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Oliga [24]
2 years ago
6

The base class Pet has private fields petName, and petAge. The derived class Dog extends the Pet class and includes a private fi

eld for dogBreed. Complete main() to: create a generic pet and print information using printInfo(). create a Dog pet, use printInfo() to print information, and add a statement to print the dog's breed using the getBreed() method.

Computers and Technology
1 answer:
Lapatulllka [165]2 years ago
5 0

Answer:

Hello attached is the Java program written to solve the problem

The Pet.java and Dog.java files are unaltered

Explanation:

The input and output  codes are attached as well i.e the second image is the input while the third image is the output code

You might be interested in
What is the distance rn between the point of application of n⃗ and the axis of rotation? what is the distance rw between the poi
Mandarinka [93]

Complete Question.

A. Suppose that you are holding a pencil balanced on its point. If you release the pencil and it begins to fall, what will be the angular acceleration when it has an angle of 10.0 degrees from the vertical?

B. What is the distance rn between the point of application of n⃗ and the axis of rotation?

C. What is the distance rw between the point of application of w⃗ and the axis? enter your answers in meters separated by a comma?

Answer:

A. â = - 17rad/s²

B. rn = 0, meters

C. rw = 0.074, meters

Explanation:

Where;

l = length of the pencil

θ = angle between the vertical line and the pencil

â = angular acceleration.

The torque due to weight about the pivotal point is negative since it's clockwise;

t = - mg(l/2)sinØ ..... equation 1

torque with respect to angular acceleration;

t = Iâ ..... equation 2

The moment of Inertia of the pencil about one end;

I = (ml²)/3

Substituting I into equation 2;

t = [(ml²)/3]×â

Equating the equation, we have;

[(ml²)/3]×â = - mg(l/2)sinØ

â = (-3gsinØ)/2l

â = (-3*9.8*sin10°)/(2*0.15)

Angular acceleration, â = -17rad/s²

B. The normal force is acting at the normal force only, so the distance r n between the point of application of n ⃗ and the axis of rotation is zero because the axis of pencil is passing the same point of application w.

C. rn = (lcosØ)/2

rn = (0.15 * cos10°)/2

rn = (0.15 * 0.9848)/2

rn = (0.1477)/2

rn = 0.074m.

Since the gravity acts at exactly half of the length of pencil, distance r w between the point of application of w⃗ and the axis equals 0.074m.

8 0
2 years ago
In cell M2, enter a formula using a nested IF function as follows to determine first if a student has already been elected to of
aleksley [76]

Answer:

Following are the code to this question:

code:

=IF(EXACT(I2,"Yes"),"Elected",IF(EXACT(K2,"Yes"),"Yes","No"))

Explanation:

In the given the data is not defined so we explain only  the above code, but before that, we briefly define working of if the function that can be defined as follows:

  • The If() function, is used only when one of the logical functions and its value must return the value true. or we can say it only return a true value.
  • In the above function, a column "Elected" is used that uses other column values to check this column value is equal if this condition is true it will return "yes" value.

8 0
2 years ago
What is the name of the item that supplies the exact or near exact voltage at the required wattage to all of the circuitry insid
gizmo_the_mogwai [7]

Answer:

It's the <u><em>power supply</em></u>

Explanation:

The power supply is what essentially enables the computer to operate. It is able to do that by converting the incoming alternating current (AC) to direct current (DC) at the correct wattage rating that is required by the computer to function. The power supply is a metal box that is generally placed in the corner of the case.

4 0
2 years ago
For any element in keysList with a value greater than 50, print the corresponding value in itemsList, followed by a space. Ex: I
VARVARA [1.3K]

Answer:

import java.util.Scanner;  //mentioned in the question.

public class ArraysKeyValue {  //mentioned in the question.

public static void main (String [] args) {  //mentioned in the question.

final int SIZE_LIST = 4; //mentioned in the question.

int[] keysList = new int[SIZE_LIST]; //mentioned in the question.

int[] itemsList = new int[SIZE_LIST]; //mentioned in the question.

int i; //mentioned in the question.

keysList[0] = 13; //mentioned in the question.                                

keysList[1] = 47; //mentioned in the question.

keysList[2] = 71; //mentioned in the question.

keysList[3] = 59; //mentioned in the question.

itemsList[0] = 12; //mentioned in the question.

itemsList[1] = 36; //mentioned in the question.

itemsList[2] = 72; //mentioned in the question.

itemsList[3] = 54;//mentioned in the question.

// other line to complete the solution is as follows--

for(i=0;i<(keysList.length);i++) //Loop to access all the element of keysList array variable.

    {// open for loop braces

        if(keysList[i]>50) // compare the value of keysList array variable with 50.

        System.out.println(itemsList[i]); // print the value

   }// close for loop braces.

}//  close main function braces.

} //close the class braces.

Output:

72

54

Explanation:

In the solution part--

  1. One loop is placed which moves from 0 to (size-1) of the 'keyslist' array.
  2. Then the 'if' statement is used to check the element of 'keyslist' array, which is greater than 50 or not.
  3. If it is greater, then print the element of item_list array of 'i' index which is also the index of greater value of keyslist array element.

5 0
2 years ago
When an author produce an index for his or her book, the first step in this process is to decide which words should go into the
Igoryamba

Answer:

import string

dic = {}

book=open("book.txt","r")

# Iterate over each line in the book

for line in book.readlines():

   tex = line

   tex = tex.lower()

   tex=tex.translate(str.maketrans('', '', string.punctuation))

   new = tex.split()

   for word in new:

       if len(word) > 2:

           if word not in dic.keys():

               dic[word] = 1

           else:

               dic[word] = dic[word] + 1

for word in sorted(dic):

   print(word, dic[word], '\n')

                 

book.close()

Explanation:

The code above was written in python 3.

<em>import string </em>

Firstly, it is important to import all the modules that you will need. The string module was imported to allow us carry out special operations on strings.

<em>dic = {} </em>

<em>book=open("book.txt","r") </em>

<em> </em>

<em># Iterate over each line in the book</em>

<em>for line in book.readlines(): </em>

<em> </em>

<em>    tex = line </em>

<em>    tex = tex.lower() </em>

<em>    tex=tex.translate(str.maketrans('', '', string.punctuation)) </em>

<em>    new = tex.split() </em>

<em />

An empty dictionary is then created, a dictionary is needed to store both the word and the occurrences, with the word being the key and the occurrences being the value in a word : occurrence format.

Next, the file you want to read from is opened and then the code iterates over each line, punctuation and special characters are removed from the line and it is converted into a list of words that can be iterated over.

<em />

<em> </em><em>for word in new: </em>

<em>        if len(word) > 2: </em>

<em>            if word not in dic.keys(): </em>

<em>                dic[word] = 1 </em>

<em>            else: </em>

<em>                dic[word] = dic[word] + 1 </em>

<em />

For every word in the new list, if the length of the word is greater than 2 and the word is not already in the dictionary, add the word to the dictionary and give it a value 1.

If the word is already in the dictionary increase the value by 1.

<em>for word in sorted(dic): </em>

<em>    print(word, dic[word], '\n') </em>

<em>book.close()</em>

The dictionary is arranged alphabetically and with the keys(words) and printed out. Finally, the file is closed.

check attachment to see code in action.

7 0
2 years ago
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