In creating C++ applications, you have the ability to utilize various formatting functions in the iostream library. What are some of the formatting vulnerabilities that can be encountered in using the iostream library in C++?

Answer:

The maximum load the rod can take before it starts to permanently elongate =  = 6086.84 lbf or 27075.59 N

The maximum energy that the  rod would store =2.536 ft·lbf or 3.439 J

Explanation:

We list out the variables to the question as follows

Rod diameter = 0.5 in

Length of rod = 5 in

Young's modulus of elasticity for the material = 15.5 Msi

The maximum load rthe rd can take before it starts to permanently elongate is the yield stress and it can vbe calclulated by applyng the 0.2% offset rule by taking the yeild strain to be equal to 2%

Thus Young's Modulus at yield point =[tex]\frac{Stress}{Strain}[/tex] so that the yield stess is

Yield stress = strain × Young's Modulus

= 2/100 × E = 0.002 × 15.5 Msi = 0.031 Msi = 31 ksi

The maximum load that the rod would take before it starts to permanently elongate = F = Stress×Area

where area = π·D²/4 = 0.196 in²

F = 31 kSi × 0.196 in² = 6086.84 lbf or 27075.59 N

To calculate the strain energy stored in the rod we apply the strain energy formula thus

U = V×σ²/2·E

To calculate the volume we have V = L ×π·D²/4 =5 in × 0.196 in² = 0.98 in³

then U = 0.98 × (31000)²/(2·15.5×10⁶)

= 30.43 in³·psi = 2.536 ft·lbf

or 3.439 J

The maximum energy that the rod would store = 2.536 ft·lbf

Answer:

Please find detail answer given below

Explanation:

(a) Stiffness

It is defined as the property of a material that is rigid and difficult to bend. The example of stiffness is the rubber band. If an elastic band is stretched with two fingers, the stiffness is less and the flexibility is greater. Similarly, if we use the rubber band assembly and stretch it with two fingers, the stiffness will be more rigid and the flexibility will be less.

(b) Strength

Force measures how much stress can be applied to an element before it is permanently deformed or fractured.

(c) ductility

Ductility is a measure of a metal's ability to resist tensile stress: any force that separates the two ends of an object. The tug of war game provides a good example of the tension of tension that is applied to a rope. Ductility is the plastic deformation that occurs in the metal as a result of such types of deformation. The term "ductile" literally means that a metallic substance is capable of stretching on a thin wire without weakening or becoming more fragile in the process.

(d) Yielding

The performance of a material is explained as the tension at which a material begins to deform irreversibly. Before the elastic limit, the material will deform elastically, which means that it will return to its original shape when the applied tension is eliminated (that is, without permanent and visible change in the shape of the material).

(e) toughness

The ability of a metal to deform plastically and to absorb energy in the process before the fracture is called resistance. The emphasis of this definition should be placed on the ability to absorb energy before the fracture.

(f) strain hardening

The process of strain hardening is at what time when metal is strained beyond the yield point. An intensifying pressure is required to produce extra plastic deformation and the metal apparently becomes stronger and more difficult to deform.

The number of electrons that move past a fixed reference point in 2.55 ps when the current is 7.3  extmu A is approximately 116, after calculating the total charge and then dividing it by the charge of an electron.

The number of electrons moving past a fixed reference point in a given time interval when the electric current is known can be calculated by first determining the total charge that passes through the reference point and then dividing that by the charge per electron. The total charge ( extit{Q}) that moves through a fixed point is the product of the current ( extit{I}) and the time interval ( extit{t}), which is expressed by the formula  extit{Q} =  extit{I}  imes  extit{t}. Given that the charge of an electron is approximately -1.60  imes 10^{-19} C (coulombs), one can calculate the number of electrons by dividing the total charge by the electronic charge.

