As a talented grad of CMPEN 270, your new job as a digital designer is to design a "greedy" vending machine that accepts only ni
ckels, dimes, and quarters. It is "greedy" because it does not give back any change! There is only one type of snack inside the vending machine which sells for 20 cents. To simplify the problem, assume only one coin is accepted at a time and generates the following inputs (XY) to the state machine. (USE T-Flip Flops ONLY)
Condition XY
No coins were put in the vending machine 00
One nickel was put in the vending machine 01
One dime was put in the vending machine 10
One quarter was put in the vending machine 11
Condition XY
No coins were put in the vending machine 00
One nickel was put in the vending machine 01
One dime was put in the vending machine 10
One quarter was put in the vending machine 11
1 answer:
Answer:
See explaination
Explanation:
In the below answer "Z" is the output. please kindly check attachment for further details.
When z=1 , the vending machine gives snacks
z=0 , the vending machine doesn't give snacks.
And z becomes 1( high) only when the machine gets 20 cents or more than 20 cents. Since the machine doesn't give change, after giving snacks it will restart from initial state A (where no coins were put).
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Answer:
a) 2622.903 N/m^3
b) 1.38233
c)4.878811765
Explanation:
Find the void ratio using the formula:
Here;
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is unit weight of water = 998 kg/m^3
is the moisture content = 0.17
is the void ratio
is the unit weight of soil = 14.9KN/m^3
Saturation Ratio Formula:
S is saturation rate
Substitute Eq 2 into Eq 1
Specific gravity of soil solids
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Answer:
Kindly note that, you're to replace "at" with shift 2 as the brainly text editor can't take the symbol
Explanation:
import javafx.application.Application;
import javafx.stage.Stage;
import javafx.scene.Group;
import javafx.scene.Scene;
import javafx.scene.layout.VBox;
import javafx.scene.layout.HBox;
import javafx.scene.control.TextField;
import javafx.scene.control.Button;
public class Calculator extends Application {
public static void main(String[] args) {
// TODO Auto-generated method stub
launch(args);
}
"at"Override
public void start(Stage primaryStage) throws Exception {
// TODO Auto-generated method stub
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VBox mainBox = new VBox();
HBox inpBox = new HBox();
TextField txtInput = new TextField ();
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txtInput.setStyle("-fx-font: 20 mono-spaced;");
txtInput.setText("0.0");
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txtInput.setMinWidth(200);
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mainBox.getChildren().add(inpBox);
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btn8.setMinHeight(50);
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btn9.setMinHeight(50);
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btn4.setMinHeight(50);
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btn5.setMinWidth(50);
btn5.setMinHeight(50);
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btn6.setMinHeight(50);
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btn1.setMinHeight(50);
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btn2.setMinHeight(50);
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btn3.setMinHeight(50);
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Answer:
hello below is missing piece of the complete question
minimum size = 0.3 cm
answer : 0.247 N/mm2
Explanation:
Given data :
section span : 10.9 and 13.4 cm
minimum load applied evenly to the top of span : 13 N
maximum load for each member ; 4.5 N
lets take each member to be 4.2 cm
Determine the max value of P before truss fails
Taking average value of section span ≈ 12 cm
Given minimum load distributed evenly on top of section span = 13 N
we will calculate the value of by applying this formula
= = 1.56 * 10^-5
next we will consider section ; 4.2 cm * 0.3 cm
hence Z (section modulus ) = BD^2 / 6
= ( 0.042 * 0.003^2 ) / 6 = 6.3*10^-8
Finally the max value of P( stress ) before the truss fails
= M/Z = ( 1.56 * 10^-5 ) / ( 6.3*10^-8 )
= 0.247 N/mm2