The velocity field of a flow is given by where x and y are in feet. Determine the fluid speed at points along the x axis; along
2 answers:
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Answer:
Explanation:
Previous concepts
Angular momentum. If we consider a particle of mass m, with velocity v, moving under the influence of a force F. The angular momentum about point O is defined as the “moment” of the particle’s linear momentum, L, about O. And the correct formula is:
Applying Newton’s second law to the right hand side of the above equation, we have that r ×ma = r ×F =
MO, where MO is the moment of the force F about point O. The equation expressing the rate of change of angular momentum is this one:
MO = H˙ O
Principle of Angular Impulse and Momentum
The equation MO = H˙ O gives us the instantaneous relation between the moment and the time rate of change of angular momentum. Imagine now that the force considered acts on a particle between time t1 and time t2. The equation MO = H˙ O can then be integrated in time to obtain this:
Solution to the problem
For this case we can use the principle of angular impulse and momentum that states "The mass moment of inertia of a gear about its mass center is ".
If we analyze the staritning point we see that the initial velocity can be founded like this:
And if we look the figure attached we can use the point A as a reference to calculate the angular impulse and momentum equation, like this:
And if we integrate the left part and we simplify the right part we have
And if we solve for we got:
Answer:
=> base transport factor = 0.98.
=> emitter injection efficiency = 0.99.
=> common-base current gain = 0.97.
=> common-emitter current gain = 32.34.
=> ICBO = 1 × 10^-6 A.
=> base transit time = 0.325.
=> lifetime = 1.875.
Explanation:
(Kindly check the attachment for the diagram showing the dominant current components, with proper arrows, for directions in the normal active mode).
The following parameters or data are given for a p-n-p BJT with NE 7 NB 7 NC and they are: IEp = 10 mA, IEn = 100 mA, ICp = 9.8 mA, and ICn = 1 mA.
(1). The base transport factor = ICp/IEp=9.8/10 = 0.98.
(2). emitter injection efficiency =IEp/ IEp + ICn = 10/10 + 0.1 = 0.99.
(3).common-base current gain = 0.98 × 0.99 = 0.9702.
(4).common-emitter current gain =0.97 / 1- 0.97 = 32.34.
(5). Icbo = Ico = 1 × 10^-6 A.
(6). base transit time = 1248 × 10^-2 × (1.38× 10^-23/1.603 × 10^-19). = 0.325.
(7).lifetime;
= > 2 = √0.325 + √ lifetime.
= Lifetime = 2.875.
Answer:
Here is the JAVA program:
import java.util.Scanner; // to get input from user
public class DrawHalfArrow{ // start of the class half arrow
public static void main(String[] args) { // starts of main() function body
Scanner scnr = new Scanner(System.in); //reads input
int arrowBaseHeight = 0; // stores the height of arrow base
int arrowBaseWidth = 0; // holds width of arrow base
int arrowHeadWidth = 0; // contains the width of arrow head
// prompts the user to enter arrow base height, width and arrow head width
System.out.println("Enter arrow base height: ");
arrowBaseHeight = scnr.nextInt(); // scans and reads the input as int
System.out.println("Enter arrow base width: ");
arrowBaseWidth = scnr.nextInt();
/* while loop to continue asking user for an arrow head width until the value entered is greater than the value of arrow base width */
while (arrowHeadWidth <= arrowBaseWidth) {
System.out.println("Enter arrow head width: ");
arrowHeadWidth = scnr.nextInt(); }
//start of the nested loop
//outer loop iterates a number of times equal to the height of the arrow base
for (int i = 0; i < arrowBaseHeight; i++) {
//inner loop prints the stars asterisks
for (int j = 0; j <arrowBaseWidth; j++) {
System.out.print("*"); } //displays stars
System.out.println(); }
//temporary variable to hold arrowhead width value
int k = arrowHeadWidth;
//outer loop to iterate no of times equal to the height of the arrow head
for (int i = 1; i <= arrowHeadWidth; i++)
{ for(int j = k; j > 0; j--) {//inner loop to print stars
System.out.print("*"); } //displays stars
k = k - 1;
System.out.println(); } } } // continues to add more asterisks for new line
Explanation:
The program asks to enter the height of the arrow base, width of the arrow base and the width of arrow head. When asking to enter the width of the arrow head, a condition is checked that the arrow head width arrowHeadWidth should be less than or equal to width of arrow base arrowBaseWidth. The while loop keeps iterating until the user enters the arrow head width larger than the value of arrow base width.
The loop is used to output an arrow base of height arrowBaseHeight. So point (1) is satisfied.
The nested loop is being used which as a whole outputs an arrow base of width arrowBaseWidth. The inner loop draws the stars and forms the base width of the arrow, and the outer loop iterates a number of times equal to the height of the arrow. So (2) is satisfied.
A temporary variable k is used to hold the original value of arrowHeadWidth so that it keeps safe when modification is done.
The last nested loop is used to output an arrow head of width arrowHeadWidth. The inner loop forms the arrow head and prints the stars needed to form an arrow head. So (3) is satisfied.
The value of temporary variable k is decreased by 1 so the next time it enters the nested for loop it will be one asterisk lesser.
The screenshot of output is attached.
