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3 years ago

15

Same rule: If both players spend the same number of coins, player 2 gains 1 coin. Off-by-one rule: If the players do not spend t

he same number of coins and the positive difference between the number of coins spent by the two players is 1, player 2 is awarded 1 coin. Off-by-two rule: If the players do not spend the same number of coins and the positive difference between the number of coins spent by the two players is 2, player 1 is awarded 2 coins. The following is an example of a game played with a starting value of 10 coins and a game length of 5 rounds

1 answer:

Galina-37 [17]3 years ago

3 0

Answer:

Check the explanation

Explanation:

1 -

public int getPlayer2Move(int round)

{

int result = 0;

//If round is divided by 3

if(round%3 == 0) {

result= 3;

}

//if round is not divided by 3 and is divided by 2

else if(round%3 != 0 && round%2 == 0) {

result = 2;

}

//if round is not divided by 3 or 2

else {

result = 1;

}

return result;

}

2-

public void playGame()

{

//Initializing player 1 coins

int player1Coins = startingCoins;

//Initializing player 2 coins

int player2Coins = startingCoins;

for ( int round = 1 ; round <= maxRounds ; round++) {

//if the player 1 or player 2 coins are less than 3

if(player1Coins < 3 || player2Coins < 3) {

break;

}

//The number of coins player 1 spends

int player1Spends = getPlayer1Move();

//The number of coins player 2 spends

int player2Spends = getPlayer2Move(round);

//Remaining coins of player 1

player1Coins -= player1Spends;

//Remaining coins of player 2

player2Coins -= player2Spends;

//If player 2 spends the same number of coins as player 2 spends

if ( player1Spends == player2Spends) {

player2Coins += 1;

continue;

}

//positive difference between the number of coins spent by the two players

int difference = Math.abs(player1Spends - player2Spends) ;

//if difference is 1

if( difference == 1) {

player2Coins += 1;

continue;

}

//If difference is 2

if(difference == 2) {

player1Coins += 2;

continue;

}

}

// At the end of the game

//If player 1 coins is equal to player two coins

if(player1Coins == player2Coins) {

System.out.println("tie game");

}

//If player 1 coins are greater than player 2 coins

else if(player1Coins > player2Coins) {

System.out.println("player 1 wins");

}

//If player 2 coins is grater than player 2 coins

else if(player1Coins < player2Coins) {

System.out.println("player 2 wins");

}

}

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Number of turns in primary wire N₁ = 900

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there fore

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Now we substitute

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Tensile Strength (MPa) Number-Average Molecular Weight (g/mol)

IceJOKER [234]

Answer:

$\mathbf{T_{S \infty } \ \approx 215.481 \ MPa}$

$\mathbf{M_n = 49163.56431 \ g/mol }$

Explanation:

The question can be well structured in a table format as illustrated below:

Tensile Strength (MPa) Number- Average Molecular Weight (g/mol)

82 12,700

156 28,500

The tensile strength and number-average molecular weight for two polyethylene materials given above.

Estimate the number-average molecular weight that is required to give a tensile strength required above. Using the data given find TS (infinity) in MPa.

<u>SOLUTION:</u>

We know that :

$T_S = T_{S \infty} - \dfrac{A}{M_n}$

where;

$T_S$ = Tensile Strength

$T_{S \infty}$ = Tensile Strength (Infinity)

$M_n$ = Number- Average Molecular Weight (g/mol)

SO;

$82= T_{S \infty} - \dfrac{A}{12700} ---- (1)$

$156= T_{S \infty} - \dfrac{A}{28500} ---- (2)$

From equation (1) ; collecting the like terms; we have :

$T_{S \infty} =82+ \dfrac{A}{12700}$

From equation (2) ; we have:

$T_{S \infty} =156+ \dfrac{A}{28500}$

So; $T_{S \infty} = T_{S \infty}$

Then;

$T_{S \infty} =82+ \dfrac{A}{12700} =156+ \dfrac{A}{28500}$

Solving by L.C.M

$\dfrac{82(12700) + A}{12700} =\dfrac{156(28500) + A}{28500}$

$\dfrac{1041400 + A}{12700} =\dfrac{4446000 + A}{28500}$

By cross multiplying ; we have:

$({4446000 + A})* {12700} ={28500} *({1041400 + A})$

$(5.64642*10^{10} + 12700A) =(2.96799*10^{10}+ 28500A)$

Collecting like terms ; we have

$(5.64642*10^{10} - 2.96799*10^{10} ) =( 28500A- 12700A)$

$2.67843*10^{10} = 15800 \ A$

Dividing both sides by 15800:

$\dfrac{ 2.67843*10^{10} }{15800} =\dfrac{15800 \ A}{15800}$

A = 1695208.861

From equation (1);

$82= T_{S \infty} - \dfrac{A}{12700} ---- (1)$

Replacing A = 1695208.861 in the above equation; we have:

$82= T_{S \infty} - \dfrac{1695208.861}{12700}$

$T_{S \infty}= 82 + \dfrac{1695208.861}{12700}$

$T_{S \infty}= \dfrac{82(12700) +1695208.861 }{12700}$

$T_{S \infty}= \dfrac{1041400 +1695208.861 }{12700}$

$T_{S \infty}= \dfrac{2736608.861 }{12700}$

$\mathbf{T_{S \infty } \ \approx 215.481 \ MPa}$

From equation(2);

$156= T_{S \infty} - \dfrac{A}{28500} ---- (2)$

Replacing A = 1695208.861 in the above equation; we have:

$156= T_{S \infty} - \dfrac{1695208.861}{28500}$

$T_{S \infty}= 156 + \dfrac{1695208.861}{28500}$

$T_{S \infty}= \dfrac{156(28500) +1695208.861 }{28500}$

$T_{S \infty}= \dfrac{4446000 +1695208.861 }{28500}$

$T_{S \infty}= \dfrac{6141208.861}{28500}$

$\mathbf{T_{S \infty } \ \approx 215.481 \ MPa}$

We are to also estimate the number- average molecular weight that is required to give a tensile strength required above.

If the Tensile Strength (MPa) is 82 MPa

Definitely the average molecular weight will be = 12,700 g/mol

If the Tensile Strength (MPa) is 156 MPa

Definitely the average molecular weight will be = 28,500 g/mol

But;

Let us assume that the Tensile Strength (MPa) = 181 MPa for example.

Using the same formula:

$T_S = T_{S \infty} - \dfrac{A}{M_n}$

Then:

$181 = 215.481- \dfrac{1695208.861 }{M_n}$

Collecting like terms ; we have:

$\dfrac{1695208.861 }{M_n} = 215.481- 181$

$\dfrac{1695208.861 }{M_n} =34.481$

1695208.861= 34.481 $M_n$

Dividing both sides by 34.481; we have:

$M_n$ = $\dfrac{1695208.861}{34.481}$

$\mathbf{M_n = 49163.56431 \ g/mol }$

5 0

3 years ago