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1 year ago

Which type of Artificial Intelligence (AI) can repeatedly perform tasks of limited scope?

Engineering

1 answer:

seraphim [82]1 year ago

7 0

Narrow AI is the type of Artificial Intelligence (AI) can repeatedly perform tasks of limited scope.

Natural Language Understanding (NLU) is used to describe extracting information from unstructured text using algorithms.

<h3>What is Narrow AI ?</h3>

This is known to be a form or a type of AI that is said to be able to carry out a dedicated task as a result of its intelligence.

Note that Narrow AI is the type of Artificial Intelligence (AI) can repeatedly perform tasks of limited scope.

Natural Language Understanding (NLU) is used to describe extracting information from unstructured text using algorithms.

Learn more about Artificial Intelligence from

brainly.com/question/25523571

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Design a half-wave recti er which provides a peak voltage of 15 V, and anaverage voltage of 3.8 V when driven by a 120 V (rms) a

nirvana33 [79]

Answer:

You need a 120V to 24V commercial transformer (transformer 1:5), a 100 ohms resistance, a 1.5 K ohms resistance and a diode with a minimum forward current of 20 mA (could be 1N4148)

Step by step design:

Because you have a 120V AC voltage supply you need an efficient way to reduce that voltage as much as possible before passing to the rectifier, for that I recommend a standard 120V to 24V transformer. 120 Vrms = 85 V and 24 Vrms = 17V = Vin
Because 17V is not 15V you still need a voltage divider to step down that voltage, for that we use R1 = 100Ω and R2 = 1.3KΩ. You need to remember that more than 1 V is going to be in the diode, so for our calculation we need to consider it. Vf = (V*R2)/(R1+R2), V = Vin - 1 = 17-1 = 16V and Vf = 15, Choosing a fix resistance R1 = 100Ω and solving the equation we find R2 = 1.5KΩ
Finally to select the diode you need to calculate two times the maximum current and that would be the forward current (If) of your diode. Imax = Vf/R2 = 10mA and If = 2*Imax = 20mA

Our circuit meet the average voltage (Va) specification:

Va = (15)/(pi) = 4.77V considering the diode voltage or 3.77V without considering it

6 0

3 years ago

A pumping test was made in pervious gravels and sands extending to a depth of 50 ft. ,where a bed of clay was encountered. The n

Vikki [24]

Answer:per minute from the pumping well, a steady state was attained in about 24 hr. The draw-down at a distance of 10 ft. was 5.5 ft. and at 25 ft. was 1.21 ft.

Explanation:

6 0

2 years ago

if both the ram air input and drain hole of the pitot system become blocked, the indicated airspeed will

egoroff_w [7]

If both the ram air input and drain hole of the pitot system become blocked, the indicated airspeed will: a) increase during a climb.

<h3>What is a ram air input?</h3>

A ram air input can be defined as an air intake system which is designed and developed to use the dynamic air pressure that is created due to vehicular motion, or ram pressure, in order to increase the static air pressure within the intake manifold of an internal combustion engine of an automobile.

This ultimately implies that, a ram air input allows a greater mass-flow of air through the engine of an automobile, thereby, increasing the engine's power.

In conclusion, indicated airspeed will increase during a climb when both the ram air input and drain hole of the pitot system become blocked.

Read more on pilots here: brainly.com/question/10381526

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Complete Question:

If both the ram air input and drain hole of the pitot system become blocked, the indicated airspeed will

a) increase during a climb

b) decrease during a climb

c) remain constant regardless of altitude change

6 0

1 year ago

Which physical quantity measures fraction of the input energy a machine actually concerts into output energy ?

Dvinal [7]

The answer is Kinetic Power.

4 0

2 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;

}

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

2 years ago