In this case, the current is 7.3  extmu A (microamperes), which is equal to 7.3  imes 10^{-6} A (amperes), and the time interval is 2.55 ps (picoseconds), or 2.55  imes 10^{-12} s (seconds). Using the formula  extit{Q} =  extit{I}  imes  extit{t}, the total charge moved past the reference point is calculated as follows:

Q = 7.3  imes 10^{-6} A  imes 2.55  imes 10^{-12} s = 1.8615  imes 10^{-17} C

Next, to find the number of electrons that move past the reference point, we divide the total charge by the charge of an electron:

Number of electrons =  rac{Q}{e} =  rac{1.8615  imes 10^{-17} C}{1.60  imes 10^{-19} C/e^-}

This yields approximately 116.34 electrons. Since the question asks for an integer value, we round this to 116 electrons as the number of electrons that move past a fixed reference point every 2.55 ps if the current is 7.3  extmu A.

To find the number of electrons moving past a fixed reference point with a given current, you can use the formula: Current (i) = Charge (q) / Time (t). By substituting the known values and calculating, you can determine the number of electrons passing the point at a specific time.

To find the number of electrons moving past a fixed reference point with given current:

Given: Current (i) = 7.3 μA = 7.3 x 10^-6 A

Formula: Current (i) = Charge (q) / Time (t)

Calculation: Number of electrons = (Current x Time) / Charge of an electron

Substitute values and calculate.

Answer:

//This Program is written in C++

// Comments are used for explanatory purpose

#include <iostream>

using namespace std;

enum mailbox{open, close};

int box[149];

void closeAllBoxes();

void OpenClose();

void printAll();

int main()

{

closeAllBoxes();

OpenClose();

printAll();

return 0;

}

void closeAllBoxes()

{

for (int i = 0; i < 150; i++) //Iterate through from 0 to 149 which literarily means 1 to 150

{

box[i] = close; //Close all boxes

}

}

void OpenClose()

{

for(int i = 2; i < 150; i++) {

for(int j = i; j < 150; j += i) {

if (box[j] == close) //Open box if box is closed

box[j] = open;

else

box[j] = close; // Close box if box is opened

}

}

// At the end of this test, all boxes would be closed

}

void printAll()

{

for (int x = 0; x < 150; x++) //use this to test

{

if (box[x] = 1)

{

cout << "Mailbox #" << x+1 << " is closed" << endl;

// Print all close boxes

}

}

}

Explanation:

Answer:

a) t = √(2*h/g)

b) v₀ = 67*√(g/(2*h))

c) ∅ = tan⁻¹(2*h/67)

Explanation:

Suppose that the height of the building is known (h), then we have

a) y = h = g*t²/2

then the time of flight is

t = √(2*h/g)

b) The initial speed (v₀) can be obtained as follows

x = v₀*t    ⇒    v₀ = x / t

where

x = xmax = 67 m    and  t = √(2*h/g)

So, we have

v₀ = 67 / √(2*h/g) = 67*√(g/(2*h))

c) In order to get the speed and the angle with respect to the horizontal of the velocity vector at impact, we obtain vx and vy as follows

vx = v₀ = 67*√(g/(2*h))

and

vy = gt = g*√(2*h/g)   ⇒   vy = √(2*h*g)

then we apply

v = √(vx² + vy²) = √((67*√(g/(2*h)))² + (√(2*h*g))²)

⇒  v = √(g*(4489 + 4*h²)/(2*h))

The angle is obtained as follows

tan ∅ = vy / vx     ⇒    tan ∅ = √(2*h*g) / 67*√(g/(2*h))

⇒ tan ∅ = 2*h / 67    ⇒   ∅ = tan⁻¹(2*h/67)

Answer:

(d) b and c

Explanation:

Given;

Function (true, 0, 2.6)

From the above, the function Function receives three arguments;

=>First argument (i.e true), is of type boolean

=>Second argument (i.e 0), is of type int, float or double since integers are inherently floating point numbers. Therefore 0 can be int, float or double.

=>Third argument (i.e 2.6), is a floating point number and could be of type float or double.

