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KonstantinChe [14]
2 years ago
15

Draw an ERD for each of the following situations. (If you believe that you need to make additional assumptions, clearly state th

em for each situation.) Draw the same situation using the tool you have been told to use in the course. a. A company has a number of employees. The attributes of EMPLOYEE include Employee ID (identifier), Name, Address, and Birthdate. The company also has several projects. Attributes of PROJECT include Project ID (identifier), Project Name, and Start Date. Each employee may be assigned to one or more projects or may not be assigned to a project. A project must have at least one employee assigned and may have any number of employees assigned. An employee's billing rate may vary by project, and the company wishes to record the applicable billing rate (Billing Rate) for each employee when assigned to a particular project. Do the attribute names in this description follow the guidelines for naming attributes? If not, suggest better names. Do you have any associative entities on your ERD? If so, what are the identifiers for those associative entities? Does your ERD allow a project to be created before it has any employees assigned to it? Explain. How would you change your ERD if the Billing Rate could change in the middle of a project? b. A laboratory has several chemists who work on one or more projects. Chemists also may use certain kinds of equipment on each project. Attributes of CHEMIST include Employee ID (identifier), Name, and Phone No. Attributes of PROJECT include Project ID (identifier) and Start Date. Attributes of EQUIPMENT include Serial No and Cost The organization wishes to record Assign Date—that is, the date when a given equipment item was assigned to a particular chemist working on a specified project A chemist must be assigned to at least one project and one equipment item. A given equipment item need not be assigned, and a given project need not be assigned either a chemist or an equipment item. Provide good definitions for all of the relationships in this situation. c. A college course may have one or more scheduled sections or may not have a scheduled section. Attributes of COURSE include Course ID, Course Name, and Units. Attributes of SECTION include Section Number and Semester ID. Semester ID is composed of two parts: Semester and Year. Section Number is an integer (such as 1 or 2) that distinguishes one section from another for the same course but does not uniquely identify a section. How did you model SECTION? Why did you choose this way versus alternative ways to model SECTION? d. A hospital has a large number of registered physicians. Attributes of PHYSICIAN include Physician ID (the identifier) and Specialty. Patients are admitted to the hospital by physicians. Attributes of PATIENT include Patient ID (the identifier) and Patient Name. Any patient who is admitted must have exactly one admitting physician. A physician may optionally admit any number of patients. Once admitted, a given patient must be treated by at least one physician. A particular physician may treat any number of patients, or may not treat any patient& Whenever a patient is treated by a physician, the hospital wishes to record the details of the treatment (Treatment Detail). Components of Treatment Detail include Date, Time, and Results. Did you draw more than one relationship between physician and patient? Why or why not? Did you include hncnithi ac an antitv type? Why or why not?
Engineering
1 answer:
Julli [10]2 years ago
3 0

Answer:

it wqas red

Explanation:

red

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A gas metal arc welder is also known as a _____ welder.
dusya [7]

Answer:

Metal inert gas or Metal Active Gas

Explanation:

Gas Metal Arc Welder(GMAW) also is also termed as Metal Inert Gas(MIG) welder or Metal Active Gas (MAG) welder.

Gas Metal Arc welding is a process in which 2 metal pieces are melted through the use of electricity and joined together. A continuous wire is fed through the welding gun, which heats up the metal. This type of welding was introduced back in 1949 and was used for welding of Aluminium.

7 0
2 years ago
CHALLENGE ACTIVITY 2.8.1: Using constants in expressions. The cost to ship a package is a flat fee of 75 cents plus 25 cents per
mihalych1998 [28]

Answer:

Weight(lb): 10

Flat fee(cents): 75

Cents per pound: 25

Shipping cost(cents): 325

Explanation:

we run this as a jave programming language

import java.util.Scanner;

public class Shipping Calculator {

   public static void main (String [] args) {

       int shipWeightPounds = 10;

       int shipCostCents = 0;

       final int FLAT_FEE_CENTS = 75;

       

        final int CENTS_PER_POUND = 25;

       shipCostCents = FLAT_FEE_CENTS + CENTS_PER_POUND * shipWeightPound

       /* look up the solutioin above */

       System.out.println("Weight(lb): " + shipWeightPounds);

       System.out.println("Flat fee(cents): " + FLAT_FEE_CENTS);

       System.out.println("Cents per pound: " + CENTS_PER_POUND);

       System.out.println("Shipping cost(cents): " + shipCostCents);

   }

}

7 0
2 years ago
An optical mouse originally cost $31.85. Before it was removed from the store, it underwent the following changes in price. 27%
saul85 [17]

Answer:

b.

$43.21

Explanation:

8 0
2 years ago
Read 2 more answers
a. (24 points) Describe the microstructure present in a 10110 steel after each step in each of the following heat treatments (no
Mrac [35]

Answer:

Explanation:

Please check the below file for the attached file

6 0
2 years ago
Determine the angular acceleration of the uniform disk if (a) the rotational inertia of the disk is ignored and (b) the inertia
lukranit [14]

Answer:

α = 7.848 rad/s^2  ... Without disk inertia

α = 6.278 rad/s^2  .... With disk inertia

Explanation:

Given:-

- The mass of the disk, M = 5 kg

- The right hanging mass, mb = 4 kg

- The left hanging mass, ma = 6 kg

- The radius of the disk, r = 0.25 m

Find:-

Determine the angular acceleration of the uniform disk without and with considering the inertia of disk

Solution:-

- Assuming the inertia of the disk is negligible. The two masses ( A & B )  are hung over the disk in a pulley system. The disk is supported by a fixed support with hinge at the center of the disk.

- We will make a Free body diagram for each end of the rope/string ties to the masses A and B.

- The tension in the left and right string is considered to be ( T ).

- Apply newton's second law of motion for mass A and mass B.

                      ma*g - T = ma*a

                      T - mb*g = mb*a

Where,

* The tangential linear acceleration ( a ) with which the system of two masses assumed to be particles move with combined constant acceleration.

- g: The gravitational acceleration constant = 9.81 m/s^2

- Sum the two equations for both masses A and B:

                      g* ( ma - mb ) = ( ma + mb )*a

                      a =  g* ( ma - mb ) / ( ma + mb )

                      a = 9.81* ( 6 - 4 ) / ( 6 + 4 ) = 9.81 * ( 2 / 10 )

                      a = 1.962 m/s^2  

- The rope/string moves with linear acceleration of ( a ) which rotates the disk counter-clockwise in the direction of massive object A.

- The linear acceleration always acts tangent to the disk at a distance radius ( r ).

- For no slip conditions, the linear acceleration can be equated to tangential acceleration ( at ). The correlation between linear-rotational kinematics is given below :

                     a = at = 1.962 m/s^2

                     at = r*α      

Where,

           α: The angular acceleration of the object ( disk )

                    α = at / r

                    α = 1.962 / 0.25

                    α = 7.848 rad/s^2                                

- Take moments about the pivot O of the disk. Apply rotational dynamics conditions:

             

                Sum of moments ∑M = Iα

                 ( Ta - Tb )*r = Iα

- The moment about the pivots are due to masses A and B.

 

               Ta: The force in string due to mass A

               Tb: The force in string due to mass B

                I: The moment of inertia of disk = 0.5*M*r^2

                   ( ma*a - mb*a )*r = 0.5*M*r^2*α

                   α = ( ma*a - mb*a ) / ( 0.5*M*r )

                   α = ( 6*1.962 - 4*1.962 ) / ( 0.5*5*0.25 )

                   α = ( 3.924 ) / ( 0.625 )

                   α = 6.278 rad/s^2

6 0
2 years ago
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