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Bezzdna [24]
2 years ago
8

In a parallel circuit, as more resistances are added, what happens to the total circuit current?

Engineering
1 answer:
Mazyrski [523]2 years ago
5 0

Answer:

the equivalent resistance of the circuit decreases and the total current of the circuit increases. 

Explanation:

hope this helps

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2. One of the many methods used for drying air is to cool the air below the dew point so that condensation or freezing of the mo
REY [17]

Answer:

0.5°c

Explanation:

Humidity ratio by mass can be expressed as

the ratio between the actual mass of water vapor present in moist air - to the mass of the dry air

Humidity ratio is normally expressed in kilograms (or pounds) of water vapor per kilogram (or pound) of dry air.

Humidity ratio expressed by mass:

x = mw / ma                                  (1)

where

x = humidity ratio (kgwater/kgdry_air, lbwater/lbdry_air)

mw = mass of water vapor (kg, lb)

ma = mass of dry air (kg, lb)

It can be as:

x = 0.005 (100) / [(100 - 100)]

x = 0.005 x 100 / (100 - 100)

x = 0.005 x 100 / 0

x = 0.5°c

So the temperature to which atmospheric air must be cooled in order to have humidity ratio of 0.005 lb/lb is 0.5°c

6 0
3 years ago
Can a 1½ " conduit, with a total area of 2.04 square inches, be filled with wires that total 0.93 square inches if the maximum f
Papessa [141]

Answer:

it is not possible to place the wires in the condui

Explanation:

given data

total area = 2.04 square inches

wires total area = 0.93 square inches

maximum fill conduit =  40%

to find out

Can it is possible place wire in conduit conduit

solution

we know maximum fill is 40%

so here first we get total area of conduit that will be

total area of conduit = 40% × 2.04

total area of conduit = 0.816 square inches

but this area is less than required area of wire that is 0.93 square inches

so we can say it is not possible to place the wires in the conduit

4 0
3 years ago
2.4: Add a method called setValue(), and the description of setValue is: public int setValue(long searchKey) In this method, the
Yanka [14]

Answer:

Below is java code that must be used for the given question:

// highArray.java

// demonstrates array class with high-level interface

// to run this program: C>java HighArrayApp

////////////////////////////////////////////////////////////////

class HighArray

  {

  private long[] a;                 // ref to array a

  private int nElems;               // number of data items

  //-----------------------------------------------------------

  public HighArray(int max)         // constructor

     {

     a = new long[max];                 // create the array

     nElems = 0;                        // no items yet

     }

  //-----------------------------------------------------------

  public setValue find(long searchKey)

     {                              // find specified value

     int j;

     for(j=0; j<nElems; j++)            // for each element,

        if(a[j] == searchKey)           // found item?

           break;                       // exit loop before end

     if(j == nElems)                    // gone to end?

        return false;                   // yes, can't find it

     else

        return true;                    // no, found it

     }  // end find()

  //-----------------------------------------------------------

  public void insert(long value)    // put element into array

     {

     a[nElems] = value;             // insert it

     nElems++;                      // increment size

     }

  //-----------------------------------------------------------

  public void display()             // displays array contents

     {

     for(int j=0; j<nElems; j++)       // for each element,

        System.out.print(a[j] + " ");  // display it

     System.out.println("");

     }

  //-----------------------------------------------------------

  }  // end class HighArray

////////////////////////////////////////////////////////////////

class HighArrayApp

  {

  public static void main(String[] args)

     {

     int maxSize = 100;            // array size

     HighArray arr;                // reference to array

     arr = new HighArray(maxSize); // create the array

     arr.insert(77);               // insert 10 items

     arr.insert(99);

     arr.insert(44);

     arr.insert(55);

     arr.insert(22);

     arr.insert(88);

     arr.insert(11);

     arr.insert(00);

     arr.insert(66);

     arr.insert(33);

     arr.display();                // display items

     int searchKey = 35;           // search for item

     if( arr.find(searchKey) )

        System.out.println("Found " + searchKey);

     else

        System.out.println("Can't find " + searchKey);

     }  // end main()

  }  // end class HighArrayApp

Explanation:

6 0
3 years ago
The mathematical relationship between
11Alexandr11 [23.1K]
4) Ohms law thats the answer
7 0
3 years ago
The size of Carvins Cove water reservoir is 3.2 billion gallons. Approximately, 11 cfs of water is continuous withdrawn from thi
Zolol [24]

Answer:

471 days

Explanation:

Capacity of Carvins Cove water reservoir = 3.2 billion gallons i.e. 3.2 x 10˄9 gallons

As,  

1 gallon = 0.133 cubic feet (cf)

Therefore,  

Capacity of Carvins Cove water reservoir in cf  = 3.2 x 10˄9 x 0.133

                                                                         = 4.28 x 10˄8

 

Applying Mass balance i.e

Accumulation = Mass In - Mass out   (Eq. 01)

Here  

Mass In = 0.5 cfs

Mass out = 11 cfs

Putting values in (Eq. 01)

Accumulation  = 0.5 - 11

                         = - 10.5 cfs

 

Negative accumulation shows that reservoir is depleting i.e. at a rate of 10.5 cubic feet per second.

Converting depletion of reservoir in cubic feet per hour = 10.5 x 3600

                                                                                       = 37,800

 

Converting depletion of reservoir in cubic feet per day = 37, 800 x 24

                                                                                         = 907,200  

 

i.e. 907,200 cubic feet volume is being depleted in days = 1 day

1 cubic feet volume is being depleted in days = 1/907,200 day

4.28 x 10˄8 cubic feet volume will deplete in days  = (4.28 x 10˄8) x                    1/907,200

                                                                                 = 471 Days.

 

Hence in case of continuous drought reservoir will last for 471 days before dry-up.

8 0
3 years ago
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