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Tatiana [17]
3 years ago
8

What are the effects deforestation have on a river system?

Computers and Technology
1 answer:
Nikolay [14]3 years ago
8 0
Deforestation affects rivers in these ways
1.There is a loss of habitat for wild animals
2. Destruction of land due to floods, mudslides, 
3. Water cycle and Animal Cycle are disturbed
( ex. transpiration rate; beavers building dams)
4. Greater amount of Green House gases


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Tortise and hare race java g Modify the main class so it runs the race 100 times and reports how many times each runner wins. (T
SOVA2 [1]

Answer:

Game.java file

import java.util.Scanner;

public class Game {

/**

* t_pos and h_pos are the current positions of the Tortoise and Hare

*/

static int t_pos,h_pos;

static Tortoise tortoise;

static Hare hare;

public static void main(String[] args) {

play(); /*starting the game*/

}

public static void play(){

/**

* the method will starts the play, loop until the game is over, displays the winner

* and prompts the user if they want to play again

*/

/**

* defining Tortoise and Hare objects

*/

tortoise=new Tortoise();

hare=new Hare();

t_pos=1;

h_pos=1;

System.out.println("The race is about to start");

tortoise.printTrack();

hare.printTrack();

while(t_pos != 50 && h_pos !=50){

System.out.println("\n\n\n"); /*printing blank lines*/

t_pos=tortoise.move(); /*moving and getting the current position of tortoise*/

h_pos=hare.move();/*moving and getting the current position of hare*/

tortoise.printTrack(); /*displaying the tracks*/

hare.printTrack();

try { /*comment this part to skip the 1s break between each round; for testing*/

Thread.sleep(1000);

} catch (InterruptedException e) {

e.printStackTrace();

}

}

System.out.println("\nRace Over");

if(t_pos==50 && h_pos==50){

System.out.println("Its a tie");

}

else if(t_pos==50){

System.out.println("Tortoise wins");

}else if(h_pos==50){

System.out.println("Hare wins");

}

System.out.println("Do you want to play again? (y/n)");

Scanner scanner=new Scanner(System.in);

String ch=scanner.next();

if(ch.equalsIgnoreCase("y")){

play();

}else if(ch.equalsIgnoreCase("n")){

System.out.println("Thanks for playing, Goodbye");

}else{

System.out.println("Invalid choice, quitting..");

}

}

}

//Tortoise.java

public class Tortoise {

/**

* the current position of the tortoise

*/

int position;

/**

* track array

*/

char[] track;

/**

* speed of tortoise

*/

int speed=1;

public Tortoise() {

position=0;

track=new char[50];

for(int i=0;i<track.length;i++){

/**

* filling the track

*/

track[i]='-';

}

}

public int move(){

if(position<track.length){

position=position+speed;

}

return position+1;

}

public void printTrack(){

/**

* the current position of tortoise will be displayed by 'T' everything else will be '-'

*/

System.out.println();

for(int i=0;i<track.length;i++){

if(i==position){

System.out.print('T');

}else{

System.out.print(track[i]);

}

}

}

}

//Hare.java

import java.util.Random;

public class Hare {

int position;

int speed=10;

char[] track;

/**

* Random object to generate a random number

*/

Random random;

/**

* resting percent denotes how much time Hare will be resting

*/

int resting_percent=90;

public Hare() {

position=0;

track=new char[50];

for(int i=0;i<track.length;i++){

track[i]='-';

}

random=new Random();

}

public int move(){

int n=random.nextInt(100-1)+1; /*generating a random number between 1 and 100*/

if(n<=resting_percent){

/**

* at rest; will not move, returns the current position.

*/

return position;

}else{

/**

* not resting..

*/

if(position<track.length){

if(position+speed>=track.length){

position=track.length-1;

}else{

position=position+speed;

}

}

return position+1;

}

}

public void printTrack(){

System.out.println();

for(int i=0;i<track.length;i++){

if(i==position){

System.out.print('H');

}else{

System.out.print(track[i]);

}

}

}

}

/*Output (partial and random)*/

The race is about to start

T-------------------------------------------------

H-------------------------------------------------

-T------------------------------------------------

H-------------------------------------------------

--T-----------------------------------------------

H-------------------------------------------------

---T----------------------------------------------

----------H---------------------------------------

.

.

.

.

-----------------------------------------------T--

-------------------------------------------------H

Race Over

Hare wins

Do you want to play again? (y/n)

y

The race is about to start

T-------------------------------------------------

H-------------------------------------------------

-T------------------------------------------------

H-------------------------------------------------

.

.

.

.

-----------------------------------------------T--

--------------------H-----------------------------

------------------------------------------------T-

--------------------H-----------------------------

-------------------------------------------------T

--------------------H-----------------------------

Race Over

Tortoise wins

Do you want to play again? (y/n)

n

Explanation:

3 0
3 years ago
Explain why the intangibility of Software System poses special problems for Software Project Management
Murljashka [212]
<span>Intangible system is that system whose services are not visible</span>
4 0
3 years ago
which of the following is not a name of one of the central features of Facebook timeline activity log Graph Search for Daily New
Pani-rosa [81]
The correct answer is Graph Search for Daily News

This does not exist on Facebook, unlike the timeline and the activity log.
3 0
3 years ago
Consider bears = {"Grizzly":"angry", "Brown":"friendly", "Polar":"friendly"}. Can you replace #blank# so the code will print a g
Goshia [24]

Answer:

bears = {"Grizzly":"angry", "Brown":"friendly", "Polar":"friendly"}

for bear in bears:

   if bears[bear] == "friendly":

       print("Hello, "+bear+" bear!")

   else:

       print("odd")

Explanation:

A dictionary called bears is given. A dictionary consists of key-value pairs.

You need to check each key-value pairs in bears and find the ones that have "friendly" as value using a for loop and if-else structure. In order to access the values of the dictionary, use the dictionary name and the key (inside the loop, the key is represented as bear variable). If a key has a value of "friendly", print the greeting. Otherwise, print "odd".

6 0
3 years ago
10 Ejemplos de Software de Sistema. AYUDENME POR FAVOR SE LOS SUPLICO
bearhunter [10]

Answer:

MacOS, Linux, Android, Microsoft Windows, software de ciencia computacional, motores de juegos, automatización industrial y aplicaciones de software como servicio.

:-) (-:

6 0
2 years ago
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