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stiv31 [10]
3 years ago
10

The data shows that recently the size of the raccoon population decreased. How will the decrease in the raccoon population affec

t the other populations?
Engineering
1 answer:
LenaWriter [7]3 years ago
3 0

Answer: Increase in the population of lower organisms.

Explanation: Raccoons are mesopredators found in the middle levels of food webs and have critical impacts on the dynamics of many other species. Thus, they greatly contribute to the functioning of the ecosystem.

They are naturally equipped to checkmate the population size of other organism that are below them (i.e., their preys) in the food web. They are; predators, pathogen carriers (such as rabies), and competitors, as they compete with some specialist in the food web.

Hence, the decrease in the population of raccoons will lead to the increase in the population size of lower organisms they prey on in the food web or chain.

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Write a for loop to print all elements in courseGrades, following each element with a space (including the last). Print forwards
user100 [1]

Answer:

The first step is to;

a. import Scanner class present in java.util package for take input in arrays.

b. When final variable is initialized it can not be changed so fix the size to 4.

c. create array to hold 4 integer elements in array.

d. Now ensure users enter integer values separated by space in single line & assign elements at each index of array starting from index 0.

e. Next is to print the array elements in forward directions first.

reset i to 0 to print all elements of array from index 0.

f. Then ensure to traverse array and display each element one by one separated by space in same line using for loop.

g. Therefore to move to next line .

reset i to array.length -1. loop will go from last elements index till index 0.

h. Finally print elements in array backwards from last elements of array.

The second step is given here,

import java.util.Scanner;

public class CourseGradePrinter {

public static void main(String args) {

//using scanner to take input from user..

Scanner scnr = new Scanner(System.in);

//fix the array size to 4.

final int NUM_VALS = 4;

//array created of size 4 to hold int data

int[] courseGrades = new int[NUM_VALS];

//declare i for user input and traversing array

int i;

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

//user enters integer values

courseGrades[i] = scnr.nextInt();

}

/* Your solution goes here */

//display the array elements forwards now

i=0;

Ensure to print elements in array.

4 0
3 years ago
Put four red LED as a straight line and connect each of them to a corresponding analog output. Connect a potentiometer to a 5 Vo
N76 [4]

Explanation:

There has been no information about related to which programming language is to be used, writing code algorithm.

Defining I/O's ;

Analogue output

A01, A02, A03,AO4

Analogue Input;

AI_1   // potentiometer input

// Based on controller used, assign channels to I/O's

// code

int voltage

Voltage = AI_1;

If (Voltage > 0 && Voltage < 1.25)

 {

   A01 = voltage

   A02 = 0;

   A03= 0

   AO4= 0

 }

If (Voltage > 1.25 && Voltage < 2.5)

 {

   A01 = 1.25

   A02 = (Voltage -1.25);

   A03= 0

   AO4= 0

 }

If (Voltage > 2.5 && Voltage < 3.75)

 {

   A01 = 1.25

   A02 = 1.25

   A03=  (Voltage - 2.5);

   AO4= 0

 }

else

 {

   A01 = 1.25

   A02 = 1.25

   A03=  1.25

   AO4= (Voltage - 3.75);

 }

return

3 0
3 years ago
A tensile test was made on a tensile specimen, with a cylindrical gage section which had a diameter of 10 mm, and a length of 40
tamaranim1 [39]

Answer:

The answers are as follow:

a) 10 mm

b) 12.730 N/mm^{2}

c) 127.307 N/mm^{2}

d) 0.25

Explanation:

d1 = 10mm , L1 = 40 mm, L2 = 50 mm, reduction in area = 90% = 0.9

Force = F =1000 N

let us find initial area first, A1 = pi*r^{2} = 78.55 mm^{2}

using reduction in area formula : 0.9 = (A1 - A2 ) / A1

solving it will give,  A2 = 0.1 A1 = 7.855  mm^{2}

a) The specimen elongation is final length - initial length

50 - 40 = 10 mm

b) Engineering stress uses the original area for all stress calculations,

Engineering stress = force / original area  = F / A1 = 1000 / 78.55  

Engineering stress = 12.730 N / mm^{2}

c) True stress uses instantaneous area during stress calculations,

True fracture stress = force / final  area  = F / A2 = 1000 / 7.855

True Fracture stress = 127.30 N / mm^{2}

e) strain = change in length / original length

strain = 10 / 40  = 0.25

8 0
3 years ago
The distillation column in Figure 3 is set up for so-called boil-up (V) control. It has
11111nata11111 [884]

A distillation column is an essential item used in the distillation of liquid mixtures to separate the mixture into its component parts, or fractions, based on the differences in volatilities. Fractionating columns are used in small scale laboratory distillations as well as large scale industrial distillations.

3 0
3 years ago
Extra points!!!!! <br><br> List the general types of housing<br> (Example: condo)
Nadusha1986 [10]

Answer:

co-op, apartment, townhome, manor etc

6 0
3 years ago
Read 2 more answers
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