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nasty-shy [4]
3 years ago
14

Ayuda porfavor es para una tarea de mi capacitación de desarrollo microempresarial

Engineering
1 answer:
vovangra [49]3 years ago
3 0

Answer:

I'm a beginner I would hv helped

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HELP FAST WILL MARK BRAINLIEST (for a real answer)
Nata [24]

Answer:

B. 180 million joules

Explanation:

Apply the formula for heat transfer given as;

Q=m*c*Δt  where

Q = electrical energy consumed by the heater in joules

m= mass of air in the chamber in kg

c= specific heat of air in joules per kg degrees Celsius

Δt= change in temperatures in degrees Celsius

Given in the question;

m= 1200 kg

c= 1000 J/°C /kg

Δt = 180°-30°= 150° C

Substitute values in the equation to get Q as;

Q=m*c*Δt

Q= 1200 * 1000* 150

Q= 180000000 joules

Q = 180 million joules

<u>The correct answer option is B : 180 million joules.</u>

7 0
3 years ago
A data record of all of someone's online activity is called a
aev [14]

Answer:

The answer is computer cookies.

Explanation:

Smartness :D

4 0
3 years ago
Read 2 more answers
True or False? In the electron flow theory, electrons flow from the positively charged body to the negatively charged body.
kobusy [5.1K]

I think it's false because the theory states that electrons flow from negative to positive. Since electrons are negatively charged, it follows that they are attracted by positively charged bodies and repelled by negatively charged bodies.

3 0
3 years ago
Read 2 more answers
A sign structure on the NJ Turnpike is to be designed to resist a wind force that produces a moment of 25 k-ft in one direction.
Jlenok [28]

Solution :

Finding the cohesion of the soil(c) using the relation:

$c = \frac{q_u}{2}$

Here, $q_u$ is the unconfined compression strength of the soil;

$c = \frac{800}{2}$

   = 400 psf

∴ The cohesion value is greater than 0

So the use of the angle of internal friction is 0

Referring to the table relation between bearing capacity factors and angle of internal friction.

For the angle of inter friction $0^\circ$

    $N_c = 5.14$

   $N_q = 1.0$

   $N_r = 0$

Therefore,

$q_{ult} = (400 \times 5.14 )+(110 \times 3 \times 1.0)+(0.5 \times 100 \times 13 \times 0)$

     =  2386 psf

∴ Allowable bearing capacity $q_{a} = \frac{Q_{allow}}{A}$

                                                     $=\frac{30}{B^2}$

∴ $q_a = \frac{q_{ult}}{F.O.S}$

  $\frac{30}{B^2} = \frac{2386}{3}$

∴ B = 0.2 ft

Therefore, the dimension of the square footing is 0.2 ft x 0.2 ft

                                                                                 $=0.04 \ ft^2$

7 0
3 years ago
Assign numMatches with the number of elements in userValues that equal matchValue. userValues has NUM_VALS elements. Ex: If user
Thepotemich [5.8K]

Answer:

import java.util.Scanner;

public class FindMatchValue {

  public static void main (String [] args) {

     Scanner scnr = new Scanner(System.in);

     final int NUM_VALS = 4;

     int[] userValues = new int[NUM_VALS];

     int i;

     int matchValue;

     int numMatches = -99; // Assign numMatches with 0 before your for loop

     matchValue = scnr.nextInt();

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

        userValues[i] = scnr.nextInt();

     }

     /* Your solution goes here */

         numMatches = 0;

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

        if(userValues[i] == matchValue) {

                       numMatches++;

                }

     }

     System.out.println("matchValue: " + matchValue + ", numMatches: " + numMatches);

  }

}

8 0
3 years ago
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