Therefore the method header for the function Function could be any of the following;

i). void Function (bool b, double d, double e)

ii). void Function (bool b, int c, double d)

iii). void Function (bool b, int d, float e)

iv). void Function (bool b, float d, double e)

v). void Function (bool b, double d, float e)

vi). void Function (bool b, float d, float e)

Note:

i. The ordered list above has ii and v in bold form to show that they are part of the options specified in the question.

ii. The letters b, c, d and e specified in the method declaration variations do not matter. Since they are just dummy variables, they can be any letter.

Therefore b and c are a correct representation for the method header.

Answer:

a) V =10¹¹*(1.5q₁ + 3q₂)

b) U = 1.34*10¹¹q₁q₂

Explanation:

Given

x₁ = 6 cm

y₁ = 0 cm

x₂ = 0 cm

y₂ = 3 cm

q₁ = unknown value in Coulomb

q₂ = unknown value in Coulomb

A) V₁ = Kq₁/r₁

where   r₁ = √((6-0)²+(0-0)²)cm = 6 cm = 0.06 m

V₁ = 9*10⁹q₁/(0.06) = 1.5*10¹¹q₁

V₂ = Kq₂/r₂

where   r₂ = √((0-0)²+(3-0)²)cm = 3 cm = 0.03 m

V₂ = 9*10⁹q₂/(0.03) = 3*10¹¹q₂

The electric potential due to the two charges at the origin is

V = ∑Vi = V₁ + V₂ = 1.5*10¹¹q₁ + 3*10¹¹q₂ = 10¹¹*(1.5q₁ + 3q₂)

B) The electric potential energy associated with the system, relative to their infinite initial positions, can be obtained as follows

U = Kq₁q₂/r₁₂

where

r₁₂ = √((0-6)²+(3-0)²)cm = √45 cm = 3√5 cm = (3√5/100) m

then

U = 9*10⁹q₁q₂/(3√5/100)

⇒ U = 1.34*10¹¹q₁q₂

Answer:

See attachment below

Explanation:

To determine how much a wire stretches under a load, one must consider Hooke's Law and use Young's modulus, but the original length of the wire and the type of load distribution must be known.

The student's question relates to the concept of elasticity and specifically the calculation of how much a wire stretches under a given load within the elastic limit.

Using the provided diameter of the wire, which is 0.6 inches, and the fact that an elastic modulus (Est) of 29.0(103) ksi has been given, the stretch can be calculated using Hooke's Law, which relates stress and strain through Young's modulus. However, to calculate the stretch, we would require the initial length of the wire and the format of the distributed load (whether it's uniform or varies along the length of the wire) which have not been provided in the student’s query.

Assuming uniform distributed load and an initial length L, the stretch (delta L) can be found using  delta L =  rac{wL^2}{AE}, where w is the load per unit length, L is the initial length, A is the cross-sectional area, and E is Young's modulus. If the total weight of the load is given instead, then the formula  delta L =  rac{FL}{AE} can be used with F representing the total force or weight affecting the wire.

Answer:

# -*- coding: utf-8 -*-

# Get N from the command line

import sys

N = int(sys.argv[1])

if N > 0: #N is positive  

   positve = list(range(N,0,-1))

   print(positve)

elif N < 0: #N is negative  

   negative = list(range(N,0,1))

   print(negative)

else:

   print("Invalid in input")

Explanation:

First, you need to identify if the number entered is positive, negative or none of them, for that we use one if, one elif and one else statement:

Answer:

[tex]F=6.88 [lbf][/tex]

The bracket is strong enough.

[tex]h=20.91 [ft][/tex]

Explanation:

Let's recall that the variation of the pressure respect to displacement in a liquid incomprehensible and static will be:

[tex]\frac{dP}{dy}=-\rho g[/tex]

If we take ρ (density) as a constant and solving this differential equation, we will have:

[tex]\Delta P=\rho gh[/tex]                            

Now, the pressure at the base will be:

[tex]P_{base}=\rho gh[/tex]

We use this equation knowing that we have atmospheric pressure on the outside of the tank.

