If in Example 1.2,q = (10 - 10e2) mC, find the current at t = 0.5 s.

std::string x[n]; int mode1count = 0; // msft int mode2count = 0; // appl int mode3count = 0; // csco int mode4count = 0; // imb

rosijanka [135]

Answer:

std::string x[n];

int mode1count = 0; // msft

int mode2count = 0; // appl

int mode3count = 0; // csco

int mode4count = 0; // imb

int i;

for(i = 0; i < n; ++i) // Set the array of strings

x[i] = modeList[i];

for(i = 0; i < n; ++i)

{

if(x[i] == "msft")

++mode1count;

else if(x[i] == "appl")

++mode2count;

else if(x[i] == "csco")

++mode3count;

else if(x[i] == "imb")

++mode4count;

}

Explanation:

It is important to ensure that there is adequate space and necessary punctuation when developing codes and programming to avoid getting error messages when running the codes. In this problem, the codes required are shown in the answer section. The codes are properly writing and arranged.

7 0

3 years ago

What is the average speed of a car that travels 80km in 1.5 hours

DanielleElmas [232]

Answer:53.33km/hr.

Explanation:

Average speed =

distance traveled/time taken

Average speed =80km/1.5hr=53.33km/hrs

5 0

4 years ago

Read 2 more answers

The pressure gage on a 2.5-m^3 oxygen tank reads 500 kPa. Determine the amount of oxygen in the tank if the temperature is 28°C

s2008m [1.1K]

Answer:

19063.6051 g

Explanation:

Pressure = Atmospheric pressure + Gauge Pressure

Atmospheric pressure = 97 kPa

Gauge pressure = 500 kPa

Total pressure = 500 + 97 kPa = 597 kPa

Also, P (kPa) = 1/101.325 P(atm)

Pressure = 5.89193 atm

Volume = 2.5 m³ = 2500 L ( As m³ = 1000 L)

Temperature = 28 °C

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15

So,

T₁ = (28.2 + 273.15) K = 301.15 K

Using ideal gas equation as:

PV=nRT

where,

P is the pressure

V is the volume

n is the number of moles

T is the temperature

R is Gas constant having value = 0.0821 L.atm/K.mol

Applying the equation as:

5.89193 atm × 2500 L = n × 0.0821 L.atm/K.mol × 301.15 K

⇒n = 595.76 moles

Molar mass of oxygen gas = 31.9988 g/mol

Mass = Moles * Molar mass = 595.76 * 31.9988 g = 19063.6051 g

7 0

3 years ago

Prompt the user to enter five numbers, being five people's weights. Store the numbers in an array of doubles. Output the array's

pochemuha

Answer:

import java.util.Scanner;

public class PeopleWeights {

public static void main(String[] args) {

Scanner reader = new Scanner(System.in);

double weightOne = reader.nextDouble();

System.out.println("Enter 1st weight:");

double weightTwo = reader.nextDouble();

System.out.println("Enter 2nd weight :");

double weightThree = reader.nextDouble();

System.out.println("Enter 3rd weight :");

double weightFour = reader.nextDouble();

System.out.println("Enter 4th weight :");

double weightFive = reader.nextDouble();

System.out.println("Enter 5th weight :");

double sum = weightOne + weightTwo + weightThree + weightFour + weightFive;

double[] MyArr = new double[5];

MyArr[0] = weightOne;

MyArr[1] = weightTwo;

MyArr[2] = weightThree;

MyArr[3] = weightFour;

MyArr[4] = weightFive;

System.out.printf("You entered: " + "%.1f %.1f %.1f %.1f %.1f ", weightOne, weightTwo, weightThree, weightFour, weightFive);

double average = sum / 5;

System.out.println();

System.out.println("Total weight: " + sum);

System.out.println("Average weight: " + average);

double max = MyArr[0];

for (int counter = 1; counter < MyArr.length; counter++){

if (MyArr[counter] > max){

max = MyArr[counter];

}

System.out.println("Max weight: " + max);

}

import java.util.Scanner;

public class PeopleWeights {

public static void main(String[] args) {

Scanner reader = new Scanner(System.in);

double weightOne = reader.nextDouble();

System.out.println("Enter 1st weight:");

double weightTwo = reader.nextDouble();

System.out.println("Enter 2nd weight :");

double weightThree = reader.nextDouble();

System.out.println("Enter 3rd weight :");

double weightFour = reader.nextDouble();

System.out.println("Enter 4th weight :");

double weightFive = reader.nextDouble();

System.out.println("Enter 5th weight :");

double sum = weightOne + weightTwo + weightThree + weightFour + weightFive;

double[] MyArr = new double[5];

MyArr[0] = weightOne;

MyArr[1] = weightTwo;

MyArr[2] = weightThree;

MyArr[3] = weightFour;

MyArr[4] = weightFive;

System.out.printf("You entered: " + "%.1f %.1f %.1f %.1f %.1f ", weightOne, weightTwo, weightThree, weightFour, weightFive);

double average = sum / 5;

System.out.println();

System.out.println("Total weight: " + sum);

System.out.println("Average weight: " + average);

double max = MyArr[0];

for (int counter = 1; counter < MyArr.length; counter++){

if (MyArr[counter] > max){

max = MyArr[counter];

}

System.out.println("Max weight: " + max);

}

8 0

3 years ago

Read 2 more answers

Two stepped bar is supported at both ends.At the join point of two segments,the force F is applied(downwards).Calculate reactive

nlexa [21]

Answer:

F=200kN

Explanation:

8 0

3 years ago

If in Example 1.2,q = (10 - 10e2) mC, find the current at t = 0.5 s. ​

If in Example 1.2,q = (10 - 10e2) mC, find the current at t = 0.5 s.