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jek_recluse [69]
3 years ago
13

Light from a helium-neon laser (λ = 633 nm) illuminates two slits spaced 0.50 mm apart. A viewing screen is 2.5 m behind the sli

ts. What is the spacing between two adjacent bright fringes?
Engineering
1 answer:
serg [7]3 years ago
7 0

Answer:

Explanation:

Given

wavelength \lambda =633\ nm

distance between two slits d=0.5\ mm

Screen is placed at a distance L=2.5\ m

Location of a (n+1)th bright fringe is given by

x_{n+1}=\frac{L}{2d}(n+1)\lambda

for nth bright fringe

x_n=\frac{L}{2d}(n)\lambda

Distance between two bright fringes

x_{n+1}-x_n=\frac{L}{2d}\cdot \lambda

x_{n+1}-x_n=\frac{2.5}{2\times 0.5\times 10^{-3}}\cdot 633\times 10^{-3}

x_{n+1}-x_n=1.582\ mm

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Define the coefficient of determination and discuss the impact you would expect it to have on your engineering decision-making b
scoundrel [369]

Answer and Explanation:

The coefficient of determination also called "goodness of fit" or R-squared(R²) is used in statistical measurements to understand the relationship between two variables such that changes in one variable affects the other. The level of relationship or the degree to which one affects the other is measured by 0 to 1 whereby 0 means no relationship at all and 1 means one totally affects the other while figures in between such 0.40 would mean one variable affects 40% of the other variable.

In making a decision as an engineer while using the coefficient of determination, one would try to understand the relationship between variables under consideration and make decisions based on figures obtained from calculating coefficient of determination. In other words when there is a 0 coefficient then there is no relationship between variables and an engineer would make his decisions with this in mind and vice versa.

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3 years ago
Payment to beneficiaries who were named by<br> the insured person
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4 0
4 years ago
A sheet of steel 3-mm thick has nitrogen atomospheres on both sides at 900 C and is permitted to achieve a steady-state di usion
kati45 [8]

Answer:

X_B = 1.8 \times 10^{-3} m = 1.8 mm

Explanation:

Given data:

Diffusion constant for nitrogen is = 1.85\times 10^{-10} m^2/s

Diffusion flux = 1.0\times 10^{-7} kg/m^2-s

concentration of nitrogen at high presuure = 2 kg/m^3

location on which nitrogen  concentration is 0.5 kg/m^3   ......?

from fick's first law

J = D \frac{C_A C_B}{X_A X_B}

Take C_A as point  on which nitrogen concentration is 2 kg/m^3

x_B = X_A + D\frac{C_A -C_B}{J}

Assume X_A is zero at the surface

X_B = 0 + ( 12\times 10^{-11} ) \frac{2-0.5}{1\times 10^{-7}}

X_B = 1.8 \times 10^{-3} m = 1.8 mm

4 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");

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}

3 0
3 years ago
A motorcycle starts from rest with an initial acceleration of 3 m/s^2, and the acceleration then changes with the distance s as
katrin2010 [14]

Answer:

Follows are the solution to this question:

Explanation:

Calculating the area under the curve:  

A = as

   =\frac{1}{2}(3 +6 \frac{m}{s^2})(100 \ m)+ \frac{1}{2}(6+4 \frac{m}{s^2})(100 m) \\\\=\frac{1}{2}(9 \frac{m}{s^2})(100 \ m)+ \frac{1}{2}(10\frac{m}{s^2})(100 m) \\\\=\frac{1}{2}(900 \frac{m^2}{s^2})+ \frac{1}{2}(1,000\frac{m^2}{s^2}) \\\\=(450 \frac{m^2}{s^2})+ (500\frac{m^2}{s^2}) \\\\= 950 \ \frac{m^2}{s^2}

Calculating the kinematics equation:

\to v^2 = v^2_{o} + 2as\\\\

        =0+ \sqrt{2as}\\\\ = \sqrt{2(A)}\\\\= \sqrt{2(950 \frac{m^2}{s^2})}\\\\= 43.59 \frac{m}{s}

Calculating the value of acceleration:  

\to a= \frac{dv}{dt}

=\frac{dv}{ds}(\frac{ds}{dt}) \\\\=v\frac{dv}{ds}\\\\\to \frac{dv}{ds}=\frac{a}{v}

\to \frac{dv}{ds} =\frac{4 \frac{m}{s^2}}{43.59 \frac{m}{s}} \\\\

         =\frac{0.092}{s}

3 0
3 years ago
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