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k0ka [10]
3 years ago
13

The electrical panel schedules are located on EWR Plan number ___.

Engineering
1 answer:
Crank3 years ago
5 0
Jajajajajaja sorryyyyy
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Suppose the Bookstore is processing an input file containing the titles of books in order to remove duplicates from their list.
mash [69]

Answer:

books = []

   fp = open("bookTitles.txt")

   for line in fp.readlines():

       title = line.strip()

       if title not in books:

           books.append(title)

   fp.close()

   fout = open("noDuplicates.txt", "w")

   for title in books:

       print(tile, file=fout)

   fout.close()

except FileNotFoundError:

   print("Unable to open bookTitles.txt")

6 0
3 years ago
A pressure cylinder has an outer diameter 200 mm, maximum external pressure 4 MPa, and maximum allowable shear stress 27.5 MPa.
ludmilkaskok [199]

Answer:

The minimum value of wall thickness t=3.63 mm.

Explanation:

Given:

  D=200 mm

 P=4 MPa

t= Wall thickness

maximum shear stress=27.5 MPa

We know that

       hoop stress \sigma _{h}=\frac{Pd}{2t}

      Longitudinal stress \sigma _{l}=\frac{Pd}{4t}

So maximum shear tress in plane\tau _{max}=\dfrac{\sigma _h-\sigma _l}{2}

              \tau _{max}=\dfrac{Pd}{8t}

Now by putting the value

       27.5=\dfrac{4\times 200}{8t}

 So   t=3.36 mm

The minimum value of wall thickness t=3.63 mm.

4 0
3 years ago
java Write a program that simulates tossing a coin. Prompt the user for how many times to toss the coin. Code a method with no p
max2010maxim [7]

Answer:

The solution code is written in Java.

  1. public class Main {
  2.    public static void main(String[] args) {
  3.        Scanner inNum = new Scanner(System.in);
  4.        System.out.print("Enter number of toss: ");
  5.        int num = inNum.nextInt();
  6.        for(int i=0; i < num; i++){
  7.            System.out.println(toss());
  8.        }
  9.    }
  10.    public static String toss(){
  11.        String option[] = {"heads", "tails"};
  12.        Random rand = new Random();
  13.        return option[rand.nextInt(2)];
  14.    }
  15. }

Explanation:

Firstly, we create a function <em>toss()</em> with no parameter but will return a string (Line 14). Within the function body, create an option array with two elements, "heads" and "tails" (Line 15). Next create a Random object (Line 16) and use <em>nextInt()</em> method to get random value either 0 or 1. Please note we need to pass the value of 2 into <em>nextInx() </em>method to ensure the random value generated is either 0 or 1.  We use this generate random value as an index of <em>option </em>array and return either "heads" or "tails" as output (Line 17).

In the main program, we create Scanner object and use it to prompt user to input an number for how many times to toss the coin (Line 6 - 7). Next, we use the input num to control how many times a for loop should run (Line 9). In each round of the loop, call the function <em>toss() </em>and print the output to terminal (Line 10).  

4 0
3 years ago
Read 2 more answers
A 3.5-m3 rigid tank initially contains air whose density is 2 kg/m3 . The tank is connected to a high-pressure supply line throu
Mumz [18]

Answer:

Explanation:

First, we find the mass of the air originally in the tank.

Density is given as mass divided by volume. It is given as:

Density = \frac{mass}{volume}

Therefore, mass is:

mass = denisty *volume

Density of air = 2 kg/m^3; Volume of the tank =  3.5 m^3

=> Mass = 3.5 * 2 = 7 kg

The mass of the air initially in the tank is 7 kg.

After air is allowed to enter, the mass changes.

New density = 6.5 kg/m^3

The new mass will be:

Mass = 6.5 * 3.5 = 22.75 kg

We can now find the mass of air that has entered the tank:

Mass of air that entered tank = New mass of air - Original mass of air

M = 22.75 - 7.0 = 15.75 kg

The mass of air that entered the tank is 15.75 kg.

6 0
3 years ago
An inductor (L = 400 mH), a capacitor (C = 4.43 µF), and a resistor (R = 500 Ω) are connected in series. A 44.0-Hz AC generator
MakcuM [25]

Answer:

(A) Maximum voltage will be equal to 333.194 volt

(B) Current will be leading by an angle 54.70

Explanation:

We have given maximum current in the circuit i_m=385mA=385\times 10^{-3}A=0.385A

Inductance of the inductor L=400mH=400\times 10^{-3}h=0.4H

Capacitance C=4.43\mu F=4.43\times 10^{-3}F

Frequency is given f = 44 Hz

Resistance R = 500 ohm

Inductive reactance will be x_l=\omega L=2\times 3.14\times 44\times 0.4=110.528ohm

Capacitive reactance will be equal to X_C=\frac{1}{\omega C}=\frac{1}{2\times 3.14\times 44\times 4.43\times 10^{-6}}=816.82ohm

Impedance of the circuit will be Z=\sqrt{R^2+(X_C-X_L)^2}=\sqrt{500^2+(816.92-110.52)^2}=865.44ohm

So maximum voltage will be \Delta V_{max}=0.385\times 865.44=333.194volt

(B) Phase difference will be given as \Phi =tan^{-1}\frac{X_C-X_L}{R}=\frac{816.92-110.52}{500}=54.70

So current will be leading by an angle 54.70

5 0
3 years ago
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