Check engine light is illuminated on the instrument panel. what action should you take

Assume we have already defined a variable of type String called password with the following line of code: password' can have any

omeli [17]

Answer:

The Java code is given below with appropriate comments for better understanding

Explanation:

import java.util.Scanner;

public class ValidatePassword {

public static void main(String[] args) {

Scanner input = new Scanner(System.in);

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

String password = input.nextLine();

int count = chkPswd(password);

if (count >= 3)

System.out.println("Secure");

else

System.out.println("Not Secure");

}

public static int chkPswd(String pwd) {

int count = 0;

if (checkForSpecial(pwd))

count++;

if (checkForUpperCasae(pwd))

count++;

if (checkForLowerCasae(pwd))

count++;

if (checkForDigit(pwd))

count++;

return count;

}

// checks if password has 8 characters

public static boolean checkCharCount(String pwd) {

return pwd.length() >= 7;

}

// checks if password has checkForUpperCasae

public static boolean checkForUpperCasae(String pwd) {

for (int i = 0; i < pwd.length(); i++)

if (Character.isLetter(pwd.charAt(i)) && Character.isUpperCase(pwd.charAt(i)))

return true;

return false;

}

public static boolean checkForLowerCasae(String pwd) {

for (int i = 0; i < pwd.length(); i++)

if (Character.isLetter(pwd.charAt(i)) && Character.isLowerCase(pwd.charAt(i)))

return true;

return false;

}

// checks if password contains digit

public static boolean checkForDigit(String pwd) {

for (int i = 0; i < pwd.length(); i++)

if (Character.isDigit(pwd.charAt(i)))

return true;

return false;

}

// checks if password has special char

public static boolean checkForSpecial(String pwd) {

String spl = "[email protected]#$%^&*(";

for (int i = 0; i < pwd.length(); i++)

if (spl.contains(pwd.charAt(i) + ""))

return true;

return false;

}

5 0

3 years ago

Suppose that a wing component on an aircraft is fabricated from an aluminum alloy that has a plane-strain fracture toughness of

Orlov [11]

Answer: 133.88 MPa approximately 134 MPa

Explanation:

Given

Plane strains fracture toughness, k = 26 MPa

Stress at which fracture occurs, σ = 112 MPa

Maximum internal crack length, l = 8.6 mm = 8.6*10^-3 m

Critical internal crack length, l' = 6 mm = 6*10^-3 m

We know that

σ = K/(Y.√πa), where

112 MPa = 26 MPa / Y.√[3.142 * 8.6*10^-3)/2]

112 MPa = 26 MPa / Y.√(3.142 * 0.043)

112 = 26 / Y.√1.35*10^-2

112 = 26 / Y * 0.116

Y = 26 / 112 * 0.116

Y = 26 / 13

Y = 2

σ = K/(Y.√πa), using l'instead of l and, using Y as 2

σ = 26 / 2 * [√3.142 * (6*10^-3/2)]

σ = 26 / 2 * √(3.142 *3*10^-3)

σ = 26 / 2 * √0.009426

σ = 26 / 2 * 0.0971

σ = 26 / 0.1942

σ = 133.88 MPa

8 0

3 years ago

Consider a very long rectangular fin attached to a flat surface such that the temperature at the end of the fin is essentially t

kodGreya [7K]

Answer:

$\frac{T-20}{130-20}= e^{-14.28*0.05}$

And if we solve for T we got:

$T= 20 + 110e^{-14.28*0.05} = 73.86 C$

The answer for this case would be $T = 73.86 C$ at 5cm from the base of the fin.

Explanation:

Data given

For this case we have the following data given:

$h = 20 \frac{W}{m^2 K}$ represent the heat transfer coefficient.

$p$ represent the perimeter for this case and would be given by:

$p = 2*0.05m +2*0.001m= 0.102m$

$k = 200 \frac{W}{m C}$ represent the thermal conductivity

$w = 5cm =0.05 m$ represent the width

$h = 1mm =0.001m$ represent the thickness

$A= wh= 0.05m *0.001m = 0.00005 m^2$

Solution to the problem

For this case we assume that we have steady conditions, the temperature of the fins varies just in one direction, the heat transfer coefficient not changes with the time and the thermal properties of the fin not change.

We can determine the temperature if the fin at x=5 cm=0.05 m from the base with the following formula:

$\frac{T-T_{\infty}}{T_b -T_{\infty}} = e^{-mx}$

Where m is a coefficient given by:

$m = \sqrt{\frac{hp}{kA}}=\sqrt{\frac{20 W/m^2 C 0.102 m}{200 W/ mC 0.00005 m^2}}= 14.28 m^{-1}$

The value of x for this case represent the distance $x =5 cm =0.05m$

$T_b =130 C$ represent the base temperature

$T_{\infty}= 20$ represent the temperature of the sorroundings or the ambient.

If we replace we have this:

$\frac{T-20}{130-20}= e^{-14.28*0.05}$

And if we solve for T we got:

$T= 20 + 110e^{-14.28*0.05} = 73.86 C$

The answer for this case would be $T = 73.86 C$ at 5cm from the base of the fin.

3 0

3 years ago

What engine does the mercedes 500e have

Strike441 [17]

Answer:

5.0 m119

Explanation:

It has the 5.o m119 v8 engine which produced 322 bhp.

5 0

3 years ago

You plan to install an active, liquid-based solar heating system for hot water. There are four candidate collector systems. Your

olchik [2.2K]

Solution:

The given formula,

$x=F_{R} U_{L} \times \frac{P l}{F R_{1}} \times\left(T_{r e f}-\bar{T}_{a}\right) \Delta t \times \frac{A_{c}}{L}$

$y=F_{R}(\tau \alpha)_{n} x \frac{F_{R}^{\prime}}{F_{R}} \times \frac{(\bar{\tau} d)}{(T d)_{n}} \times \bar{H}_{T} N \times \frac{A C}{L}$

$\frac{x}{y}=\frac{ u_{L} \times\left(T_{x t}-\bar{T}_{a}\right) \times \Delta t}{\left(\tau_{x}\right)_{h} \times\left(\frac{\bar{\tau}_{d}}{\left.| \tau_{d}\right)_{n}}\right) \times \bar{H}+N}$

From the table,

$1) $\quad x=2 \cdot 87, \quad y=0.96$\\$\frac{x}{y}=\frac{2187}{0.96}$22895\\\\2) $x=3 \cdot 466 \cdot y=6 \cdot 998$\\$\frac{x}{y}=\frac{3 \cdot 466}{0.898}$$=3 \cdot 4729$$

$3$x=3 \cdot 229, y=1 \cdot 08$\\$\frac{x}{x}=\frac{3 \cdot 229}{1 \cdot 08}$\\=2.9898\)\\\\4) $x=6.525, y=1.094$\\$\frac{x}{y}=\frac{5.625}{1.094}$\\=5.0502$

8 0

4 years ago