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4 years ago

6

the two main forces acting on a computer that sits on a table are _ and _. a. weight; gravity b. weight; mass c.

weight; inertia d. weight; the normal force

Computers and Technology

2 answers:

EleoNora [17]4 years ago

8 0

It is A.weight: gravity

Flauer [41]4 years ago

5 0

I took the test and got A) Weight and gravity correct

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Exercise 8.1.9: Diving Contest Your school is hosting a diving contest, and they need a programmer to work on the scoreboard! Yo

vampirchik [111]

Answer:

def calculate_score(theTuple):

first, second, third = theTuple

if first >= 0 and first <=10 and second >= 0 and second <=10 and third >= 0 and third <=10:

print(first + second+third)

else:

print("Range must be 0 to 10")

Explanation:

This line defines the function

def calculate_score(theTuple):

This line gets the content of the function

first, second, third = theTuple

The following if condition checks if the digits are in the range 0 to 10

if first >= 0 and first <=10 and second >= 0 and second <=10 and third >= 0 and third <=10:

This calculates the sum

print(first + second+third)

else:

If number is outside range 0 and 10, this line is executed

print("Range must be 0 to 10")

4 0

4 years ago

The file command provides information about any file system object (i.e., file, directory or link) that is provided to it as an

NeTakaya

Answer:

The answer is "Option a, c, and d".

Explanation:

This command aims to define the argumentation from each entity of the file system like a disc, path, or link. It checks the usage of a measurement system, which is requested to collect the information from either the inode of the item, the description of the option as follows:

In option a, It is used to identify the file type.
In option c, This command is used to list all file types.
In option d, This command is used to list all the current directories.
In option b, It is used to describe a single entity directory.

7 0

4 years ago

Let's implement a classic algorithm: binary search on an array. Implement a class named BinarySearcher that provides one static

yKpoI14uk [10]

Answer:

Hope this helped you, and if it did , do consider giving brainliest.

Explanation:

import java.util.ArrayList;

import java.util.List;

//classs named BinarySearcher

public class BinarySearcher {

// main method

public static void main(String[] args) {

// create a list of Comparable type

List<Comparable> list = new ArrayList<>();

// add elements

list.add(1);

list.add(2);

list.add(3);

list.add(4);

list.add(5);

list.add(6);

list.add(7);

// print list

System.out.println("\nList : "+list);

// test search method

Comparable a = 7;

System.out.println("\nSearch for 7 : "+search(list,a));

Comparable b = 3;

System.out.println("\nSearch for 3 : "+search(list,b));

Comparable c = 9;

System.out.println("\nSearch for 9 : "+search(list,c));

Comparable d = 1;

System.out.println("\nSearch for 1 : "+search(list,d));

Comparable e = 12;

System.out.println("\nSearch for 12 : "+search(list,e));

Comparable f = 0;

System.out.println("\nSearch for 0 : "+search(list,f));

}

// static method named search takes arguments Comparable list and Comparable parameter

public static boolean search(List<Comparable> list, Comparable par) {

// if list is empty or parameter is null the throw IllegalArgumentException

if(list.isEmpty() || par == null ) {

throw new IllegalArgumentException();

}

// binary search

// declare variables

int start=0;

int end =list.size()-1;

// using while loop

while(start<=end) {

// mid element

int mid =(start+end)/2;

// if par equal to mid element then return

if(list.get(mid).equals(par) )

{

return true ;

}

// if mid is less than parameter

else if (list.get(mid).compareTo(par) < 0 ) {

start=mid+1;

}

// if mid is greater than parameter

else {

end=mid-1;

}

}

// if not found then retuen false

return false;

}

}import java.util.ArrayList;

import java.util.List;

//classs named BinarySearcher

public class BinarySearcher {

// main method

public static void main(String[] args) {

// create a list of Comparable type

List<Comparable> list = new ArrayList<>();

// add elements

list.add(1);

list.add(2);

list.add(3);

list.add(4);

list.add(5);

list.add(6);

list.add(7);

// print list

System.out.println("\nList : "+list);

// test search method

Comparable a = 7;

System.out.println("\nSearch for 7 : "+search(list,a));

Comparable b = 3;

System.out.println("\nSearch for 3 : "+search(list,b));

Comparable c = 9;

System.out.println("\nSearch for 9 : "+search(list,c));

Comparable d = 1;

System.out.println("\nSearch for 1 : "+search(list,d));

Comparable e = 12;

System.out.println("\nSearch for 12 : "+search(list,e));

Comparable f = 0;

System.out.println("\nSearch for 0 : "+search(list,f));

}

// static method named search takes arguments Comparable list and Comparable parameter

public static boolean search(List<Comparable> list, Comparable par) {

// if list is empty or parameter is null the throw IllegalArgumentException

if(list.isEmpty() || par == null ) {

throw new IllegalArgumentException();

}

// binary search

// declare variables

int start=0;

int end =list.size()-1;

// using while loop

while(start<=end) {

// mid element

int mid =(start+end)/2;

// if par equal to mid element then return

if(list.get(mid).equals(par) )

{

return true ;

}

// if mid is less than parameter

else if (list.get(mid).compareTo(par) < 0 ) {

start=mid+1;

}

// if mid is greater than parameter

else {

end=mid-1;

}

}

// if not found then retuen false

return false;

}

}

7 0

3 years ago

Licemer1 [7]

Answer:

yes

Explanation:

the faster the processor the better the computer loads things and works faster, there's little to no latency depending on the wifi

3 0

3 years ago

Consider the following configuration. Computer ""A"" IP address: 172.16.100.100 Subnet mask: 255.255.0.0 Computer ""B"" IP a

denpristay [2]

Answer:

There is nothing to change.

Explanation:

In the following configuration given in the problem, the necessary IP addresses and the subnet masks have been included in the configuration. The computers can be used to send and/or receive information to/from each other. The computers have been properly set-up and there is nothing to be done to the settings or configurations.

4 0

3 years ago