The force on the inspection cover will be (A=1 in²):

[tex]F=P_{base}A=\rho ghA= 62.4 [lb/ft^{3}]*32.2 [ft/s^{2}]*16 [ft]*0.00689 [ft^{2}]=221.47 [\frac{lb*ft}{s^{2}}][/tex]

We know that 1 lbf = 32.17 (lb*ft)/s², so:

[tex]F=6.88 [lbf][/tex]

The statement says that the bracket can hold a load of 9 lbf, therefore the bracket is strong enough.

We can use the equation of the force to find the depth.

[tex]F=\rho ghA[/tex]

If we solve it for h we will have:

[tex]h=\frac{F}{\rho gA}[/tex]

[tex]h=\frac{289.53}{62.4*32.2*0.00689}=20.91 [ft][/tex]

I hope it helps you!

Answer:

Delta S = 356.64J/mol and Delta H = 99620J/mol

Explanation:

The detailed steps and appropriate use of the arrehenius equation and necessary susbtitution were made as shown in the attached file.

Answer:

You must follow the steps below to deal with employees who have poor performance. I hope that you will be able to handle all the issues after reading this answer.

Explanation:

Be specific with facts in hand

It is important to confront your employees about their respective actions. But to convince them about their withdrawal from lack of interest, it is imperative to have a consistent record at hand as well. For example, if the employee has been constantly delayed for a period of time, specify the precise details about the frequency and intensity of absenteeism. Be sure not to overdo your statements or use hard phrases to reduce employee self-esteem. Just be direct and precise. Reiterate the guidelines accordingly.

Consider the needs of your employees

Poor performance is not always the result of an employee's carelessness. There can be multiple genuine reasons for lack of performance and it can vary from person to person. The first is to understand the reason and judge whether they are genuine or not. Even if they aren't, don't let the other person know. Focus on your concerns and provide solutions accordingly. For example, if your employees cannot focus on their work due to some personal stress, make appointments for the counseling sessions and make sure they can get back on track.

Focus on feedback

Everyone handles comments differently. Although it is always recommended to be direct and clear in your communication, there may be certain strategies you can adopt to communicate your comments effectively. If your employee has difficulties in achieving his goals, work with him and provide him with all the necessary help to improve his performance. The best way is to provide weekly or monthly comments to your employees, so they know what they must do to achieve their goals.

Provide Performance Support Technology

When faced with existing employees who do not meet expectations, it is a smart decision to offer them training (and / or training) and different resources to help them improve. As an example, you can combine a low-performance worker with someone to act as a mentor or offer you a manual with the procedures to follow. In addition, there are performance improvement tools: WalkMe is a very valuable technology that can help a worker be more efficient and precise, within the workflow (and not take them away from their daily tasks).

Offer rewards and recognition

Whenever you see that your employees have poor performance, it is always better to adopt a carrot and stick approach for instant and consistent improvement. It is a combination of rewards and punishments that you can induce for the best and worst weekly or monthly performance. This has proven to be one of the best ways to combat low performance problems in companies of all kinds for centuries.

Facing low-performance workers for the first time may not be too encouraging for managers as well, but having an adequate system to deal with them is essential, especially during the performance of change management. It is always better to deal with such situations, instead of ignoring them, to maintain the consistency of productivity and profitability in the business.

Answer:

[tex]Power=11.52hp[/tex]

Explanation:

Given data

[tex]p_{1}=15psia\\p_{2}=70psia\\V_{ol}=0.8ft^{3}/s[/tex]

As

[tex]m=p*V_{ol}[/tex]

Assuming in-compressible flow p is constant

The total change in system mechanical energy calculated as:

Δe=(p₂-p₁)/p

The Power can be calculated as

[tex]P=W\\P=m(delta)e\\P=p*V_{ol}*(p_{2}-p_{1})/p\\ P=V_{ol}*(p_{2}-p_{1})\\P=44(psia.ft^{3}/s )*[(\frac{1Btu}{5.404psia.ft^{3} } )-(\frac{1hp}{0.7068Btu/s } )]\\P=11.52hp[/tex]    

The power input required to pump 0.8 ft³/s of water is; P = 11.52 hP

We are given;

