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natta225 [31]
3 years ago
7

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
Engineering
1 answer:
aalyn [17]3 years ago
3 0

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);

   }

}

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Describe the greatest power in design according to Aravena?
KiRa [710]

The greatest power in design according to Aravena is "the power of synthesis”.

Hope that helps!

8 0
2 years ago
Lately, you have noticed some repetitive stress in your wrist. Which sign is most likely the cause of that stress and pain?
Cerrena [4.2K]
1, you might have been carrying things that are way too heavy for you.
2, you might have weak tendons.
3 0
3 years ago
A gas has an initial volume o.25m^3, and absolute pressure 100kPa. Its initial temperature is 290k. The gas is compressed into a
dezoksy [38]

Answer:

<h2>698.3Kpa</h2>

Explanation:

Step one:

given data

V1=0.25m^3

T1=290k

P1=100kPa

V2=0.5m^2

T2=405k

P2=? final pressure

Step two:

The combined gas equation is given as

P1V1/T1=P2V2/T2

Substituting we have

(100*0.25)/290=P2*0.05/405

25/290=0.5P2/405

0.086=0.05P2/405

cross multiply

0.086*405=0.05P2

34.9=0.05P2

divide both sides by 0.05

P2=34.9/0.05

P2=698.3Kpa

<u>Therefore the new pressure is 698.3Kpa when the gas is compressed</u>

5 0
3 years ago
A circular section of material is tested. The original specimen is 200 mm long and has a diameter of 13 mm. When loaded to its p
n200080 [17]

Answer:

modulus of elasticity = 100.45 Gpa,

proportional limit = 150.68 N/mm^2.

Explanation:

We are given the following parameters or data in the question as;

=> "The original specimen = 200 mm long and has a diameter of 13 mm."

=> "When loaded to its proportional limit, the specimen elongates by 0.3 mm."

=> " The total axial load is 20 kN"

Step one: Calculate the area

Area = π/ 4 × c^2.

Area = π/ 4 × 13^2 = 132.73 mm^2.

Step two: determine the stress induced.

stress induced = load/ area= 20 × 1000/132.73 = 150.68 N/mm^2.

Step three: determine the strain rate:

The strain rate = change in length/original length = 0.3/ 200 = 0.0015.

Step four: determine the modulus of elasticity.

modulus of elasticity = stress/strain = 150.68/0.0015 = 100453.33 N/mm^2 = 100.45 Gpa.

Step five: determine the proportional limit.

proportional limit = 20 × 1000/132.73 = 150.68 N/mm^2.

7 0
3 years ago
Read 2 more answers
The section should span between 10.9 and 13.4 cm (4.30 and 5.30 inches) as measured from the end supports and should be able to
Sergeeva-Olga [200]

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

= \frac{Wl^2}{12}  =  (0.013 * 0.0144 )/ 12  =  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

5 0
3 years ago
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