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Alina [70]
3 years ago
7

Meggers are used to measure ​

Engineering
1 answer:
svp [43]3 years ago
8 0
The Megger test is a method of testing making use of an insulation tester resistance meter that will help to verify the condition of electrical insulation. Insulation resistance quality of an electrical system degrades with time, environment condition i.e. temperature, humidity, moisture and dust particl.
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During engine operation, what force causes a cylinder head to pull tighter against the head gasket?
Svetlanka [38]

Answer:

D. Intake Stroke vacuum

Explanation:

5 0
3 years ago
The distillation column in Figure 3 is set up for so-called boil-up (V) control. It has
11111nata11111 [884]

A distillation column is an essential item used in the distillation of liquid mixtures to separate the mixture into its component parts, or fractions, based on the differences in volatilities. Fractionating columns are used in small scale laboratory distillations as well as large scale industrial distillations.

3 0
3 years ago
Base course aggregate has a target density of 121.8 lb/ft3 in place It will be laid down and compacted in a rectangular street r
irakobra [83]

Answer:

total weight of the aggregate = 3594878.28 lbs

Explanation:

given data

density = 121.8 lb/ft³

area =  1000 ft x 60 ft x 6 in = 1000 ft x 60 ft x 0.5 ft

moisture = 3.5 %

compaction = 95%

solution

we get here first volume of the space that is filled with the aggregate that is

volume = 1000 ft x 60 ft x 0.5 ft  = 30,000 cu ft

now we get fill space with aggregate that compact to 95% of dry density.

so we fill space with aggregate of density that is = 95% of 121.8

= 115.71 lb/ cu ft

so now dry weight of aggregate is

dry weight of aggregate = 30,000  × 115.71 = 3471300 lb

when we assume that moisture percentage is by weight

then weight of the moisture in aggregate will be

weight of the moisture in aggregate = 3.56 % of 3471300 lb  

weight of the moisture in aggregate = 123578.28 lbs

and

we get total weight of the aggregate to fill space that is

total weight of the aggregate = 3471300 lb +123578.28 lb

total weight of the aggregate = 3594878.28 lbs

4 0
3 years ago
Multiply each element in origList with the corresponding value in offsetAmount. Print each product followed by a space.Ex: If or
n200080 [17]

Answer:

Here is the JAVA program:  

import java.util.Scanner; // to take input from user

public class VectorElementOperations {

public static void main(String[] args) {

final int NUM_VALS = 4; // size is fixed that is 4 and assigned to NUM_VALS

int[] origList = new int[NUM_VALS];

int[] offsetAmount = new int[NUM_VALS];

int i;

//two lists origList[] and offsetAmount[] are assigned values

origList[0] = 20;

origList[1] = 30;

origList[2] = 40;

origList[3] = 50;

offsetAmount[0] = 4;

offsetAmount[1] = 6;

offsetAmount[2] = 2;

offsetAmount[3] = 8;

String product=""; // product variable to store the product of 2 lists

for(i = 0; i <= origList.length - 1; i++){

/* loop starts with i at 0th position or index and ends when the end of the origList is reached */

/* multiples each element of origList to corresponding element of offsetAmount and stores result in the form of character string in product*/

   product+= Integer.toString(origList[i] *= offsetAmount[i]) + " ";  }

 System.out.println(product); }}   //displays the product of both lists

Output:

80 180 80 400

Explanation:

If you want to print the product of origList with corresponding value in offsetAmount in vertical form you can do this in the following way:

import java.util.Scanner;

public class VectorElementOperations {

public static void main(String[] args) {

final int NUM_VALS = 4;

int[] origList = new int[NUM_VALS];

int[] offsetAmount = new int[NUM_VALS];

int i;

origList[0] = 20;

origList[1] = 30;

origList[2] = 40;

origList[3] = 50;

offsetAmount[0] = 4;

offsetAmount[1] = 6;

offsetAmount[2] = 2;

offsetAmount[3] = 8;

for(i = 0; i <= origList.length - 1; i++){

 origList[i] *= offsetAmount[i];

System.out.println(origList[i]); } }}

Output:

80                                                                                                                            

180                                                                                                                          

80                                                                                                                            

400

The program along with the output is attached as screenshot with the input given in the example.

8 0
3 years ago
Engine oil flows through a 25‐mm‐diameter tube at a rate of 0.5 kg/s. The oil enters the tube at a temperature of 25°C, while th
Elodia [21]

Answer:

a) the log mean temperature difference (Approx. 64.5 deg C)

b) the rate of heat addition into the oil.

The above have been solved for in the below workings

Explanation:

3 0
4 years ago
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