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EleoNora [17]
3 years ago
14

A car travells at 67.5 km\h in 120 km.how long will it take to reach the destination

Engineering
1 answer:
Kruka [31]3 years ago
5 0

mark me the brainiest here

average speed (in km/h) of a car stuck in traffic that drives 12 kilometers in 2 hours.

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What is a network? I'LL MARK BRAINLEST
Jobisdone [24]

Answer:

hsjeeieoj eu sou ku nahi u have UCC guide to buying it and I he was a temporary password for bees and u h ki tarah nahi to ye sab se jyada nahi hota nahi to kabhi bhi hai ki wo to sirf Tum nahi hota

7 0
2 years ago
Read 2 more answers
The two boxcars A and B have a weight of 20000lb and 30000lb respectively. If they coast freely down the incline when the brakes
kolbaska11 [484]

Answer:

T=5.98 kips

Explanation:

First, introduce forces, acting on both cars:

on car A there are 4 forces acting: gravity force mA*g, normal reaction force, friction force and force T- it represents the interaction between cars A and B. On car B, there are three forces acting: gravity force, normal reaction force and force T. Note, that force T is acting on both cars, but it has opposite direction. Force T, acting on car A has direction, opposite to the friction force, whether the T, acting on B, is directed backwards- in the same direction with the friction force. Note, that both cars have the same acceleration, which is directed backwards.

Once the forces were established, we can write components of the Second Newtons Law on vertical and horizontal axes, considering that horizontal axis is directed backwards- in the same direction with the acceleration:

For car A on the vertical axis the equation is: -mAg+NA=0

For car A on the horizontal axis, the equation is: Ffr-T=mAa

For car B, on the vertical axis the equation is: -mBg+NB=0

For car B, on the horizontal axis, the equation is: T=mBa

We need to solve these equations to find force T, knowing that Ffr=μmAg, where

After the transformations, the equations for acceleration and force in the coupling will be:

a=(μmAg)/(mA+mB)=6.43 ft/s2- note, that the given answer is not correct for the given numerical values;

and force T: T=μmAmBg/(mA+mB)=6.0 kips- note, that the force answer is in line with the given numerical value

5 0
3 years ago
Write a program that prompts for a line of text and then transforms the text based on chosen actions. Actions include reversing
nlexa [21]

Answer:

public class TextConverterDemo

{

//Method definition of action1337

public static String action1337(String current)

{

//Replace each L or l with a 1 (numeral one)

 current = current.replace('L', '1');

 current = current.replace('l', '1');

 

 //Replace each E or e with a 3 (numeral three)

 current = current.replace('E', '3');

 current = current.replace('e', '3');

 //Replace each T or t with a 7 (numeral seven)

 current = current.replace('T', '7');

 current = current.replace('t', '7');

 //Replace each O or o with a 0 (numeral zero)

 current = current.replace('O', '0');

 current = current.replace('o', '0');

 

//Replace each S or s with a $ (dollar sign)

 current = current.replace('S', '$');

 current = current.replace('s', '$');

 return current;

}

//Method definition of actionReverse

//This method is used to reverses the order of

//characters in the current string

public static String actionReverse(String current)

{

 //Create a StringBuilder's object

 StringBuilder originalStr = new StringBuilder();

 //Append the original string to the StribgBuilder's object

 originalStr.append(current);

 //Use reverse method to reverse the original string

 originalStr = originalStr.reverse();

 

 //return the string in reversed order

 return originalStr.toString();

}

//Method definition of main

public static void main(String[] args)

{

    //Declare variables

 String input, action;

 

 //Prompt the input message

 System.out.println("Welcome to the Text Converter.");

 System.out.println("Available Actions:");

 System.out.println("\t1337) convert to 1337-speak");

 System.out.println("\trev) reverse the string");

 System.out.print("Please enter a string: ");

   

 //Create a Scanner class's object

 Scanner scn = new Scanner(System.in);

 

 //Read input from the user

 input = scn.nextLine();

 do

 {

  /*Based on the action the user chooses, call the appropriate

   * action method. If an unrecognized action is entered then

   * the message "Unrecognized action." should be shown on a

   * line by itself and then the user is prompted again just

   * as they were when an action was performed.

   * */

  System.out.print("Action (1337, rev, quit): ");

  action = scn.nextLine();

  if (action.equals("1337"))

  {

   input = action1337(input);

   System.out.println(input);

  } else if (action.equals("rev"))

  {

   input = actionReverse(input);

   System.out.println(input);

  } else if (!action.equals("quit"))

  {

   System.out.println("Unrecognized action.");

  }

 } while (!action.equals("quit"));

 System.out.println("See you next time!");

 scn.close();

}

}

7 0
3 years ago
What does this work for
Anastaziya [24]

Answer:

it allows your dash board to light up you MPH RPM and all the other numbers on the spadomter

Explanat

8 0
3 years ago
Say you have a random, unordered list containing 4096 four-digit numbers. Describe the most efficient way to: sort the list and
Debora [2.8K]

Answer:

Answer explained below

Explanation:

It is given that numbers are four-digit so maximum value of a number in this list could be 9999.

So we need to sort a list of integers, where each integer lies between [0,9999].

For these given constraints we can use counting sort which will run in linear time i.e. O(n).

--------------------------------------------------------------------------------

Psuedo Code:

countSort(int numList[]) {

int count[10000];

count[i] = 0; for all i;

for(int num in numList){

count[num]+= 1;

}

return count;

}

--------------------------------------------------------------------------------

Searching in this count array will be just O(1).

E.g. Lets say we want to search if 3 was present in the original list.

Case 1: it was present in the original list:

Then the count[3] would have been incremented by our sorting algorithm. so in case element exists then count value of that element will be greater than 0.

Case 2: it was not present:

In this case count[3] will remain at 0. so in case element does not exist then count of that element will be 0.

So to search for an element, say x, we just need to check if count[x]>0.

So search is O(1).

Run times:

Sorting: O(n)

Search: O(1)

6 0
3 years ago
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