10^{-3} }$

= 1549.19 rad/s ≈ 246.56 Hz

<u>b) Bandwidth of instrument </u>

W / Wn = $\sqrt{1-2(0.4)^2}$

W / Wn = 0.8246

therefore Bandwidth ( W ) = Wn * 0.8246 = 246.56 * 0.8246 = 203.313 Hz

C ) To increase the bandwidth we have to add more springs

5 0

3 years ago

List and briefly describe two modern's materials needs

Angelina_Jolie [31]

Answer:

Modern and smart materials for making the products are improved by developing new materials and find new uses for the existing. As, modern industrialization society is increased demand and quality of the product.

Two modern's materials are:

Carbon Fiber: As, carbon fiber is a strong material and it is light in weight. Designers used it because it is five times strong as steel and two times as stiff. Carbon fiber is basically made out of very thin strands of carbon.

Fiber Optics: It is a new technology as, it is used as transparent solid to transmitted light signals.

4 0

4 years ago

Number pattern Write a recursive method called print Pattern() to output the following number pattern. Given a positive integer

lilavasa [31]

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

3 0

3 years ago

Read 2 more answers

A gas expands in a piston-cylinder assembly from p1 = 8 bar, V1 = 0.02 m3 to p2 = 2 bar. The relation between pressure and volum

Charra [1.4K]

Answer:

The heat transfer is 29.75 kJ

Explanation:

The process is a polytropic expansion process

General polytropic expansion process is given by PV^n = constant

Comparing PV^n = constant with PV^1.2 = constant

n = 1.2

(V2/V1)^n = P1/P2

(V2/0.02)^1.2 = 8/2

V2/0.02 = 4^(1/1.2)

V2 = 0.02 × 3.2 = 0.064 m^3

W = (P2V2 - P1V1)/1-n

P1 = 8 bar = 8×100 = 800 kPa

P2 = 2 bar = 2×100 = 200 kPa

V1 = 0.02 m^3

V2 = 0.064 m^3

1 - n = 1 - 1.2 = -0.2

W = (200×0.064 - 800×0.02)/-0.2 = -3.2/-0.2 = 16 kJ

∆U = 55 kJ/kg × 0.25 kg = 13.75 kJ

Heat transfer (Q) = ∆U + W = 13.75 + 16 = 29.75 kJ

7 0

3 years ago

A steady, incompressible, two-dimensional velocity field is given by the following components in the x-y plane: u=1.85+2.05x+0.6

amm1812

1) $a_x=4.287+2.772x\\a_y=-5.579+2.772y$

2) 8.418

Explanation:

1)

The two components of the velocity field in x and y for the field in this problem are:

$u=1.85+2.05x+0.656y$

$v=0.754-2.18x-2.05y$

The x-component and y-component of the acceleration field can be found using the following equations:

$a_x=\frac{du}{dt}+u\frac{du}{dx}+v\frac{du}{dy}$

$a_y=\frac{dv}{dt}+u\frac{dv}{dx}+v\frac{dv}{dy}$

The derivatives in this problem are:

$\frac{du}{dt}=0$

$\frac{dv}{dt}=0$

$\frac{du}{dx}=2.05$

$\frac{du}{dy}=0.656$

$\frac{dv}{dx}=-2.18$

$\frac{dv}{dy}=-2.05$

Substituting, we find:

$a_x=0+(1.85+2.05x+0.656y)(2.05)+(0.754-2.18x-2.05y)(0.656)=\\a_x=4.287+2.772x$

And

$a_y=0+(1.85+2.05x+0.656y)(-2.18)+(0.754-2.18x-2.05y)(-2.05)=\\a_y=-5.579+2.772y$

2)

In this part of the problem, we want to find the acceleration at the point

(x,y) = (-1,5)

So we have

x = -1

y = 5

First of all, we substitute these values of x and y into the expression for the components of the acceleration field:

$a_x=4.287+2.772x\\a_y=-5.579+2.772y$

And so we find:

$a_x=4.287+2.772(-1)=1.515\\a_y=-5.579+2.772(5)=8.281$

And finally, we find the magnitude of the acceleration simply by applying Pythagorean's theorem:

$a=\sqrt{a_x^2+a_y^2}=\sqrt{1.515^2+8.281^2}=8.418$

4 0

3 years ago