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Jobisdone [24]
3 years ago
5

How many 10" diameter circles can be cut from a semicircular shape that has a 20"

Engineering
2 answers:
Iteru [2.4K]3 years ago
5 0

Answer:

<h2> 1</h2>

  • <em>The</em><em> </em><em>diameter</em><em> </em><em>measurement</em><em> </em><em>of</em><em> </em><em>a</em><em> </em><em>semi circle</em><em> </em><em>having</em><em> </em><em>a</em><em> </em><em>measure</em><em> </em><em>of</em><em> </em><em>1</em><em>0</em><em>"</em><em> </em><em>diameter</em><em> </em><em>,</em><em> </em><em>2</em><em>0</em><em>"</em><em> </em><em>diameter</em><em> </em><em>and</em><em> </em><em>a</em><em> </em><em>length</em><em> </em><em>of</em><em> </em><em>2</em><em>5</em><em>.</em>
N76 [4]3 years ago
3 0

9514 1404 393

Answer:

  1

Explanation:

Only one such circle can be drawn. The diameter of the 10" circle will be a radius of the semicircle. In order for the 10" circle to be wholly contained, the flat side of the semicircle must be tangent to the 10" circle. There is only one position in the figure where that can happen. (see attached).

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An unknown immiscible liquid seeps into the bottom of an open oil tank. Some measurements indicate that the depth of the unknown
barxatty [35]

Answer:

The specific weight of unknown liquid is found to be 15 KN/m³

Explanation:

The total pressure in tank is measured to be 65 KPa in the tank. But, the total pressure will be equal to the sum of pressures due to both oil and unknown liquid.

Total Pressure = Pressure of oil + Pressure of unknown liquid

65 KPa = (Specific Weight of oil)(depth of oil) + (Specific Weight of unknown liquid)(depth of unknown liquid)

65 KN/m² = (8.5 KN/m³)(5 m) + (Specific Weight of Unknown Liquid)(1.5 m)

(Specific Weight of Unknown Liquid)(1.5 m) = 65 KN/m² - 42.5 KN/m²

(Specific Weight of Unknown Liquid) = (22.5 KN/m²)/1.5 m

<u>Specific Weight of Unknown Liquid = 15 KN/m³</u>  

4 0
3 years ago
Find the velocity and acceleration of box B when point A moves vertically 1 m/s and it is 5 m
Goshia [24]

Answer:

hshbdhehdjsbdjdissasoe

7 0
3 years ago
If the bending moment (M) is 4,176 ft-lb and the beam is an 1 beam, calculate the bending stress (psi) developed at a point with
SpyIntel [72]

Answer:

Bending stress at point 3.96 is \sigma_b = 1.37 psi

Explanation:

Given data:

Bending Moment M is 4.176 ft-lb = 50.12 in- lb

moment of inertia I = 144 inc^4

y = 3.96 in

\sigma_b = \frac{M}{I} \times y

putting all value to get bending stress

\sigma_b = \frac{50.112}{144} \times 3.96  

\sigma_b =  1.37 psi

Bending stress at point 3.96 is \sigma_b = 1.37 psi

3 0
3 years ago
The substance is steam (H2O). NOTE: The purpose of this problem is to illustrate that there are conditions where water vapor is
Gennadij [26K]

Answer:

See the attached pictures for detailed answer.

Explanation:

See the attached pictures for step by step explanation.

7 0
3 years ago
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
Galina-37 [17]

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");

  }

}

3 0
3 years ago
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