How are scientific discoveries used in engineering design?

Assume the availability of an existing class, ICalculator, that models an integer arithmetic calculator and contains: an instanc

shtirl [24]

We connect with computers through coding, often known as computer programming.

We connect with computers through coding, often understood as computer programming.
Coding exists similar to writing a set of instructions because it instructs a machine what to do.
You can instruct computers what to do or how to behave much more quickly by learning to write code.

class ICalculator {

int currentValue;

int add(int value) {

this.currentValue = currentValue + value;

return currentValue;

}

int sub(int value) {

this.currentValue = currentValue - value;

return currentValue;

}

int mul(int value) {

this.currentValue = currentValue * value;

return currentValue;

}

int div(int value) {

this.currentValue = currentValue / value;

return currentValue;

}

public class ICalculator2 extends ICalculator {

int negate() {

if (currentValue != 0)

this.currentValue = -currentValue;

return currentValue;

}

public static void main(String[] args) {

ICalculator2 ic = new ICalculator2();

ic.currentValue=5;

System.out.println(ic.add(2));

System.out.println(ic.sub(5));

System.out.println(ic.mul(3));

System.out.println(ic.div(3));

System.out.println(ic.negate());

}

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3 0

1 year ago

At a retirement party, a coworker described terry as dedicated

denis23 [38]

Answer:

At a retirement party, a coworker described Terry as dedicated, hardworking, and dependable. He also said that Terry was a great leader, knew the computer system, and kept the company's finances in order

8 0

3 years ago

Is santa real or nah is santa real or nah

Elena L [17]

Answer:

nah

Explanation:

3 0

2 years ago

Read 2 more answers

2. There are three drawings that architects and designers use to indicate spaces. What are these drawing?

Zarrin [17]

Answer:

Architectural plans.

Explanation:

An architectural plan is called the drawings made by architects, civil engineers or designers of spaces or interiors, in which these professionals capture their building projects, organizing the distribution of the spaces to be used, the elements to be located in them and, fundamentally, to give construction planning a projection into reality. Thus, the plans help professionals to have a better understanding of the expected end result of the projects they are carrying out.

3 0

3 years ago

A charge of +2.00 μC is at the origin and a charge of –3.00 μC is on the y axis at y = 40.0 cm . (a) What is the potential at po

Nimfa-mama [501]

a) Potential in A: -2700 V

b) Potential difference: -26,800 V

c) Work: $4.3\cdot 10^{-15} J$

Explanation:

a)

The electric potential at a distance r from a single-point charge is given by:

$V(r)=\frac{kq}{r}$

where

$k=8.99\cdot 10^9 Nm^{-2}C^{-2}$ is the Coulomb's constant

q is the charge

r is the distance from the charge

In this problem, we have a system of two charges, so the total potential at a certain point will be given by the algebraic sum of the two potentials.

Charge 1 is

$q_1=+2.00\mu C=+2.00\cdot 10^{-6}C$

and is located at the origin (x=0, y=0)

Charge 2 is

$q_2=-3.00 \mu C=-3.00\cdot 10^{-6}C$

and is located at (x=0, y = 0.40 m)

Point A is located at (x = 0.40 m, y = 0)

The distance of point A from charge 1 is

$r_{1A}=0.40 m$

So the potential due to charge 2 is

$V_1=\frac{(8.99\cdot 10^9)(+2.00\cdot 10^{-6})}{0.40}=+4.50\cdot 10^4 V$

The distance of point A from charge 2 is

$r_{2A}=\sqrt{0.40^2+0.40^2}=0.566 m$

So the potential due to charge 1 is

$V_2=\frac{(8.99\cdot 10^9)(-3.00\cdot 10^{-6})}{0.566}=-4.77\cdot 10^4 V$

Therefore, the net potential at point A is

$V_A=V_1+V_2=+4.50\cdot 10^4 - 4.77\cdot 10^4=-2700 V$

b)

Here we have to calculate the net potential at point B, located at

(x = 0.40 m, y = 0.30 m)

The distance of charge 1 from point B is

$r_{1B}=\sqrt{(0.40)^2+(0.30)^2}=0.50 m$

So the potential due to charge 1 at point B is

$V_1=\frac{(8.99\cdot 10^9)(+2.00\cdot 10^{-6})}{0.50}=+3.60\cdot 10^4 V$

The distance of charge 2 from point B is

$r_{2B}=\sqrt{(0.40)^2+(0.40-0.30)^2}=0.412 m$

So the potential due to charge 2 at point B is

$V_2=\frac{(8.99\cdot 10^9)(-3.00\cdot 10^{-6})}{0.412}=-6.55\cdot 10^4 V$

Therefore, the net potential at point B is

$V_B=V_1+V_2=+3.60\cdot 10^4 -6.55\cdot 10^4 = -29,500 V$

So the potential difference is

$V_B-V_A=-29,500 V-(-2700 V)=-26,800 V$

c)

The work required to move a charged particle across a potential difference is equal to its change of electric potential energy, and it is given by

$W=q\Delta V$

where

q is the charge of the particle

$\Delta V$ is the potential difference

In this problem, we have:

$q=-1.6\cdot 10^{-19}C$ is the charge of the electron

$\Delta V=-26,800 V$ is the potential difference

Therefore, the work required on the electron is

$W=(-1.6\cdot 10^{-19})(-26,800)=4.3\cdot 10^{-15} J$

4 0

3 years ago