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Aleks [24]
3 years ago
6

Primary U.S. interstate highways are numbered 1-99. Odd numbers (like the 5 or 95) go north/south, and evens (like the 10 or 90)

go east/west. Auxiliary highways are numbered 100-999, and service the primary highway indicated by the rightmost two digits. Thus, the 405 services the 5, and the 290 services the 90 Given a highway number, indicate whether it is a primary or auxiliary highway. If auxiliary, indicate what primary highway it serves. Also indicate if the (primary) highway runs north/south or east/west. Ex: If the input is: 90 the output is: The 90 is primary, going east/west. Ex: If the input is: 290 the output is: The 290 is auxiliary, serving the 90, going east/west Ex: If the input is G here to search The 290 is auxiliary, serving the 90, going eas Ex: If the input is: 0 or any number not between 1 and 999, the output is: 0 is not a valid interstate highway number. See Wikipedia for more info on highway numbering LAB 4.16.1: LAB: Interstate highway numbers ACTIVITY LabProgr- 1 import java.util.Scanner; 2 3 public class LabProgram public static void main(Stringl] args) t Scanner scnr new Scanner(System.in); int highwayNumber; int primaryNumber; 4 5 6 7 highwayNumber scnr.nextInt(); 1e Type your code
Engineering
1 answer:
Gekata [30.6K]3 years ago
6 0

Answer:

The Java code is given below with appropriate variable names and tags for better understanding

Explanation:

import java.util.Scanner;

public class LabProgram{

   public static void main(String[] args) {

       Scanner scnr = new Scanner(System.in);

       int highwayNumber;

       int primaryNumber;

       highwayNumber = scnr.nextInt();

       if(highwayNumber<1 || highwayNumber>999)

           System.out.println(highwayNumber+" is not a valid interstate highway number.");

       else{

           if(highwayNumber>=1 && highwayNumber<=99){

               System.out.print("The "+highwayNumber+" is primary, going ");

               if(highwayNumber%2==1)

                   System.out.println("north/south.");

               else

                   System.out.println("east/west.");

           }

           else{

               primaryNumber = highwayNumber%100;

               System.out.print("The "+highwayNumber+" is auxillary, serving the "+primaryNumber+", going ");

               if(primaryNumber%2==1)

                   System.out.println("north/south.");

               else

                   System.out.println("east/west.");

           }

       }

   }

}

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Using the Rayleigh criterion, calculate the minimum feature size that can be resolved in a system with a 0.18 NA lens when g-lin
Vladimir79 [104]

Answer:

a)

# for a g line, R = 1.211 μm

# for an I-line, R = 1.013 μm

b)

# for a g line, R = 0.726 μm

# for an I-line, R = 0.243 μm

c)

# for a g line, R = 0.605 μm

# for an I-line, R = 0.608 μm

Explanation:

We know that;

Rayleigh Resolution R = 0.5 × λ/NA

for a g line, λ = 436 nm

for an I-line λ = 365 nm

a)

Now when NA = 0.18

# for a g line, λ = 436 nm

R = 0.5 × 436/0.18 =  1.211 μm

# for an I-line λ = 365 nm

R = 0.5 × 365/0.18 =  1.013 μm

b)

when NA = 0.30

# for a g line, λ = 436 nm

R = 0.5 × 436/0.30 =  0.726 μm

# for an I-line λ = 365 nm

R = 0.5 × 365/0.30 =  0.243 μm

c)

when NA = 0.36

# for a g line, λ = 436 nm

R = 0.5 × 436/0.36 =  0.605 μm

# for an I-line λ = 365 nm

R = 0.5 × 365/0.30 =  0.608 μm

6 0
3 years ago
The process _______ boosted the production of fertilizers and strengthened the agriculture industry.
ANTONII [103]

Answer:

Process of Green Revolution

Explanation:

Green Revolution is a process which started its initiatives between the 1950s and 1960s, a technological research process, that increase agricultural productions through various systematic approach, which are different from the traditional methods. These technological approach includes:

1.  Utilization of new varieties with a high yield potential, in addition with adequate water  supply, pesticides and fertilizers.

2.  New methods of cultivation and mechanization

Hence, the process of Green Revolution boosted production of fertilizers and strengthen the agriculture industry as a whole.

6 0
3 years ago
A company buys a machine for $12,000, which it agrees to pay for in five equal annual payments, beginning one year after the dat
Yuki888 [10]

Answer:

$7,778.35

Explanation:

At year 3, the final payment of the remaining balance is equal to the present worth P of the last three payments.

