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NeTakaya
2 years ago
7

What major problems could you encounter in complex intersections?

Engineering
1 answer:
Natali5045456 [20]2 years ago
3 0

Types of Problems

Inappropriate intersection traffic control.

Inadequate visibility of the intersection or regulatory traffic control devices.

Inadequate intersection sight distance.

Inadequate guidance for motorists.

Excessive intersection conflicts within or near the intersection.

Vehicle conflicts with non-motorists.

You might be interested in
Consider a fan located in a 3 ft by 3 ft square duct. Velocities at various points at the outlet are measured, and the average f
natulia [17]

Answer:

minimum electric power consumption of the fan motor is 0.1437 Btu/s

Explanation:

given data

area = 3 ft by 3 ft

air density = 0.075 lbm/ft³

to find out

minimum electric power consumption of the fan motor

solution

we know that energy balance equation that is express as

E in - E out  = \frac{dE \ system}{dt}    ......................1

and at steady state  \frac{dE \ system}{dt} = 0

so we can say from equation 1

E in = E out

so

minimum power required is

E in = W = m \frac{V^2}{2} = \rho A V \frac{V^2}{2}  

put here value

E in =  \rho A V \frac{V^2}{2}  

E in =  0.075 *3*3* 22 \frac{22^2}{2}  

E in = 0.1437 Btu/s

minimum electric power consumption of the fan motor is 0.1437 Btu/s

5 0
2 years ago
Your Java program will be reading input from a file name strInput.txt. Each record contains String firstname String lastName Str
stiks02 [169]

Answer:

The program requires that you have the specified input files and it reads from each file at a time and processes salary in digits, states the city, state and bonus with respective first and last name as requested in the question. Note that you must have access to the mentioned output files for the program to work properly. Below is the java version of the program.

import java.io.File;

import java.io.FileNotFoundException;

import java.io.PrintWriter;

import java.util.Scanner;

class Driver

{

public static void main(String[] args) throws FileNotFoundException

{

Scanner sc = new Scanner(new File("strInput.txt"));

PrintWriter pd = new PrintWriter(new File("strOutputD"));

PrintWriter prf = new PrintWriter(new File("strOutputRF"));

String firstname = "", lastname = "", strSalary = "", status = "", cityState = "", city = "", state = "";

double salary = 0, bonus = 0;

int incorrectRecords = 0;

int dRecords = 0;

int fRecords = 0;

while(sc.hasNextLine())

{

firstname = sc.next();

lastname = sc.next();

strSalary = sc.next();

status = sc.next();

cityState = sc.next();

if(!status.equals("D") && !status.equals("F"))

{

System.out.println("Records is neither D nor F. Skipping this...");

incorrectRecords++;

continue;

}

else if(status.equals("D") || status.equals("F"))

{

char c = ' ';

int i = 0;

for(i=0; i<strSalary.length() && c != '.'; i++)

{

c = strSalary.charAt(i);

if(!Character.isDigit(c))

{

System.out.println("Char at position " + (i+1) + " in salary is not a digit");

incorrectRecords++;

continue;

}

}

if(c == '.')

{

if(i+1 == strSalary.length()-1)

{

if(!Character.isDigit(strSalary.charAt(i)))

{

System.out.println("Char at position " + (i+1) + " in salary is not a digit");

incorrectRecords++;

continue;

}

if(!Character.isDigit(strSalary.charAt(i+1)))

{

System.out.println("Char at position " + (i+1+1) + " in salary is not a digit");

incorrectRecords++;

continue;

}

}

else

{

System.out.println("Period is in the wrong position. Expected at " + (strSalary.length()-3) + " but found at " + (i+1));

continue;

}

}

city = cityState.split(",")[0];

state = cityState.split(",")[1];

salary = Double.parseDouble(strSalary);

if(status.equals("D"))

{

bonus = salary * 0.125;

dRecords++;

pd.write(firstname + " " + lastname + " " + status + " " + salary + " " + bonus + " " + city + " " + state);

}

else

{

bonus = salary * 0.18;

fRecords++;

prf.write(firstname + " " + lastname + " " + status + " " + salary + " " + bonus + " " + city + " " + state);

}

}

}

System.out.println("No of D records : " + dRecords);

System.out.println("No of F records : " + fRecords);

System.out.println("No of incorrect records : " + incorrectRecords);

}

}

6 0
2 years ago
The greater the force applied to an object, the _____ the change in speed or direction of the object.
storchak [24]

Answer:

b

Explanation:

8 0
2 years ago
10. An engineer is designing a total hip implant. She intends to make the femoral stem out of titanium because it forms a good i
creativ13 [48]

Answer:

Yes. She should be worried about corrosion. The 18-8 stainless exhibits intergranular corrosion due to high (0.08%) carbon content and gross pitting due to low molybdenum content.

Explanation: lol

8 0
3 years ago
An ideal gas initially at 300 K and 1 bar undergoes a three-step mechanically reversible cycle in a closed system. In step 12, p
Veseljchak [2.6K]

Answer:

Ts =Ta E)- 300(

569.5 K

5

Q12-W12 = -4014.26

Mol

AU2s = Q23= 5601.55

Mol

AUs¡ = Ws¡ = -5601.55

Explanation:

A clear details for the question is also attached.

(b) The P,V and T for state 1,2 and 3

P =1 bar Ti = 300 K and Vi from ideal gas Vi=

10

24.9x10 m

=

P-5 bar

Due to step 12 is isothermal: T1 = T2= 300 K and

VVi24.9 x 10x-4.9 x 10-3 *

The values at 3 calclated by Uing step 3l Adiabatic process

B-P ()

Since step 23 is Isochoric: Va =Vs= 4.99 m* and 7=

14

Ps-1x(4.99 x 103

P-1x(29x 10)

9.49 barr

And Ts =Ta E)- 300(

569.5 K

5

(c) For step 12: Isothermal, Since AT = 0 then AH12 = AU12 = 0 and

Work done for Isotermal process define as

8.314 x 300 In =4014.26

Wi2= RTi ln

mol

And fromn first law of thermodynamic

AU12= W12 +Q12

Q12-W12 = -4014.26

Mol

F'or step 23 Isochoric: AV = 0 Since volume change is zero W23= 0 and

Alls = Cp(L3-12)=5 x 8.311 (569.5 - 300) = 7812.18-

AU23= C (13-72) =5 x 8.314 (569.3 - 300) = 5601.53

Inol

Now from first law of thermodynamic the Q23

AU2s = Q23= 5601.55

Mol

For step 3-1 Adiabatic: Since in this process no heat transfer occur Q31= 0

and

AH

C,(T -Ts)=x 8.314 (300- 569.5)= -7842.18

mol

AU=C, (T¡-T)= x 8.314 (300

-5601.55

569.5)

mol

Now from first law of thermodynamie the Ws1

J

mol

AUs¡ = Ws¡ = -5601.55

3 0
3 years ago
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