If you are setting up a race car. What is the cross weight? Does it matter?
1 answer:
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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.
The total evaporation loss of water is 87.873 × lbm /day.
<u>Explanation:</u>
Assume A is the water and B is air.
A is diffusing to non diffusing B.
By the table 6.2 - 1 at 25°C, the diffusivity of air and water system is .
Total pressure P = 1 atm = 101.325 KPa
= 23.76 mm Hg
=
= 0.03126 atm
= 3.167 K Pa
When air surrounded is dry air, then = 0 mm Hg
R = 8.314
=
= 101.325 - 3.167
= 98.158 K Pa
= 101.325 - 0
= 101.325 K Pa
= 99.733 K Pa
Z = 1 ft = 0.3048 m
T = 298 K
=
= 0.11077 ×
mol/
.s
= 0.11077 ×
× 18 × (60×60×24)
= 0.1723
/
.day
Area of individual pipe is
A = 0.00051
= 0.1723 × 0.00051
= 0.000087873 lbm/day
In 1000 ft length of ditch,there will be a 10 pipes. The amount of evaporation water is
= 10 × 0.000087873 = 0.00087873 lbm /day
The total evaporation loss of water is 87.873 × lbm /day.
Answer:
6.5 × 10¹⁵/ cm³
Explanation:
Thinking process:
The relation
With the expression Ef - Ei = 0.36 × 1.6 × 10⁻¹⁹
and ni = 1.5 × 10¹⁰
Temperature, T = 300 K
K = 1.38 × 10⁻²³
This generates N₀ = 1.654 × 10¹⁶ per cube
Now, there are 10¹⁶ per cubic centimeter
Hence,
Answer:
0.506N
Explanation:
In this question, we are asked to calculate the total drag force on a plate which is oriented parallel to an air flow at a particular temperature and atmospheric pressure.
Please check attachment for complete solution, plate diagram and step-by-step explanation
