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Temka [501]
4 years ago
9

Two cars A and B leave an intersection at the same time. Car A travels west at an average speed of x miles per hour and car B tr

avels south at an average speed of y miles per hour. Write a program that prompts the user to enter the average speed of both the cars and the elapsed time (in hours and minutes) and outputs the (shortest) distance between the cars.
Engineering
1 answer:
svet-max [94.6K]4 years ago
7 0

Answer:

Distance between A & B, dAB, is hypotenus.

dAB²=dA²+dB²

A'=70 mph, B'=55mph, (convert t to hours)

dA=A'•t, dB=B'•t

dAB=√((70t)²+(55t)²)

Explanation:

Distance between A & B, dAB, is hypotenus.

dAB²=dA²+dB²

A'=70 mph, B'=55mph, (convert t to hours)

dA=A'•t, dB=B'•t

dAB=√((70t)²+(55t)²)

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A 10.2 mm diameter steel circular rod is subjected to a tensile load that reduces its cross- sectional area to 52.7 mm^2. Determ
VMariaS [17]

Answer:

The percentage ductility is 35.5%.

Explanation:

Ductility is the ability of being deform under applied load. Ductility can measure by percentage elongation and percentage reduction in area. Here, percentage reduction in area method is taken to measure the ductility.

Step1

Given:

Diameter of shaft is 10.2 mm.

Final area of the shaft is 52.7 mm².

Calculation:

Step2

Initial area is calculated as follows:

A=\frac{\pi d^{2}}{4}

A=\frac{\pi\times(10.2)^{2}}{4}

A = 81.713 mm².

Step3

Percentage ductility is calculated as follows:

D=\frac{A_{i}-A_{f}}{A_{i}}\times100

D=\frac{81.713-52.7}{81.713}\times100

D = 35.5%.

Thus, the percentage ductility is 35.5%.

5 0
3 years ago
Anaircraft component is fabricated from an aluminum alloy that has a plane-strain fracture toughness of 40 MPa 1/2.It has been d
navik [9.2K]

Answer:

Yes, fracture will occur since toughness (42.4 MPa) is greater than the toughness of the material, 40MPa

Explanation:

Given

Toughness, k = 40Mpa

Stress, σ = 300Mpa

Length, l = 4mm = 4 * 10^-3m

Under which fracture occurred (i.e., σ= 300 MPa and 2a= 4.0 mm), first we solve for parameter Y (The dimensionless parameter)

Y = k/(σπ√a)

Where a = ½ of the length in metres

Y = 40/(300 * π * √(4/2 * 10^-3))

Y = 1.68 ---- Approximated

To check if fracture will occur of not; we apply the same formula.

Y = k/(σπ√a)

Then we solve for k, where

σ = 260Mpa and a = ½ * 6 * 10^-3

So,.we have

1.68 = k/(260 * π * √(6*10^-3)/2)

k = 1.68 * (260 * π * (6*10^-3)/2)

k = 42.4 MPa --- Approximately

Therefore, fracture will occur since toughness (42.4 MPa) is greater than the toughness of the material, 40 MPa

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3 years ago
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Answer:

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Explanation:

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3 years ago
import java.util.Scanner; public class FindSpecialValue { public static void main (String [] args) { Scanner scnr = new Scanner(
Hitman42 [59]

Answer:

Java program explained below

Explanation:

FindSpecialNumber.java

import java.util.Scanner;

public class FindSpecialNumber {

public static void main(String[] args) {

//Declaring variable

int number;

/*

* Creating an Scanner class object which is used to get the inputs

* entered by the user

*/

Scanner sc = new Scanner(System.in);

//getting the input entered by the user

System.out.print("Enter a number :");

number = sc.nextInt();

/* Based on user entered number

* check whether it is special number or not

*/

if (number == -99 || number == 0 || number == 44) {

System.out.println("Special Number");

} else {

System.out.println("Not Special Number");

}

}

}

_______________

Output#1:

Enter a number :-99

Special Number

Output#2:

Enter a number :49

Not Special Number

7 0
3 years ago
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