Initial pressure; P1 = 15 psia

Final pressure; P2 = 70 psia

Volume flow rate; V' = 0.8 ft³/s

The formula for the mass flow rate is;

m' = ρV'

The total change in the mechanical energy of the system is;

△E = (P2 - P1)/ρ

Now, formula for power is;

P = m' × △E

P = ρV' × (P2 - P1)/ρ

P = V'(P2 - P1)

P = 0.8(70 - 15)

P = 44 psia.ft³/s

Converting to Btu/s gives;

P = 8.142 btu/s

Converting Btu/s to HP from conversion tables gives; P = 11.52 hP

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Answer:

Explanation: 100 and 110 Hz

Here fL=2kHz, fH=2.1kHz, B=0.1kHz, fL=2kHz,  

n≤fH/(fH-fL)

n≤(2.1*1000)/((2.1-2)*1000)

n≤20

2fH/n  ≤ fs

(2*2.1*1000)/20 ≤ fs

210≤ fs

Answer:

dP/dt =  0.3003003003

Explanation:

speed of submarine = 3m/s

rate of change of pressure with depth = (105 Pa) per 10 meter of depth

to get time taken by the submarine to cover 10m ; t = 10/3 = 3.33s

time rate of change of pressure = dP/dt =  105 Pa/3.33

= 30030.03003or 0.3003003003

Answer:

using System.IO;

using System;

class Program

{

static void Main()

{

int credits, year;

string inputString;

double tuition;

const int LOWCREDITS = 12;

const int HIGHCREDITS = 18;

const double HOURFEE = 15000;

const double DISCOUNT = 0.15;

const double FLAT = 1900.00;

const double RATE = 100.00;

const int SENIORYEAR = 4;

Console.WriteLine("How many credits? ");

inputString = Console.ReadLine();

credits = Convert.ToInt32(inputString);

Console.WriteLine("Year in school? ");

inputString = Console.ReadLine();

year = Convert.ToInt32(inputString);

if(credits > LOWCREDITS)

tuition = HOURFEE * credits;

else if(credits == HIGHCREDITS)

tuition = FLAT;

else

tuition = FLAT + (credits - HIGHCREDITS) * RATE;

if(year < SENIORYEAR)

tuition = tuition - (tuition * DISCOUNT);

Console.WriteLine("For year {0}, with {1} credits",year, credits);

Console.WriteLine("Tuition is {0}", tuition.ToString("C"));

}

You haven't declared the type for the year variable in Main. There's a typo in Readline(); it should be ReadLine().

Here's the corrected version of your code:

using System;

using static System.Console;

class DebugFour3

{

   static void Main()

   {

       int credits, year; // Declare type for year

       string inputString;

       double tuition;

       const int LOWCREDITS = 12;

       const int HIGHCREDITS = 18;

       const double HOURFEE = 150.00; // Corrected constant

       const double DISCOUNT = 0.15;

       const double FLAT = 1900.00;

       const double RATE = 100.00;

       const int SENIORYEAR = 4;

       WriteLine("How many credits? ");

       inputString = ReadLine();

       credits = Convert.ToInt32(inputString);

       WriteLine("Year in school? ");

       inputString = ReadLine();

       year = Convert.ToInt32(inputString);

       if (credits > LOWCREDITS)

           tuition = HOURFEE * credits;

       else if (credits <= HIGHCREDITS) // Corrected condition

           tuition = FLAT;

       else

           tuition = FLAT + (credits - HIGHCREDITS) * RATE;

       if (year < SENIORYEAR)

       {

           // Corrected calculation for discount

           tuition = tuition - (tuition * DISCOUNT);

       }

       WriteLine("For year {0}, with {1} credits", year, credits);

       WriteLine("Tuition is {0}", tuition.ToString("C"));

   }

}

Answer:

Explanation:

[tex]\dot x (t) - ax(t)=bu(t)\\\\e^{-at}\dot x = e^{-at}ax = \frac{d}{dt} (e^{-at})=e^{-at}bu\\\\\int\limits^t_0 \frac{d}{d\tau}(e^{-a\tau}x(\tau))d\tau=e^{-at}x(t)-x(0)=\int\limits^t_0 e^{-a\tau}bu(\tau))d\tau\\\\x(t)=e^{at}x(0)+ \int\limits^t_0 e^{-a(t-\tau)}bu(\tau))d\tau[/tex]

