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Rufina [12.5K]
3 years ago
13

The section of the area to be examined is shown circumscribed by broken lines with circles at

Engineering
1 answer:
Aloiza [94]3 years ago
3 0

This question is about Circle Geometry. it evaluates connected and broken lines with respect to circles.

<h3>What is Circle Geometry?</h3>

This refers to the body of knowledge in mathematics that has to do with the various problems associated with the Circle.

In real-world scenarios, circle geometry is used in technologies involving:

  • Camera lenses
  • Circular Architectural structures
  • Steering Wheels
  • Buttons etc.

Learn more about Circle Geometry at:
brainly.com/question/24375372

You might be interested in
A device that helps increase field worker productivity by providing reliable location and time
frutty [35]

Global Positioning System (GPS) is the device that helps increase field worker productivity by providing reliable location and time.

<u>Explanation:</u>

GPS, a satellite based and radio navigation oriented system. It can be accessible from anywhere in the world irrespective of obstructions in weather and extremely used by Air force.

With advancing technologies, the uses of GPS can be extended to improve the productivity of the workforce by identifying location services and field operation insights.

Today, GPS chips are built in various devices including smartphones, tablet, and other gadgets. It doe not need users to send data as it can work on internet reception.

3 0
3 years ago
1) Find the time in seconds to reach full charge in an RL circuit with L = 5 H and R = 100 ohms
pav-90 [236]

The time constant to reach full charge in an RL circuit is 0.05 ms.

Explanation:

To find the time constant,

The time constant for an RL circuit is defined by τ = L/R.

The given data is

L= 5 H

R= 100 ohms

by using the formula,

τ = L/R

  = 5/100

  = 0.05 ms

τ = 0.05 ms

Thus, the time constant to reach full charge in an RL circuit is 0.05 ms.

8 0
3 years ago
On Beverly's last project, the team identified only a few lessons learned. Which approach to lessons learned
Afina-wow [57]

Option C (making lessons learned a regular part of meetings) is the correct approach.

  • As nothing more than a general rule, typically construction companies only plan lessons that have been learned exercises or initiatives towards the end of a particular endeavor or segment.
  • As almost a result of the team knowing, valuable lessons are intended to increase the comprehensive implementation of quality management practices as well as deadlines.

Aside from this, none of the choices are viable methods to learning lessons or gaining knowledge from the past. As a result, the methodology outlined above is the appropriate one.

Learn more about project teamwork here:

brainly.com/question/14279121

7 0
3 years ago
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
Two vertical, parallel clean glass plates are spaced a distance of 2mm apart. if the plates are placed in water, how high will t
Ulleksa [173]

Answer with Explanation:

The capillary rise in 2 parallel plates immersed in a liquid is given by the formula

h=\frac{2\sigma cos(\alpha )}{\rho gd}

where

\sigma is the surface tension of the liquid

\alpha is the contact angle of the liquid

\rho is density of liquid

'g' is acceleratioj due to gravity

'd' is seperation between thje plates

Part a) When the liquid is water:

For water and glass we have

\sigma =7.28\times 10^{-2}N/m

\alpha =0

\rho _{w}=1000kg/m^3

Applying the values we get

h=\frac{2\times 7.28\times 10^{-2}cos(0)}{1000\times 9.81\times 2\times 10^{-3}}=7.39mm

Part b) When the liquid is mercury:

For mercury and glass we have

\sigma =485.5\times 10^{-3}N/m

\alpha =138^o

\rho _{w}=13.6\times 10^{3}kg/m^3

Applying the values we get

h=\frac{2\times 485.5\times 10^{-3}cos(138)}{13.6\times 1000\times 9.81\times 2\times 10^{-3}}=-2.704mm

The negative sign indicates that there is depression in mercury in the tube.

4 0
3 years ago
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