First, calculate the uniform payments A:

A = 12000(A/P, 4%, 5)

= 12000(0.2246) = 2695.2  (from the calculator)

Then take the last three payments as its own cash flow.

To calculate the new P:

P = 2695.2 + 2695.2(P/A, 4%, 2) = 2695.2 + 2695.2(1.886) = 7778.35

Therefore, the final payment is $7,778.35

4 0
4 years ago
Which of the following is NOT a provision of the Agricultural Adjustment Act? insurance for farmers alternative energy programs
Arisa [49]

Answer:

Farm equipment

Explanation:

Most people have heard claims that the US government pays farmers not to grow crops. The Agricultural Adjustment Act is the legislation that started this program. It was the first “Farm Bill.” The current farm bill provides for the following:

Subsidies for farmers

Insurance for farmers

Price supports

Food assistance for economically challenged Americans (the largest portion of the Farm Bill)

Forestry conservation programs

Alternative energy programs.

4 0
3 years ago
A 50 Hz, four pole turbo-generator rated 100 MVA, 11 kV has an inertia constant of 8.0 MJ/MVA. (a) Find the stored energy in the
raketka [301]

Given Information:

Frequency = f = 60 Hz

Complex rated power = G = 100 MVA

Intertia constant = H = 8 MJ/MVA

Mechanical power = Pmech = 80 MW

Electrical power = Pelec = 50 MW

Number of poles = P = 4

No. of cycles = 10

Required Information:

(a) stored energy = ?

(b) rotor acceleration = ?

(c) change in torque angle = ?

(c) rotor speed = ?

Answer:

(a) stored energy = 800 Mj

(b) rotor acceleration = 337.46 elec deg/s²

(c) change in torque angle (in elec deg) = 6.75 elec deg

(c) change in torque angle (in rmp/s) = 28.12 rpm/s

(c) rotor speed = 1505.62 rpm

Explanation:

(a) Find the stored energy in the rotor at synchronous speed.

The stored energy is given by

E = G \times H

Where G represents complex rated power and H is the inertia constant of turbo-generator.

E = 100 \times 8 \\\\E = 800 \: MJ

(b) If the mechanical input is suddenly raised to 80 MW for an electrical load of 50 MW, find rotor acceleration, neglecting mechanical and electrical losses.

The rotor acceleration is given by

$ P_a = P_{mech} - P_{elec} = M \frac{d^2 \delta}{dt^2}  $

Where M is given by

$ M = \frac{E}{180 \times f} $

$ M = \frac{800}{180 \times 50} $

M = 0.0889 \: MJ \cdot s/ elec \: \: deg

So, the rotor acceleration is

$ P_a = 80 - 50 = 0.0889 \frac{d^2 \delta}{dt^2}  $

$  30 = 0.0889 \frac{d^2 \delta}{dt^2}  $

$   \frac{d^2 \delta}{dt^2} = \frac{30}{0.0889}  $

$   \frac{d^2 \delta}{dt^2} = 337.46 \:\: elec \: deg/s^2 $

(c) If the acceleration calculated in part(b) is maintained for 10 cycles, find the change in torque angle and rotor speed in revolutions per minute at the end of this period.

The change in torque angle is given by

$ \Delta  \delta = \frac{1}{2} \cdot \frac{d^2 \delta}{dt^2}\cdot (t)^2 $

Where t is given by

1 \: cycle = 1/f = 1/50 \\\\10 \: cycles = 10/50 = 0.2  \\\\t = 0.2 \: sec

So,

$ \Delta  \delta = \frac{1}{2} \cdot 337.46 \cdot (0.2)^2 $

$ \Delta  \delta = 6.75 \: elec \: deg

The change in torque in rpm/s is given by

$ \Delta  \delta = \frac{337.46 \cdot 60}{2 \cdot 360\circ  }   $

$ \Delta  \delta =28.12 \: \: rpm/s $

The rotor speed in revolutions per minute at the end of this period (10 cycles) is given by

$ Rotor \: speed = \frac{120 \cdot f}{P}  + (\Delta  \delta)\cdot t  $

Where P is the number of poles of the turbo-generator.

$ Rotor \: speed = \frac{120 \cdot 50}{4}  + (28.12)\cdot 0.2  $

$ Rotor \: speed = 1500  + 5.62  $

$ Rotor \: speed = 1505.62 \:\: rpm

4 0
4 years ago
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