The larger the Bi number, the more accurate the lumped system analysis. a)-True b)- False

Answer:

the relative compaction is 105.88 %

Explanation:

Given;

dry unit weight of field compaction, = 18 kN/m³

maximum dry unit weight measured, = 17 kN/m³

Relative compaction (RC) of the site is given as the ratio of dry unit weight of field compaction and maximum dry unit weight measured

Relative compaction (RC) = dry unit weight of field compaction / maximum dry unit weight measured

substitute the given values;

RC (%) = 105.88 %

Therefore, the relative compaction is 105.88 %

Answer:

The solution is given in the attachments.

Answer:

Time period  = 41654.08 s

Explanation:

Given data:

Internal volume is 210 m^3

Rate of air infiltration  

length of cracks 62 m

air density = 1.186 kg/m^3

Total rate of air infiltration

total volume of air  infiltration

Time period

The answer is "45.3 NM".

There at end of the movement, the forging force is given by

h is the final height.

The ultimate radius is determined by following a volume constancy law, which states that volumes before deformation measured amount following distortion.

You may deduce from the graph flow that , thus we use the formula.

Therefore, the answer is "45.3 NM".

Learn more:

brainly.com/question/17139328

Answer:

45.3 MN

Explanation:

The forging force at the end of the stroke is given by

F = Y.π.r².[1 + (2μr/3h)]

The final height, h is given as h = 100/2

h = 50 mm

Next, we find the final radius by applying the volume constancy law

volumes before deformation = volumes after deformation

π * 75² * 2 * 100 = π * r² * 2 * 50

75² * 2 = r²

r² = 11250

r = √11250

r = 106 mm

E = In(100/50)

E = 0.69

From the graph flow, we find that Y = 1000 MPa, and thus, we apply the formula

F = Y.π.r².[1 + (2μr/3h)]

F = 1000 * 3.142 * 0.106² * [1 + (2 * 0.2 * 0.106/ 3 * 0.05)]

F = 35.3 * [1 + 0.2826]

F = 35.3 * 1.2826

F = 45.3 MN

Answer:

The amount of heat transferred to the air is 340.24 kJ

Explanation:

From P-V diagram,

Initial temperature T1 = 27°C

Initial pressure P1 = 100 kPa

final pressure P3 = P2 = 300 kPa

volume at point 2, V2 = V1 = 0.4 m³

final temperature T2 = T3 = 1200 K

To determine the final pressure V3, use ideal gas equation

PV = mRT

Where R is the specific gas constant = 0.2870 KPa m³ kg K

But,

from initial condition, mass m = PV/RT

m = (P1*V1)/R*T1

T1 = 27+273 = 300K

m = (100*0.4)/(0.2870*300) = 0.4646 kg

Then;

Final volume V3 = mRT3/P3

V3 = (0.4646*0.2870*1200)/300

V3 = 0.5334 m³

Total work done W is determined where there is volume change which from point 2 to 3.

W = P3*(V3-V2)

W = 300*(0.5334-0.4) = 40.02 kJ

To get the internal energy, the heat capacity at room temperature Cv is 0.718 kJ/kg K

∆U = m*Cv*(T2-T1)

∆U = 0.4646*0.718(1200-300)

∆U = 300.22 kJ

The heat transfer Q = W + ∆U

Q = 40.02 + 300.22 = 340.24 kJ

Determine the amount of heat transferred to the air, in kJ, while increasing the temperature to 1200 K is 340.24 kJ

The attached file shows the Pressure - Volume relationship (P -V graph)

Answer:

Output:-

Enter the five digit lottery number

Enter the digit 1 : 23

Enter the digit 2 : 44

Enter the digit 3 : 43

Enter the digit 4 : 66

Enter the digit 5 : 33

YOU LOSS!!

Computer Generated Lottery Number :

|12|38|47|48|49|

Lottery Number Of user:

|23|33|43|44|66|

Number Of digit matched: 0

Code:-

import java.util.Arrays;

import java.util.Random;

import java.util.Scanner;

public class Lottery {

int[] lotteryNumbers = new int[5];

public int[] getLotteryNumbers() {

return lotteryNumbers;

}

Lottery() {

Random randomVal = new Random();

for (int i = 0; i < lotteryNumbers.length; i++) {

lotteryNumbers[i] = randomVal.nextInt((50 - 1) + 1);

}

}

int compare(int[] personLottery) {

int count = 0;

Arrays.sort(lotteryNumbers);

Arrays.sort(personLottery);

for (int i = 0; i < lotteryNumbers.length; i++) {

if (lotteryNumbers[i] == personLottery[i]) {

count++;

}

}

return count;

}

public static void main(String[] args) {

int[] personLotteryNum = new int[5];

int matchNum;

Lottery lnum = new Lottery();

Scanner input = new Scanner(System.in);

System.out.println("Enten the five digit lottery number");

for (int i = 0; i < personLotteryNum.length; i++) {

System.out.println("Enter the digit " + (i + 1) + " :");

personLotteryNum[i] = input.nextInt();

}

matchNum = lnum.compare(personLotteryNum);

if (matchNum == 5)

System.out.println("YOU WIN!!");

else

System.out.println("YOU LOSS!!");

System.out.println("Computer Generated Lottery Number :");

System.out.print("|");

for (int i = 0; i < lnum.getLotteryNumbers().length; i++) {

System.out.print(lnum.getLotteryNumbers()[i] + "|");

}

System.out.println("\n\nLottery Number Of user:");

System.out.print("|");

for (int i = 0; i < personLotteryNum.length; i++) {

System.out.print(personLotteryNum[i] + "|");

}

System.out.println();

System.out.println("Number Of digit matched: " + matchNum);

}

}

Explanation: