List the RTL (Register Transfer Language) sequence of micro-operations needed to execute the instruction STORE X

from the MARIE instruction set architecture. Then write the corresponding signal sequence to perform these micro-operations and to reset the clock cycle counter.

You may refer to the provided "MARIE Architecture and Instruction Set" file in the Front Matter folder.

Answers

Answer 1
Answer:

Answer:

So these are the RTL representation:

MAR<----X

MBR<-----AC

M[MAR]<------MBR

Control signal sequence are:

P3T0:MAR<----X

P2P3 P4T1:MBR<-----AC

P0P1 P3T2:M[MAR]<------MBR

Explanation:

STORE X instruction is used for storing the value of AC to the memory address pointed by X. This operation can be done by using the Register Transfers at System Level and this can be represented by using a notation called Register Transfer Language RTL. Let us see what are the register transfer operations happening at the system level.

1. First of all the address X has to be tranfered on to the Memory Address Register MAR.  

MAR<----X

2. Next we have to tranfer the contents of AC into the Memory Buffer Register MBR

MBR<-----AC

3. Store the MBR into memory where MAR points to.

M[MAR]<------MBR

So these are the RTL representation:

MAR<----X

MBR<-----AC

M[MAR]<------MBR

Control signal sequence are:

P3T0:MAR<----X

P2P3 P4T1:MBR<-----AC

P0P1 P3T2:M[MAR]<------MBR


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Write two scnr.nextInt statements to get input values into birthMonth and birthYear. Then write a statement to output the month, a slash, and the year. End with newline. The program will be tested with inputs 1 2000 and then with inputs 5 1950. Ex: If the input is 1 2000, the output is: 1/2000

Answers

Answer:

import java.util.Scanner;

public class InputExample {

   public static void main(String[] args) {

       Scanner scnr = new Scanner(System.in);

       int birthMonth;

       int birthYear;

       birthMonth = scnr.nextInt();

       birthYear = scnr.nextInt();

       System.out.println(birthMonth+"/"+birthYear);

   }

}

import java.util.Scanner;

public class InputExample {

   public static void main(String[] args) {

       Scanner scnr = new Scanner(System.in);

       int birthMonth;

       int birthYear;

       // Get input values for birthMonth and birthYear from the user

       birthMonth = scnr.nextInt();

       birthYear = scnr.nextInt();

       // Output the month, a slash, and the year

       System.out.println(birthMonth + "/" + birthYear);

   }

}

When the program is tested with inputs 1 2000, the output will be:

1/2000

And when tested with inputs 5 1950, the output will be:

5/1950

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A power plant burns natural gas to supply heat to a heat engine which rejects heat to the adjacent river. The power plant produces 800 MW of electrical power and has a thermal efficiency of 38%. Determine the heat transfer rates from the natural gas and to the river, in MW.

Answers

A. The heat transfer rate from natural gas is 2105.26 MW

B. The heat transfer rate to river is 1305.26 MW

Efficiency formula

Efficiency = (power output / power input) × 100

A. How to determine the heat transfer from natural gas

  • Efficiency = 38%
  • Power output = 800 MW
  • Power input =?

Power input = Power input / efficiency

Power input = 800 / 38%

Power input = 800 / 0.38

Power input = 2105.26 MW

Thus, the heat transfer from natural gas is 2105.26 MW

B. How to determine the heat transfer to the river

  • Total heat = 2105.26 MW
  • Heat used by plant = 800 MW
  • Heat to the river =?

Heat to the river = 2105.26 – 800

Heat to the river = 1305.26 MW

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Answer:

heat transfer from natural gas is 2105.26 MW

heat transfer to river is 1305.26 MW

Explanation:

given data

power output Wn = 800 MW

efficiency = 38%

solution

we know that efficiency is express as

\eta = (Wn)/(Qin)    ......................1

put here value we get

38% = (800)/(Qin)  

Qin  = 2105.26 MW

so heat supply is 2105.26

so we can say

Wn = Qin - Qout

800 = 2105.26 - Qout

Qout = 2105.26 - 800

Qout = 1305.26 MW

so heat transfer from natural gas is 2105.26 MW

and heat transfer to river is 1305.26 MW

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b.600x10-6V



c.800x10-6V



d.1000 x 10-6 V

Answers

Answer:

a.400 x 10-6 V

Explanation:

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Answers

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An environmental engineer is considering three methods for disposing of a nonhazardous chemical sludge: land application, fluidized-bed incineration, and private disposal contract. The estimates for each method are below. (a) Determine which has the least cost on the basis of an annual worth comparison at 10% per year. (b) Determine the equivalent present worth value of each alternative using its AW value

Answers

Answer:

Explanation:

The annual worth of land application = $ 110,000 ( A/P , 10% , 5 years ) + $ 95,000 - $ 15,000 (A/F , 10% , 5 years )

The annual worth of land application = $ 110,000 X 0.263797 + $ 95,000 - $ 15,000 X 0.163797

The annual worth of land application = $ 29,017.54 + $ 95,000 - $ 2,456.95

The annual worth of land application = $ 121,560.59

The annual worth of land Incineration = $ 800,000 ( A/P , 10% , 6 years ) + $ 60,000 - $ 250,000 X (A/F , 10% , 6 years )

The annual worth of land Incineration = $ 800,000 X 0.229607 + $ 60,000 - $ 250,000 X 0.129607

The annual worth of land Incineration = $ 211,283.85

The annual worth of contract = $ 190,000

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Answers

Answer:

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