Similarly,

[tex]e^{-At}\dot x(t) -e^{-At}Ax(t) = \frac{d}{dt} (e^{-At}x(t))=e^{-At}Bu(t)\\\\\int\limits^t_0 \frac{d}{d\tau}(e^{-A\tau}x(\tau))d\tau=e^{-At}x(t)-e^{-A.0}x(0)=\int\limits^t_0 e^{-A\tau}Bu(\tau))d\tau\\\\since\:\:\:e^{-A.0}=I \:\:\: and\:\:\:[e^{-At}]^{-1}=e^{At}\\\\x(t)=e^{At}x(0)+ e^{At}\int\limits^t_0 e^{-A\tau}Bu(\tau))d\tau\\\\x(t)=e^{At}x(0)+ \int\limits^t_0 e^{A(t-\tau)}Bu(\tau))d\tau[/tex]

For initial condition x(0) = 0

[tex]x(t)= \int\limits^t_0 e^{A(t-\tau)}Bu(\tau))d\tau[/tex]

The magnitude of the velocity and the acceleration is 0.06 m/s^2., the velocity is 11 m/s

Given the information, we can calculate the velocity and acceleration of the car as it moves one-third of the way around the circular track.

First, we need to find the time t when the car has traveled one-third of the way around the track. The distance traveled by the car in one-third of the way around the track is d = r/3, where r is the radius of the track. The velocity of the car at time t is v = 5 + 0.06t. The distance traveled by the car can be found using the equation d = vt. Setting these two equations equal to each other, we have:

d = r/3 = 5t + 0.03t^2

Solving for t, we find that t = 20 seconds.

Next, we can find the velocity of the car at time t = 20 seconds using the equation v = 5 + 0.06t = 5 + 0.06 * 20 = 11 m/s.

Finally, the acceleration of the car can be found using the equation a = dv/dt = 0.06 m/s^2.

So, when the car has traveled one-third of the way around the track, its velocity is 11 m/s and its acceleration is 0.06 m/s^2.

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#SPJ1

To solve this problem we will use the values of the density on surface, which relates the total load and the area of the surface. From there we will get the total charge, which will allow us to find the electric flow.

Surface density is defined as,

[tex]\mu= \frac{Q_{tot}}{A_{surface}}[/tex]

Where,

[tex]A_{surface} = 4\pi r^2[/tex]

Replacing,

[tex]4*10^{-9} = \frac{Q_{tot}}{(4\pi 0.02^2)}[/tex]

[tex]Q_{tot}= 20.11*10^{-2} C[/tex]

At the same time electric flux can be defined as,

[tex]\Phi = \frac{Qtot}{\epsilon_0}[/tex]

Here,

[tex]\epsilon_0 =[/tex] Permittivity vacuum constant

Replacing,

[tex]\Phi =\frac{(20.11 * 10-12 )}{(8.85 * 10-12)}[/tex]

[tex]\Phi =2.27N \cdot m^2 \cdot C^{-1}[/tex]

Therefore the total electric flux through the concentric spherical surface is [tex]2.27Nm^2/C[/tex]

Through conc. spherical surface, the total electric flux will be:

"2.27 N.m².C⁻¹".

Uniform surface density charge, [tex]A_{surface}[/tex] = 4.0 nC/m²

Radius, r = 2.0 cm, or

               = 0.02

We know the relation,

→ Surface density, μ = [tex]\frac{Q_{tot}}{A_{surface}}[/tex]

Here, [tex]A_{surface}[/tex] = 4πr²

                [tex]Q_{tot}[/tex] = 20.11 × 10⁻² C

Now,

→ Electric flux, [tex]\Phi[/tex] = [tex]\frac{Q_{tot}}{\epsilon_0}[/tex]

By substituting the values,

                           = [tex]\frac{20.11\times 10^{-12}}{8.85\times 10^{-12}}[/tex]

                           = 2.27 N.m².C⁻¹

Thus the above response is correct.

