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LekaFEV [45]
2 years ago
8

Personal computer hard disk platters typically have storage capacities ranging from 40 gb to ____.

Computers and Technology
1 answer:
Doss [256]2 years ago
7 0
2tb I think I'm not that sure
You might be interested in
You would like the user of a program to enter a customer’s last name. Write a statement thaUse the variables k, d, and s so that
mojhsa [17]

Answer:

1st question:

Use the variables k, d, and s so that they can read three different values from standard input an integer, a float, and a string respectively. On one line, print these variables in reverse order with exactly one space in between each. On a second line, print them in the original order with one space in between them.

Solution:

In Python:

k = input()  #prompts user to input value of k i.e. integer value

d = input()  #prompts user to input value of d i.e. float value

s = input()  #prompts user to input value of s i.e. a string

print (s, d, k)  #displays these variable values in reverse order

print (k, d, s)#displays these variable values in original order

In C++:

#include <iostream>    // to use input output functions

using namespace std;   //to identify objects like cin cout

int main() {    //start of main function

  int k;   //declare int type variable to store an integer value

  float d; //  declare float type variable to store a float value

  string s;   //  declare string type variable to store an integer value

  cin >> k >> d >> s;    //reads the value of k, d and s

  cout << s << " " << d << " " << k << endl;     //displays these variables values in reverse order

  cout << k << " " << d << " " << s << endl;   } // displays these variable values in original order

Explanation:

2nd question:

You would like the user of a program to enter a customer’s last name. Write a statement that asks user "Last Name:" and assigns input to a string variable called last_name.

Solution:

In Python:

last_name = input("Last Name:")

# input function is used to accept input from user and assign the input value to last_name variable

In C++:

string last_name;  //declares a string type variable named last_name

cout<<"Last Name: ";  // prompts user to enter last name by displaying this message Last Name:

cin>>last_name; // reads and assigns the input value to string variable last_name

The programs alongwith their outputs are attached.

6 0
2 years ago
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
Mary has been locked out of her account after failing to correctly enter her password three times. As the system administrator,
Digiron [165]
Oh hey lol yea yea lol I got the money back to me put on the phone so you know
7 0
2 years ago
Consider a short, 10-meter link, over which a sender can transmit at a rate of 150 bits/sec in both directions. Suppose that pac
Katarina [22]

Answer:

The Tp value 0.03 micro seconds as calculated in the explanation below is negligible. This would lead to a similar value of time delay for both persistent HTTP and non-persistent HTTP.

Thus, persistent HTTP is not faster than non-persistent HTTP with parallel downloads.

Explanation:

Given details are below:

Length of the link = 10 meters

Bandwidth = 150 bits/sec

Size of a data packet = 100,000 bits

Size of a control packet = 200 bits

Size of the downloaded object = 100Kbits

No. of referenced objects = 10

Ler Tp to be the propagation delay between the client and the server, dp be the propagation delay and dt be the transmission delay.

The formula below is used to calculate the total time delay for sending and receiving packets :

d = dp (propagation delay) + dt (transmission delay)

For Parallel downloads through parallel instances of non-persistent HTTP :

Bandwidth = 150 bits/sec

No. of referenced objects = 10

For each parallel download, the bandwith = 150/10

  = 15 bits/sec

10 independent connections are established, during parallel downloads,  and the objects are downloaded simultaneously on these networks. First, a request for the object was sent by a client . Then, the request was processed by the server and once the connection is set, the server sends the object in response.

Therefore, for parallel downloads, the total time required  is calculated as:

(200/150 + Tp + 200/150 + Tp + 200/150 + Tp + 100,000/150 + Tp) + (200/15 + Tp + 200/15 + Tp + 200/150 + Tp + 100,000/15 + Tp)

= ((200+200+200+100,00)/150 + 4Tp) + ((200+200+200+100,00)/15 + 4Tp)

= ((100,600)/150 + 4Tp) + ((100,600)/15 + 4Tp)

= (670 + 4Tp) + (6706 + 4Tp)

= 7377 + 8 Tp seconds

Thus, parallel instances of non-persistent HTTP makes sense in this case.

Let the speed of propogation  of the medium be 300*106 m/sec.

Then, Tp = 10/(300*106)

               = 0.03 micro seconds

The Tp value 0.03 micro seconds as calculated above is negligible. This would lead to a similar value of time delay for both persistent HTTP and non-persistent HTTP. Thus, persistent HTTP is not faster than non-persistent HTTP with parallel downloads.

4 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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