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Tema [17]
3 years ago
13

How do i do this? if y’all don’t mind helping lol

Engineering
1 answer:
Nookie1986 [14]3 years ago
7 0
Get a ruler and measure it and for every 1/8 of an inch you multiply by 1 to get your footage
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In a slowly cooled hypereutectoid iron-carbon steel, the pearlite colonies are
jek_recluse [69]

In a slowly cooled hypereutectoid iron-carbon steel, the pearlite colonies are normally separated from each other by a more or less continuous boundary layer of cementite done by Slower cooling reasons coarse Pearlite, even as rapid cooling reasons first-rate pearlite to form.

<h3>What levels is in Hypereutectoid metal?</h3>

Hypoeutectoid steels can, upon preliminary cooling from the austenite single segment field, exist as extraordinary levels, eutectoid ferrite and austenite, every with extraordinary carbon contents.

At room temperature, hypereutectoid steels have a pearlitic primary microstructure (ferrite grains with embedded cementite lamellae) with moreover induced cementite on the grain boundaries! The micrograph under suggests a hypereutectoid metal with 1.0 Carbon (C100).

Read more about the cementite:

brainly.com/question/24924853

#SPJ1

7 0
2 years ago
Which of the following explains the difference between rangeland management specialists and conservation biologists?
tekilochka [14]

Answer:

b

Explanation:

8 0
3 years ago
Read 2 more answers
This program will store roster and rating information for a soccer team. Coaches rate players during tryouts to ensure a balance
Flauer [41]

Answer:

#include <iostream>

#include <vector>

using namespace std;

int main() {

   vector<int> jerseyNumber;

   vector<int> rating;

   int temp;

   for (int i = 1; i <= 5; i++) {

       cout << "Enter player " << i

            << "'s jersey number: ";

       cin >> temp;

       jerseyNumber.push_back(temp);

       cout << "Enter player " << i

            << "'s rating: ";

       cin >> temp;

       rating.push_back(temp);

       cout << endl;

   }

   cout << "ROSTER" << endl;

   for (int i = 0; i < 5; i++)

       cout << "Player " << i + 1 << " -- "

            << "Jersey number: " << jerseyNumber.at(i)

            << ", Rating: " << rating.at(i) << endl;

   char option;

   '

   while (true) {

       cout << "MENU" << endl;

       cout << "a - Add player" << endl;

       cout << "d - Remove player" << endl;

       cout << "u - Update player rating" << endl;

       cout << "r - Output players above a rating"

            << endl;

       cout << "o - Output roster" << endl;

       cout << "q - Quit" << endl << endl;

       cout << "Choose an option: ";

       cin >> option;

       switch (option) {

           case 'a':

           case 'A':

               cout << "Enter a new player's"

                    << "jersey number: ";

               cin >> temp;

               jerseyNumber.push_back(temp);

               cout << "Enter the player's rating: ";

               cin >> temp;

               rating.push_back(temp);

               break;

           case 'd':

           case 'D':

               cout << "Enter a jersey number: ";

               cin >> temp;

               int i;

               for (i = 0; i < jerseyNumber.size();

                    i++) {

                   if (jerseyNumber.at(i) == temp) {

                       jerseyNumber.erase(

                               jerseyNumber.begin() + i);

                       rating.erase(rating.begin() + i);

                       break;

                   }

               }

               break;

           case 'u':

           case 'U':

               cout << "Enter a jersey number: ";

               cin >> temp;

               for (int i = 0; i < jerseyNumber.size();

                    i++) {

                   if (jerseyNumber.at(i) == temp) {

                       cout << "Enter a new rating "

                            << "for player: ";

                       cin >> temp;

                       rating.at(i) = temp;

                       break;

                   }

               }

               break;

           case 'r':

           case 'R':

               cout << "Enter a rating: ";

               cin >> temp;

               cout << "\nABOVE " << temp << endl;

               for (int i = 0; i < jerseyNumber.size();

                    i++)

                   if (rating.at(i) > temp)

                       cout << "Player " << i + 1

                            << " -- "

                            << "Jersey number: "

                            << jerseyNumber.at(i)

                            << ", Rating: "

                            << rating.at(i) << endl;

               break;

           case 'o':

           case 'O':

               cout << "ROSTER" << endl;

               for (int i = 0; i < jerseyNumber.size();

                    i++)

                   cout << "Player " << i + 1 << " -- "

                        << "Jersey number: "

                        << jerseyNumber.at(i) << ", Rating: "

                        << rating.at(i) << endl;

               break;

           case 'q':

               return 0;

           default:

               cout << "Invalid menu option."

                    << " Try again." << endl;

       }

   }

}

Explanation:

4 0
3 years ago
A glass plate is subjected to a tensile stress of 40 MPa. If the specific surface energy is 0.3 J/m2 and the modulus of elastici
Pavlova-9 [17]

Answer:

8.24μm

Explanation:

The theory of brittle fracture was used to solve this problem.

And if you follow through with the attachment made a the subject of the formula

Such that,

a = 2x(69x10⁹)x0.3/pi(40x10⁶)²

= 4.14x10¹⁰/5.024x10¹⁵

= 8.24x10^-06

= 8.24μm

This is the the maximum length of the surface flaw

4 0
3 years ago
An AC power generator produces 50 A (rms) at 3600 V. The voltage is stepped up to 100 000 V by an ideal transformer and the ener
RSB [31]

Given:

I_{rms} = 50 A

voltage, V = 3600V

step-up voltage, V' = 100000 V

Resistance of line, R = 100\ohm

Solution:

To calculate % heat loss in long distance power line:

Power produced by AC generator, P = 50\times 3600 W

P = 180000 W = 180 kW

At step-up voltage, V = 100000V or 100 kV

current, I = \frac{P}{V'}

I = \frac{1800000}{100000}

I = 1.8 A

Power line voltage drop is given by:

V_{drop} = I\times R

V_{drop} = 1.8\times 100

V_{drop} = 180 V

Power dissipated in long transmission line P_{dissipated} = V_{drop}\times I

Power dissipated in long transmission line P_{dissipated} = 180\times 1.8 = 324 W

% Heat loss in power line, P_{loss} = \frac{P_{dissipated}}{P}\times 100

% Heat loss in power line, P_{loss} = \frac{324}{180000}\times 100

P_{loss} = 0.18%

 

5 0
3 years ago
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