Find out more information about electric flux here:

https://brainly.com/question/1592046

Answer:

attached below

Explanation:

Full Question

Write a static method called quadrant that takes as parameters a pair of real numbers representing an (x, y) point and that returns the quadrant number for that point.

Recall that quadrants are numbered as integers from 1 to 4 with the upper-right quadrant numbered 1 and the subsequent quadrants numbered in a counter-clockwise fashion. Notice that the quadrant is determined by whether the x and y coordinates are positive or negative numbers. If a point falls on the x-axis or the y-axis, then the method should return 0.

Answer

public int quadrant(double x, double y) // Line 1

{

if(x > 0 && y > 0){ //Line 2

return 1;

}

if(x < 0 && y > 0){ // Line 3

return 2;

}

if(x < 0 && y < 0) {// Line 4

return 3;

}

if(x > 0 && y < 0) {// Line 5

return 4;

}

return 0;

}

Explanation:

Line 1 defines the static method.

The method is named quadrant and its of type integer.

Along with the method definition, two variables x and y of type double are also declared

Line 2 checks if x and y are greater than 0.

If yes then the program returns 1

Line 3 checks if x is less than 0 and y is greater than 0

If yes then the program returns 2

Line 4 checks if x and y are less than 0

If yes then the program returns 3

Line 5 checks is x is greater than 0 and y is less than 0

If yes then the program returns 4

The theory that explains this phenomenon is linked to the convergence of tectonic plates when there is immersion over the crazy oceanic ones. This movement could be referred to as an oceanic-oceanic convergence where the immersion of two ocean slabs occurs and one descends below another plate and initiates volcanic activity by the same mechanism that operates in all subduction zones.

In this way the movement of the plate on the Martian surface could be relatively much faster than the occurrence of the movement of the plate on Earth. Giant volcanoes form because the area of ​​the most oceanic crust converges faster.

Answer:

The bulb will be [tex]\frac{1}{4}[/tex] times as bright as it is in Europe.

Explanation:

Data provided in the question:

Power of bulb in Europe = 75 W

Voltage provided in Europe = 240 V

Voltage provided in United Stated = 120 V

Now,

We know,

Power = Voltage² ÷ Resistance

Therefore,

75 = 240² ÷ Resistance

or

Resistance = 240² ÷ 75

or

Resistance = 768 Ω

Therefore,

Power in United States = Voltage² ÷ Resistance

= 120² ÷ 768

= 18.75 W

Therefore,

Ratio of powers

[tex]\frac{\text{Power in united states}}{\text{Power in Europe}}=\frac{18.75}{75}[/tex]

= [tex]\frac{1}{4}[/tex]

Hence,

The bulb will be [tex]\frac{1}{4}[/tex] times as bright as it is in Europe.

Answer:

178 kJ

Explanation:

Assuming no heat transfer out of the cooling device, and if we can neglect the energy stored in the aluminum can, the energy transferred by the canned drinks, would be equal to the change in the internal energy of the canned drinks, as follows:

ΔU = -Q = -c*m*ΔT (1)

where c= specific heat of water = 4180 J/kg*ºC

           m= total mass = 6*0.355 Kg = 2.13 kg

           ΔT = difference between final and initial temperatures = 20ºC

Replacing by these values in (1), we can solve for Q as follows:

Q = 4180 J/kg*ºC * 2.13 kg * -20 ºC = -178 kJ

So, the amount of heat transfer from the six canned drinks is 178 kJ.

