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 system with a mass of 8 kg, initially moving horizontally with a velocity of 40 m/s, experiences a constant horizontal force of 25 N opposing the direction of motion. As a result, the system comes to rest.Determine the amount of energy transfer by work, in kJ, for this process and the total distance, in m, that the system travels.
Answers
Answer:
V₂=1.76 m³
P=222.03 KPa
Explanation:
Given that
For tank 1
V₁=1 m³
T₁= 10°C = 283 K
P₁=350 KPa
For tank 2
m₂=3 kg
T₂=35°C = 308 K
P₂=150 KPa
We know that for air
P V = m R T
P=pressure ,V= Volume,R= gas constant ,T= temperature ,m =mass
for tank 2
P₂ V₂ = m₂ R T₂
By putting the values
150 x V₂ = 3 x 0.287 x 308
V₂=1.76 m³
Final mass = m₁+m₂
m =m₁+m₂
The final volume V= V₂+V₁
V= 1.76 + 1 m³
V= 2.76 m³
The final temperature T= 19.5°C
T= 292.5 K
m =m₁+m₂
m =4.3 + 3 = 7.3 kg
Now at final state
P V = m R T
P x 2.76 = 7.3 x 0.287 x 292.5
P=222.03 KPa
Answers
Answer:
Explanation:
From the question we are told that:
Voltage
Power
Initial Power factor
Final Power factor
Generally the equation for Reactive Power is mathematically given by
Q=P(tan \theta_2-tan \theta_1)
Since
And
Therefore
Therefore
The size of the capacitor in vars that is necessary to raise the power factor to 0.9 lagging is
Answers
Answer:
Explanation:
Given that,
Average fiber diameter is 0.01mm
d = 0.01mm = 1 × 10^-5m
The average fiber length is 2.5mm
L = 2.5mm = 0.0025m
Volume of the fraction of fibers is 0.40
Vf = 0.40
Fiber matrix bond strengths is 75MPa
τ = 75 MPa
The fraction strength of the fibers is 3500 Mpa
σf = 3500 MPa
The matrix street fiber is 8 MPa
σm = 8 MPa
We need to find the critical fiber length and compare it to original fiber length
Ic = σf•d / 2τ
Ic = 3500 × 0.01 / 75 × 2
Ic = 0.233 mm
Since the critical fiber length of 0.233 mm is much less than the provided length of the fiber (2.5mm) , so we can use the following equation to find the longitudinal tensile strength
σcd = σf•Vf(1—Ic / 2L) + σm(1—Vf)
σcd = 3500×0.4[1—0.233/(2 × 2.5)] + 8(1—0.4)
σcd = 1400(1—0.0467) + (8 × 0.6)
σcd = 1334.67 + 4.8
σcd = 1339.47 MPa
The longitudinal tensile strength of the aligned glass fiber-epoxy matrix composite is 1339.47 MPa
Answers
Answer:
0.556 Watts
Explanation:
w = Weight of object = 762 N
s = Distance = 5 m
t = Time taken = 29 seconds
a = Acceleration
g = Acceleration due to gravity = 9.81 m/s²
Equation of motion
Mass of the body
Force required to move the body
Velocity of object
Power
∴ Amount of power required to move the object is 0.556 Watts
b. The change in diameter of the specimen. Indicate an increase in diameter with a positive number and a decrease with a negative number.
Answers
Answer:
1)ΔL = 0.616 mm
2)Δd = 0.00194 mm
Explanation:
We are given;
Force; F = 52900 N
Initial length; L_o = 207 mm = 0.207 m
Diameter; d_o = 19.2 mm = 0.0192 m
Elastic modulus; E = 61.4 GPa = 61.4 × 10^(9) N/m²
Now, from Hooke's law;
E = σ/ε
Where; σ is stress = force/area = F/A
A = πd²/4 = π × 0.0192²/4
A = 0.00009216π
σ = 52900/0.00009216π
ε = ΔL/L_o
ε = ΔL/0.207
Thus,from E = σ/ε, we have;
61.4 × 10^(9) = (52900/0.00009216π) ÷ (ΔL/0.207)
Making ΔL the subject, we have;
ΔL = (52900 × 0.207)/(61.4 × 10^(9) × 0.00009216π)
ΔL = 0.616 × 10^(-3) m
ΔL = 0.616 mm
B) Poisson's ratio is given as;
υ = ε_x/ε_z
ε_x = Δd/d_o
ε_z = ΔL/L_o
Thus;
υ = (Δd/d_o) ÷ (ΔL/L_o)
Making Δd the subject gives;
Δd = (υ × d_o × ΔL)/L_o
We are given Poisson's ratio to be 0.34.
Thus;
Δd = (0.34 × 19.2 × 0.616)/207
Δd = 0.00194 mm
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