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stich3 [128]
4 years ago
12

Why do the holes at the top of parachutes make it go slower

Computers and Technology
1 answer:
Zina [86]4 years ago
8 0
The more air resistance they encounter, the slower they fall. Eventually an equilibrium is reached between the downward acceleration caused by gravity and the upward push caused by air resistance. All because of the drag of the parachute.
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All file types have unique _____________________ that determine which program to use to open a particular file and to access its
pychu [463]

Answer:

File format xdd

Explanation:

8 0
3 years ago
A technician is troubleshooting a printer connected to a computer via USB. The printer will print using the controls at the prin
ki77a [65]

Verify the USB cable is attached

Answer: Option A.

<u>Explanation:</u>

The USB cable utilized for printers is known as a USB AB cable, named for the attachments on each end. The USB-An end is a level, rectangular fitting; the USB-B end is a square attachment with two bended edges, which goes into the printer.

A USB link associates your printer to your PC, so you have an immediate association each time you print. Most of printers are good with a USB 2.0 A/B link. This is a standard link that is generally accessible and it can likewise be utilized to interface a few mice and cameras to your PC as well.

7 0
3 years ago
Write a method that accepts a String object as an argument and returns a copy of the string with the first character of each sen
emmainna [20.7K]

Answer:

The programming language is not stated; However, the program written in C++ is as follows: (See Attachment)

#include<iostream>

using namespace std;

string capt(string result)

{

result[0] = toupper(result[0]);

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

 if(result[i]=='.' || result[i]=='?' ||result[i]=='!')  {

 if(result[i+1]==' ') {

  result[i+2] = toupper(result[i+2]);

 }

 if(result[i+2]==' ')  {

  result[i+3] = toupper(result[i+3]);

 }

 } }

return result;

}

int main(){

string sentence;

getline(cin,sentence);

cout<<capt(sentence);

return 0;

}

Explanation:

<em>The method to capitalize first letters of string starts here</em>

string capt(string result){

<em>This line capitalizes the first letter of the sentence</em>

result[0] = toupper(result[0]);

<em>This iteration iterates through each letter of the input sentence</em>

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

This checks if the current character is a period (.), a question mark (?) or an exclamation mark (!)

if(result[i]=='.' || result[i]=='?' ||result[i]=='!')  {

 if(result[i+1]==' '){  ->This condition checks if the sentence is single spaced

  result[i+2] = toupper(result[i+2]);

-> If both conditions are satisfied, a sentence is detected and the first letter is capitalized

 }

 if(result[i+2]==' '){ ->This condition checks if the sentence is double spaced

  result[i+3] = toupper(result[i+3]);

-> If both conditions are satisfied, a sentence is detected and the first letter is capitalized

 }

}

}  <em>The iteration ends here</em>

return result;  ->The new string is returned here.

}

The main method starts here

int main(){

This declares a string variable named sentence

string sentence;

This gets the user input

getline(cin,sentence);

This passes the input string to the method defined above

cout<<capt(sentence);

return 0;

}

Download cpp
7 0
4 years ago
Kelly is evaluating the following image. Which of the statements that she makes about the photograph is true?
Stells [14]

Photo not included

Plz add photo

7 0
3 years ago
A drowsy cat spots a flowerpot that sails first up and then down past an open window. The pot is in view for a total of 0.50 sec
GaryK [48]

Answer:

2.342m

Explanation:

Given

Time = 0.5 s

Height of Window = 2m

Because the pot was in view for a total of 0.5 seconds, we can assume that it took the cat 0.25 seconds to go from the bottom of the window to the top

Using this equation of motion

S = ut - ½gt²

Where s = 2

u = initial velocity = ?

t = 0.25

g = 9.8

So, we have.

2 = u * 0.25 - ½ * 9.8 * 0.25²

2 = 0.25u - 0.30625

2 + 0.30625 = 0.25u

2.30625 = 0.25u

u = 2.30625/0.25

u = 9.225 m/s ------------ the speed at the bottom of the pot

Using

v² = u² + 2gs to calculate the height above the window

Where v = final velocity = 0

u = 9.225

g = 9.8

S = height above the window

So, we have

0² = 9.225² - 2 * 9.8 * s

0 = 85.100625 - 19.6s

-85.100625 = -19.6s

S = -85.100625/19.6

S = 4.342

If 4.342m is the height above the window and the window is 2m high

Then 4.342 - 2 is the distance above the window

4.342 - 2 = 2.342m

6 0
3 years ago
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