-178.92 kJ

The amount or quantity (Q) of heat transferred during a chemical process such as cooling is the product of the mass (m) of the substance involved, the specific heat capacity (c) of the substance and the change in temperature (ΔT) of the substance. i.e

Q = m x c x Δ T   ----------------------(i)

From the question;

m = total mass of the canned drinks = 6 x 0.355kg = 2.13kg

ΔT = final temperature - initial temperature = 3°C - 23°C = -20°C

Known constant;

c = specific heat capacity of water = 4200J/kg°C

Substitute all these values into equation (i) as follows;

Q = 2.13 x 4200 x (-20)

Q = -178920 J

Divide the result by 1000 to convert it to kJ as follows;

Q = (-178920 / 1000) kJ

Q = -178.92 kJ

Therefore, the quantity of heat transferred from the six canned drinks is -178.92 kJ

Note;

The -ve sign shows that the heat transferred is actually the heat lost in cooling the drinks from 23°C to 3°C

Answer:

Q=58.716 W

Explanation:

Given that

mass ,m  = 15 g

The decrease in the temperature ,ΔT = 37.3 - 20.5  °C

ΔT = 16.8°C

We know that specific heat of silver ,Cp = 0.233 J/g°C

The energy change of the silver  is given as

Q = m Cp ΔT

Now by putting the values we get

[tex]Q= 15 \times 0.233\times 16.8\ W[/tex]

Q=58.716 W

Therefore the change in the energy will be 58.716 W.

Answer:

the volume density of atoms = 1.75 x 10^22 /cm3

Explanation:

The detailed steps is as shown in the attached file.

Answer:

A Not Periodic

B Periodic

C Periodic

D Not Periodic

E Periodic

F Periodic

Explanation:

The detailed steps to ascertain the periodicity or the non periodicity of each functions is as shown in the attached file.

Answer:

2 μT

Explanation:

Assuming that the diameter of the current carrying wire is very small compared with its length, we can treat it as an infinite wire.

For an infinite wire, the magnetic field, at a distance r from the wire, is given by Biot-Savart law equation, as follows:

[tex]B = \frac{\mu0*I}{2*\pi*r}[/tex]

where:

μ₀ = 4*π*10⁻⁷

I = current carried by the wire

r = distance from the wire.

So, B₁ can be calculated as follows:

[tex]B1 = \frac{\mu0*I}{2*\pi*0.02m} = 4e-6 T[/tex]

If we change the distance where we want to know the value of B, all other factors remain constant, we can write the following expression for the new value of B, B₂:

[tex]B2 = \frac{\mu0*I}{2*\pi*0.04m}[/tex]

Dividing B1 by B2, and simplifying common terms, we have:

[tex]\frac{B1}{B2} = \frac{0.04m}{0.02m} = 2[/tex]

⇒ [tex]B2 = \frac{B1}{2} = \frac{4e-6T}{2} = 2e-6 T[/tex]

⇒ B₂ = 2 μT

Answer:

a) 740 W b) 6.2 A c) 8.1%

Explanation:

We need first to get the total energy spent during the 30 days, that can be calculated as follows:

1 month = 30 days = 720 hr

If the total cost amounts $80 (for 720 hr), and the cost per kwh is 0.15, we have:

80 $/mo  =  0.15 $/Kwh*x Kwh/mo

Solving for the total energy spent in the month:

[tex]x (kwh/mo) = \frac{80}{0.15} = 533.3 kwh[/tex]

Assuming that the power delivered is constant over the entire 30 days, as power is the rate of change of energy, we can find the power as follows:

[tex]P = \frac{E}{t} = \frac{533.3 kWh}{720 h} = 0.74 kW = 740 W[/tex]

b) If the power is supplied by a voltage of 120 V, we can find the current I as follows:

[tex]I =\frac{P}{V} =\frac{740W}{120V} = 6.2A[/tex]

c) If part of the electrical load is a 60-W light, we can substract this power from the one we have just found, as follows:

P = 740 W - 60 W = 680 W

The new value of the energy spent during the entire month will be as follows:

E = 0.68 kW*24(hr/day)*30(days/mo) = 490 kWh/mo

The reduction in percentage regarding the total energy spent can be calculated as follows:

ΔE = [tex]\frac{533.3-490}{533.3} = 0.081*100= 8.1%[/tex]

⇒ ΔE(%) = 8.1%

[tex]%(E) =\frac{533.3-490}{533.3} = 0.081 * 100 = 8.1%[/tex]