Studies show that chlorofluorocarbons (CFCs) such as CFC-12, are contributing to the depletion of the

A closed system of mass 10 kg undergoes a process during which there is energy transfer by work from the system of 0.147 kJ per

mr_godi [17]

Answer:

-50.005 KJ

Explanation:

Mass flow rate = 0.147 KJ per kg

mass= 10 kg

Δh= 50 m

Δv= 15 m/s

W= 10×0.147= 1.47 KJ

Δu= -5 kJ/kg

ΔKE + ΔPE+ ΔU= Q-W

0.5×m×(30^2- 15^2)+ mgΔh+mΔu= Q-W

Q= W+ 0.5×m×(30^2- 15^2) +mgΔh+mΔu

= 1.47 +0.5×1/100×(30^2- 15^2)-9.7×50/1000-50

= 1.47 +3.375-4.8450-50

Q=-50.005 KJ

7 0

3 years ago

Read 2 more answers

Which of the following is not an electronic device ?

PolarNik [594]

Answer:

B

Explanation:

it's does not transmit any energy

6 0

2 years ago

The dam cross section is an equilateral triangle, with a side length, L, of 50 m. Its width into the paper, b, is 100 m. The dam

lisabon 2012 [21]

Answer:

Explanation:

In an equilateral trinagle the center of mass is at 1/3 of the height and horizontally centered.

We can consider that the weigth applies a torque of T = W*b/2 on the right corner, being W the weight and b the base of the triangle.

The weigth depends on the size and specific gravity.

W = 1/2 * b * h * L * SG

Then

Teq = 1/2 * b * h * L * SG * b / 2

Teq = 1/4 * b^2 * h * L * SG

The water would apply a torque of elements of pressure integrated over the area and multiplied by the height at which they are apllied:

$T1 = \int\limits^h_0 {p(y) * sin(30) * L * (h-y)} \, dy$

The term sin(30) is because of the slope of the wall

The pressure of water is:

p(y) = SGw * (h - y)

Then:

$T1 = \int\limits^h_0 {SGw * (h-y) * sin(30) * L * (h-y)} \, dy$

$T1 = \int\limits^h_0 {SGw * sin(30) * L * (h-y)^2} \, dy$

$T1 = SGw * sin(30) * L * \int\limits^h_0 {(h-y)^2} \, dy$

$T1 = SGw * sin(30) * L * \int\limits^h_0 {h^2 - 2*h*y + y^2} \, dy$

T1 = SGw * sin(30) * L * (h^2*y - h*y^2 + 1/3*y^3)(evaluated between 0 and h)

T1 = SGw * sin(30) * L * (h^2*h - h*h^2 + 1/3*h^3)

T1 = SGw * sin(30) * L * (h^3 - h^3 + 1/3*h^3)

T1 = 1/3 * SGw * sin(30) * L * h^3

To remain stable the equilibrant torque (Teq) must be of larger magnitude than the water pressure torque (T1)

1/4 * b^2 * h * L * SG > 1/3 * SGw * sin(30) * L * h^3

In an equilateral triangle h = b * cos(30)

1/4 * b^3 * cos(30) * L * SG > 1/3 * SGw * sin(30) * L * b^3 * (cos(30))^3

SG > SGw * 4/3* sin(30) * (cos(30))^2

SG > 1/2 * SGw

For the dam to hold, it should have a specific gravity of at leas half the specific gravity of water.

This is avergae specific gravity, including holes.

6 0

3 years ago

Find a negative feedback controller with at least two tunable gains that (1) results in zero steady state error when the input i

IceJOKER [234]

Answer:

Gc(s) = $\frac{0.1s + 0.28727}{s}$

Explanation:

comparing the standard approximation with the plot attached we can tune the PI gains so that the desired response is obtained. this is because the time requirement of the setting is met while the %OS requirement is not achieved instead a 12% OS is seen from the plot.

attached is the detailed solution and the plot in Matlab

8 0

3 years ago

Your program will be a line editor. A line editor is an editor where all operations are performed by entering commands at the co

soldier1979 [14.2K]

Answer:

Java program given below

Explanation:

import java.util.*;

import java.io.*;

public class Lineeditor

{

private static Node head;

class Node

{

int data;

Node next;

public Node()

{data = 0; next = null;}

public Node(int x, Node n)

{data = x; next =n;}

}

public void Displaylist(Node q)

{if (q != null)

{

System.out.println(q.data);

Displaylist(q.next);

}

public void Buildlist()

{Node q = new Node(0,null);

head = q;

String oneLine;

try{BufferedReader indata = new

BufferedReader(new InputStreamReader(System.in)); // read data from terminals

System.out.println("Please enter a command or a line of text: ");

oneLine = indata.readLine(); // always need the following two lines to read data

head.data = Integer.parseInt(oneLine);

for (int i=1; i<=head.data; i++)

{System.out.println("Please enter another command or a new line of text:");

oneLine = indata.readLine();

int num = Integer.parseInt(oneLine);

Node p = new Node(num,null);

q.next = p;

q = p;}

}catch(Exception e)

{ System.out.println("Error --" + e.toString());}

}

public static void main(String[] args)

{Lineeditor mylist = new Lineeditor();

mylist.Buildlist();

mylist.Displaylist(head);

}

7 0

3 years ago