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Mrrafil [7]
3 years ago
13

What does the dashed line in the beaker separating the two sides represent?______________________?

Computers and Technology
2 answers:
Papessa [141]3 years ago
8 0

Answer:

What does the dashed line in the beaker separating the two sides represent <u>membrane potential</u>?

Explanation:

<u>Membrane Potential:</u>

A gradient is membrane, through which the different molecules move from one portions of the body or system into the another. It is due to the potential difference between the two sides, as there are different factors which involves in the movement from point of a system into another.

As, the diffusion occurs through the difference in potential between the sides, along with that it can also lead to the process of osmosis. As osmosis is the movement of the particles across any membrane through a small aperture present in the membrane.

kodGreya [7K]3 years ago
7 0

I guess the correct answer is A membrane

A mеmbranе is a sеlеctivе barriеr; it allοws sοmе things tο pass thrοugh but stοps οthеrs. Such things may bе mοlеculеs, iοns, οr οthеr small particlеs.

You might be interested in
Why would you use a billeted list in a slide presentation?
Damm [24]

Answer:

You would use a Bullet Point on a slide presentation because its concise and to the point.

Explanation:

The correct way to use a Bullet point is to keep it simple. Its not made to be a full sentence. You need to create a bullet point you can expand upon. when you keep it plain and simple you can get more info on your points.

7 0
3 years ago
Create a class called Hokeemon that can be used as a template to create magical creatures called Hokeeemons. Hokeemons can be of
Viefleur [7K]

Answer:

See Explaination

Explanation:

/*****************************************Hokeemon.java********************************/

import java.io.File;

import java.io.FileNotFoundException;

import java.util.Scanner;

public class Hokeemon {

private String name;

private String type;

private int age;

public Hokeemon(String name, String type, int age) {

super();

this.name = name;

this.type = type;

this.age = age;

}

public Hokeemon() {

this.name = "";

this.type = "";

this.age = 0;

}

public String getName() {

return name;

}

public void setName(String name) {

this.name = name;

}

public String getType() {

return type;

}

public void setType(String type) {

this.type = type;

}

public int getAge() {

return age;

}

public void setAge(int age) {

this.age = age;

}

public String liveIn() {

if (this.type.equalsIgnoreCase("dwarf")) {

return "Mountain";

} else if (this.type.equalsIgnoreCase("elf")) {

return "Dale";

} else if (this.type.equalsIgnoreCase("fairy")) {

return "Forest";

} else {

return "Shire";

}

}

public boolean areFriends(Hokeemon other) {

if (this.type.equalsIgnoreCase(other.type)) {

return true;

} else if ((this.type.equalsIgnoreCase("dwarf") && other.type.equalsIgnoreCase("elf"))

|| (this.type.equalsIgnoreCase("elf") && other.type.equalsIgnoreCase("dwarf"))) {

return true;

} else if ((this.type.equalsIgnoreCase("hobbit") && other.type.equalsIgnoreCase("fairy"))

|| (this.type.equalsIgnoreCase("fairy") && other.type.equalsIgnoreCase("hobbit"))) {

return true;

} else {

return false;

}

}

public static Hokeemon[] getData(String file) {

Hokeemon[] hokeemons = new Hokeemon[8];

int i = 0;

try {

Scanner scan = new Scanner(new File(file));

while (scan.hasNextLine() && i < hokeemons.length) {

String line = scan.nextLine();

String[] data = line.split("\\s+");

String name = data[0];

String type = data[1];

int age = Integer.parseInt(data[2]);

hokeemons[i] = new Hokeemon(name, type, age);

i++;

}

scan.close();

} catch (FileNotFoundException e) {

System.err.println(e);

}

return hokeemons;

}

public static String getBio(Hokeemon[] hokeemons) {

String s = "";

for (Hokeemon hokeemon : hokeemons) {

s += "I am " + hokeemon.getName() + ": Type " + hokeemon.getType() + ": Age=" + hokeemon.getAge()

+ ", I live in " + hokeemon.liveIn() + "\n";

s += "My friends are: ";

for (Hokeemon hokeemon2 : hokeemons) {

if (hokeemon.areFriends(hokeemon2) && !hokeemon.equals(hokeemon2)) {

s += (hokeemon2.getName() + " ");

}

}

s += "\n";

}

return s;

}

atOverride

public String toString() {

return "Name " + name + ": Type " + type + ": Age=" + age + "\n";

}

}

/*******************************HokeemonMain.java**************************/

import java.util.Arrays;

public class HokeemonMain {

public static void main(String[] args) {

Hokeemon[] hokeemons = Hokeemon.getData("data.txt");

System.out.println(Arrays.toString(hokeemons));

System.out.println(Hokeemon.getBio(hokeemons));

}

}

/**********************************data.txt********************/

Noddy Dwarf 4

Legolas Elf 77

Minerva Fairy 33

Samwise Hobbit 24

Merry Hobbit 29

Warhammer Dwarf 87

Ernedyll Elf 19

Frodo Hobbit 18

3 0
3 years ago
For Internet Protocol (IP) v6 traffic to travel on an IP v4 network, which two technologies are used
qaws [65]

Answer:

These technologies are: Dual Stack Routers, Tunneling, and NAT Protocol Translation.

Explanation:

Dual Stack Routers -  This is the process in which a router’s interface is attached with IPv4 and IPv6 addresses

Tunneling -Tunneling is used as a medium to help the different IP versions  communicate with the transit network.

NAT Protocol Translation -  This helps the IPv4 and IPv6 networks communicate with each other since they do not understand the IP addresses of each other since, they are different IP versions.

3 0
3 years ago
When creating an electronic slide presentation, Lee should avoid
Amanda [17]

I believe the answer is <u>Using sound effects between slides.</u>

Using sound effects between slides can cause for a distraction, and if you are in college, your professor may not score your presentation as well as if it were made without sound effects. Hope this helps!

5 0
3 years ago
Assume that a single page of printed text requires 52 lines of text, and that each line of text averaged 80 characters. If each
zysi [14]

Answer:

4160000 bytes

Explanation:

One page = 52 lines of text

One line of text = 80 characters

=> One page = 80 x 52 = 4160 characters

Therefore, 500 pages of text will have 4160 x 500 characters = 2080000 characters.

Since 1 character takes up 2 bytes of computer memory, it impleies that a 500 page novel will take up 2080000 x 2 bytes = 4160000 bytes.

5 0
4 years ago
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