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2 years ago

Help me!!

Engineering

1 answer:

kati45 [8]2 years ago

6 0

Answer: System Consists Of 1 Kg Of CO2 (Cp = 46.4 J Moll K:') Gas Initially At 1 Bar And 300K. The System Undergo

Explanation:

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When the outside temperature is 5.2 ⁰C, a steel beam of cross-sectional area 52 cm2 is installed in a building with the ends of

il63 [147K]

Multiply the coefficient by the change in temperature:

1.1*10^-5 x (37-5.2) = 0.0003498

Multiply Young's modulus by the area by the above answer:

2*10^11 x 52 * 0.0003498 x (1/100)^2 = 3.63792 x 10^5 N

6 0

3 years ago

Accidents occur as a result of ____ and ____.

telo118 [61]

BRIGHT HEADLIGHTS

AND SEVERE WEATHER CONDITIONS

5 0

2 years ago

The mean of 10 numbers is 9, then the sum (total) of these numbers will be

qwelly [4]

Answer:

Explanation:

mean is basically taking the sum of all numbers and then dividing the sum with the number of all given numbers..

here, the mean is 9, total numbers are 10.. so the sum will be 9 multiplied by 10, that is 90.

5 0

2 years ago

Read 2 more answers

Fix the code so the program will run correctly for MAXCHEESE values of 0 to 20 (inclusive). Note that the value of MAXCHEESE is

GarryVolchara [31]

Answer:

Code fixed below using Java

Explanation:

<u>Error.java </u>

import java.util.Random;

public class Error {

public static void main(String[] args) {

final int MAXCHEESE = 10;

String[] names = new String[MAXCHEESE];

double[] prices = new double[MAXCHEESE];

double[] amounts = new double[MAXCHEESE];

// Three Special Cheeses

names[0] = "Humboldt Fog";

prices[0] = 25.00;

names[1] = "Red Hawk";

prices[1] = 40.50;

names[2] = "Teleme";

prices[2] = 17.25;

System.out.println("We sell " + MAXCHEESE + " kind of Cheese:");

System.out.println(names[0] + ": $" + prices[0] + " per pound");

System.out.println(names[1] + ": $" + prices[1] + " per pound");

System.out.println(names[2] + ": $" + prices[2] + " per pound");

Random ranGen = new Random(100);

// error at initialising i

// i should be from 0 to MAXCHEESE value

for (int i = 0; i < MAXCHEESE; i++) {

names[i] = "Cheese Type " + (char) ('A' + i);

prices[i] = ranGen.nextInt(1000) / 100.0;

amounts[i] = 0;

System.out.println(names[i] + ": $" + prices[i] + " per pound");

}

7 0

3 years ago

Determine the combined moment about O due to the weight of the mailbox and the cross member AB. The mailbox weighs 3.2 lb and th

koban [17]

Answer:

Attached is the complete question but the weight of the mailbox and cross bar differs from the given values which are : weight of mail box = 3.2 Ib, weight of the uniform cross member = 10.3 Ib

Answer : moment of inertia = 186.7 Ib - in

Explanation:

Given data

weight of the mailbox = 3.2 Ib

weight of the uniform cross member = 10.3 Ib

The origin is of mailbox and cross member is 0

The perpendicular distance from Y axis of centroid of the mailbox

= 4 + (25/2) = 16.5"

The centroid of the bar =( ( 1 + 25 + 4 + 4 ) / 2 ) - 4 = 13"

therefore The moment of Inertia( Mo) = (3.2 * 16.5) + ( 10.3 * 13)

= 52.8 + 133.9 = 186.7 Ib-in

8 0

3 years ago