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3 years ago

8

The flow rate on an arterial is 1,800 veh/h, evenly distributed over two lanes. If the average speed in these lanes is 40 mi/h,

what is the density in veh/h/ln?

1 answer:

ss7ja [257]3 years ago

8 0

Using

q = k*v,

where q = traffic flow rate, k = density, v = space mean speed.

so, 1800 = k x 40

so, k = 1800/40 = 45 veh/mie on two lanes,

so k = 45/2 = 22.5veh/h/in

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Brainly and points if you want

Tju [1.3M]

Answer:

thank you

Explanation:

have a nice day

8 0

2 years ago

Read 2 more answers

If gain of the critically damped system is increased, the system will behave as a) Under damped b) Over damped c) Critically dam

Ganezh [65]

Answer:

a) Under damped

Explanation:

Given that system is critically damped .And we have to find out the condition when gain is increased.

As we know that damping ratio given as follows

$\zeta =\dfrac{C}{C_c}$

Where C is the damping coefficient and Cc is the critical damping coefficient.

$C_c=2\sqrt{mK}$

So from above we can say that

$\zeta =\dfrac{C}{2\sqrt{mK}}$

$\zeta \alpha \dfrac{1}{\sqrt K}$

From above relationship we can say when gain (K) is increases then system will become under damped system.

7 0

3 years ago

The cubic capacity of a four-stroke over-square spark-ignition engine is 245cc. The over-square ratio is (1.1) The clearance vol

pashok25 [27]

Answer:

Bore = 7 cm

stroke = 6.36 cm

compression ratio = 10.007

Explanation:

Given data:

Cubic capacity of the engine, V = 245 cc

Clearance volume, v = 27.2 cc

over square-ratio = 1.1

thus,

D/L = 1.1

where,

D is the bore

L is the stroke

Now,

V = $\frac{\pi}{4}D^2L$

or

V = $\frac{\pi}{4}\frac{D^3}{1.1}$

on substituting the values, we have

245 = $\frac{\pi}{4}\frac{D^3}{1.1}$

or

D = 7.00 cm

Now,

we have

D/L = 1.1

thus,

L = D/1.1

L = 7/1.1

or

L= 6.36 cm

Now,

the compression ratio is given as:

$\textup{compression ratio}=\frac{V+v}{v}$

on substituting the values, we get

$\textup{compression ratio}=\frac{245+27.2}{27.2}$

or

Compression ratio = 10.007

4 0

3 years ago

The 240-ft structure is used to provide various support services to launch vehicles prior to liftoff. In a test, a 12-ton weight

alina1380 [7]

Answer:

hello your question lacks the required question attached below is the missing diagram

Forces in GJ = -4.4444 i.e. 4.4444 tons

Forces in IG = 15.382 tons ( T )

Explanation:

Forces in GJ = -4.4444 i.e. 4.4444 tons

Forces in IG = 15.382 tons ( T )

attached below is the detailed solution

3 0

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

2 years ago