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Ksenya-84 [330]
3 years ago
12

Consider that a system has two entities, Students, Instructors and Course. The Student has the following properties: student nam

e, ID, and GPA.The instructor has the following properties: name, ID, Salary and department name. Finally the Course has Name, ID , Instructor and array of 3 students.
Implement the above system taking into account the following requirements:

Define array of 4 Courses with instantiating all its properties.[5 points]
Assume that instructor1 and instructor2 are two objects of type Instructor, write a code that enables you to write: if (instructor1.isequal (instructor2)). The statement is true if the two objects have the same name and ID. [3 points]
Define a function with the following signature void PrintCSalary(Course []), The function prints the index and the name of the course that has Instructor with Max Salary. [3 points]
Define a function with the following signature void PrintMaxGPA(Course []), The function prints the index and the name of the course that has Max average of its Students GPA.[3 points]
Define a function that has the following signature Instructor * getInstByID(Course [], int id). The function returns a pointer to one of the instructors in the array of courses, who has the same id (sent to the function).[3 points]
Write a getCourseList(int ID) function that returns a list of courses that the student registered in.[3 points]
Engineering
1 answer:
tekilochka [14]3 years ago
5 0

Answer:

There's no answer ?

Explanation:

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2 years ago
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At a point on the free surface of a stressed body, the normal stresses are 20 ksi (T) on a vertical plane and 30 ksi (C) on a ho
victus00 [196]

Answer:

The principal stresses are σp1 = 27 ksi, σp2 = -37 ksi and the shear stress is zero

Explanation:

The expression for the maximum shear stress is given:

\tau _{M} =\sqrt{(\frac{\sigma _{x}^{2}-\sigma _{y}^{2}  }{2})^{2}+\tau _{xy}^{2}    }

Where

σx = stress in vertical plane = 20 ksi

σy = stress in horizontal plane = -30 ksi

τM = 32 ksi

Replacing:

32=\sqrt{(\frac{20-(-30)}{2} )^{2} +\tau _{xy}^{2}  }

Solving for τxy:

τxy = ±19.98 ksi

The principal stress is:

\sigma _{x}+\sigma _{y} =\sigma _{p1}+\sigma _{p2}

Where

σp1 = 20 ksi

σp2 = -30 ksi

\sigma _{p1}  +\sigma _{p2}=-10 ksi (equation 1)

\tau _{M} =\frac{\sigma _{p1}-\sigma _{p2}}{2} \\\sigma _{p1}-\sigma _{p2}=2\tau _{M}\\\sigma _{p1}-\sigma _{p2}=32*2=64ksi equation 2

Solving both equations:

σp1 = 27 ksi

σp2 = -37 ksi

The shear stress on the vertical plane is zero

4 0
3 years ago
Identify the right components for gsm architecture that consists of the hardware or physical equipment such as digital signal pr
sergiy2304 [10]

The right components for gsm architecture that consists of the hardware or physical equipment such as digital signal processors, radio transceiver, display, battery, case and sim card is the Mobile station.

<h3>What are the 4 main components?</h3>

In GSM, a cell station includes 4 fundamental additives: Mobile termination (MT) - gives not unusualplace features consisting of: radio transmission and handover, speech encoding and decoding, blunders detection and correction, signaling and get right of entry to to the SIM. The IMEI code is connected to the MT.

Under the GSM framework, a cell tele cell smartphone is called a Mobile Station and is partitioned into  wonderful additives: the Subscriber Identity Module (SIM) and the Mobile Equipment (ME).

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6 0
2 years ago
Investigating how slime molds reproduce is an example of applied research.<br> True<br> False
azamat

Answer:

false

Explanation:

3 0
3 years ago
A 4-L pressure cooker has an operating pressure of 175 kPa. Initially, one-half of the volume is filled with liquid and the othe
vodomira [7]

Answer:

the highest rate of heat transfer allowed is 0.9306 kW

Explanation:

Given the data in the question;

Volume = 4L = 0.004 m³

V_f = V_g = 0.002 m³

Using Table ( saturated water - pressure table);

at pressure p = 175 kPa;

v_f = 0.001057 m³/kg

v_g = 1.0037 m³/kg

u_f = 486.82 kJ/kg

u_g 2524.5 kJ/kg

h_g = 2700.2 kJ/kg

So the initial mass of the water;

m₁ = V_f/v_f + V_g/v_g

we substitute

m₁ = 0.002/0.001057  + 0.002/1.0037

m₁ = 1.89414 kg

Now, the final mass will be;

m₂ = V/v_g

m₂ = 0.004 / 1.0037

m₂ = 0.003985 kg

Now, mass leaving the pressure cooker is;

m_{out = m₁ - m₂

m_{out = 1.89414  - 0.003985

m_{out = 1.890155 kg

so, Initial internal energy will be;

U₁ = m_fu_f + m_gu_g

U₁ = (V_f/v_f)u_f  + (V_g/v_g)u_g

we substitute

U₁ = (0.002/0.001057)(486.82)  + (0.002/1.0037)(2524.5)

U₁ = 921.135288 + 5.030387

U₁ = 926.165675 kJ

Now, using Energy balance;

E_{in -  E_{out = ΔE_{sys

QΔt - m_{outh_{out = m₂u₂ - U₁

QΔt - m_{outh_g = m₂u_g - U₁

given that time = 75 min = 75 × 60s = 4500 sec

so we substitute

Q(4500) - ( 1.890155 × 2700.2 ) = ( 0.003985 × 2524.5 ) - 926.165675

Q(4500) - 5103.7965 = 10.06013 - 926.165675

Q(4500) = 10.06013 - 926.165675 + 5103.7965

Q(4500) = 4187.690955

Q = 4187.690955 / 4500

Q = 0.9306 kW

Therefore, the highest rate of heat transfer allowed is 0.9306 kW

5 0
3 years ago
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