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Keith_Richards [23]
2 years ago
8

Technician A states that air tools generally produce more noise than electric tools, so wear ear protection when using air tools

. Technician B states that you should always use impact sockets with impact guns. Who is correct?
Engineering
1 answer:
dusya [7]2 years ago
5 0

Answer:

Both Technician A and Technician B are correct

Explanation:

Air tools and electric tools are both power tools as they are used to make work easier. Air tools generally use an air compressor that powers the motor of the tool making it possible to use it while electric tools as the name implies are powered by an electric source which in this case is batteries. An example of an air tool is the nail gun which can be used by furniture makers to drive nails and they are often louder than electric tools because of vibrations caused by the compressor making it necessary to use ear protection when using the tool for ear safety.

Technician B  is also correct because it is always advisable to use impact sockets while using impact guns due to the ability of the impact sockets to withstand the force caused by operating impact guns and make work neater when nuts and bolts are being loosened or tightened.

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A storm sewer is carrying snow melt containing 1.2 g/L of sodium chloride into a small stream. The stream has a naturally occurr
galina1969 [7]

Answer:

Given Data:

concentration of sewer Csewer = 1.2 g/L

converting into mg/L = Csewer = 1.2 g/L x 1000 mg/g = 1200 mg/L

flow rate of sewer Qsewer = 2000 L/min

concentration of sewer Cstream = 20 mg/L

flow rate of sewer Qstream = 2m3/s

converting Q into L/min = 2m3/s x 1000 x 60 = 120000 L/min

mass diagram is

6 0
3 years ago
According to the decreasing order of toughness. list the following materials (note: the steels are assumed to have no cold work
fiasKO [112]

Answer:

1090 Steel >1040 Steel > Pure aluminium >Diamond.

Explanation:

Toughness:

  Toughness can be define as the are of load -deflection diagram up to fracture point.

Modulus of toughness can be defines as the area of stress-strain diagram up to fracture point.Modulus of toughness is the property of material.

So the decreasing order of toughness can be given as follows

1090 Steel >1040 Steel > Pure aluminium >Diamond.

8 0
3 years ago
The angle of twist can be computed using the material’s shear modulus if and only if: (a)- The shear stress is still in the elas
ollegr [7]

Answer:

The angle of twist can be computed using the material’s shear modulus if and only if the shear stress is still in the elastic region

Explanation:

The shear modulus (G) is the ratio of shear stress to shear strain. Like the modulus of elasticity, the shear modulus is governed by Hooke’s Law: the relationship between shear stress and shear strain is proportional up to the proportional limit of the material. The angle of twist can be computed using the material’s shear modulus if and only if the shear stress is still in the elastic region.

3 0
3 years ago
In part A you are asked to write the pseudocode for the program. In part B you are asked to write the syntax of the code for the
Naya [18.7K]

Answer:

C++.

Explanation:

#include <iostream>

#include <string>

using namespace std;

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

int main() {

   string quote, book;

   int page;

   

   cout<<"What is your favorite quote from a book?"<<endl;

   getline(cin, quote);

   cout<<endl;

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

   cout<<"What book was that quote from?"<<endl;

   getline(cin, book);

   cout<<endl;

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

   cout<<"What page was that quote from?"<<endl;

   cin>>page;

   cout<<endl;

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

   int no_of_upper_characters = 0;

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

       if (isupper(quote[i]))

          no_of_upper_characters++;

   }

   

   cout<<"No. of upper case characters: "<<no_of_upper_characters<<endl;

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

   int no_of_characters = quote.length();

   cout<<"No. of characters: "<<no_of_characters<<endl;

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

   bool isDog = false;

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

       if (isDog == true)

           break;

       else if (quote[i] == 'd') {

           for (int j=i+1; j<quote.length(); j++) {

               if (isDog == true)

                   break;

               else if (quote[j] == 'o') {

                   for (int z=j+1; z<quote.length(); z++) {

                       if (quote[z] == 'g') {

                           isDog = true;

                           break;

                       }

                   }

               }

           }

       }

   }

   

   if (isDog == true)

       cout<<"This includes 'd' 'o' 'g' in the quote";

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

   return 0;

}

3 0
3 years ago
Problem definition
LekaFEV [45]

Answer:

ummm thats alot

Explanation:

8 0
3 years ago
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