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3 years ago

Show from the first principles that, for a perfect gas with constant specific heat capacities

Engineering

1 answer:

iVinArrow [24]3 years ago

5 0

The given question is incomplete as the equation given is not making sense.

From internet source i have taken the complete question and its been attached below.

The solution to the question is also attached in next image with complete detail.

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A westbound section of freeway currently has three 12-ft wide lanes, a 6-ft right shoulder, and no ramps within 3 miles upstream

Tresset [83]

Answer:

•Estimated density = 39.685Pc/mi/en

•Level of service, LOS frequency = LOSC

Explanation:

We are given:

•Freeway current lane width,B = 12ft

• freeway current shoulder width,b = 6ft

• percentage of heavy vehicle, Ptb = 10℅

• peak hour factor, PHF = 0.9

Let's consider,

•Number of lanes N = 4

• flow of traffic V = 7500vph

• percentage of Rv = 0, therefore the freeflow speed in freeway FFS = 70mph

• cars equivalent for recreational purpose Er= 2

•cars to be used for trucks and busses Etb= 2.5

Let's first calculate for the heavy adjustment factor.

We have:

$F_H_v = \frac{1}{1+P_t_b(E_t_b-1)+Pr(Er-1)}$

Substituting figures in the equation we have:

$= \frac{1}{1+0.1(2.5-1)+0(2-1)}$

= 0.75

Let's now calculate equivalent flow rate of the car using:

$Vp = \frac{V}{(P_H_F)*N)*(F_H_v)*(F_p)}$

$= \frac{7500}{0.9*4*0.75*1}$

= 2777.7 pc/h/en

Calculating for traffic density, we have:

$D = \frac{Vp}{FFS}$

$D = \frac{2777.7}{70}$

D = 39.685 Pc/mi/en

Using the table for LOS criteria of basic frequency segment, the level of service LOS of frequency is LOSC

4 0

3 years ago

Number pattern Write a recursive method called print Pattern() to output the following number pattern. Given a positive integer

lilavasa [31]

Answer:

See explaination

Explanation:

Code;

import java.util.Scanner;

public class NumberPattern {

public static int x, count;

public static void printNumPattern(int num1, int num2) {

if (num1 > 0 && x == 0) {

System.out.print(num1 + " ");

count++;

printNumPattern(num1 - num2, num2);

} else {

x = 1;

if (count >= 0) {

System.out.print(num1 + " ");

count--;

if (count < 0) {

System.exit(0);

}

printNumPattern(num1 + num2, num2);

}

public static void main(String[] args) {

Scanner scnr = new Scanner(System.in);

int num1;

int num2;

num1 = scnr.nextInt();

num2 = scnr.nextInt();

printNumPattern(num1, num2);

}

See attachment for sample output

3 0

3 years ago

Read 2 more answers

Give me the description of - Feedforward control loops with an example.

ValentinkaMS [17]

Answer:

Feedforward basically configured and used mainly to avoid errors in a control system entering or disrupting a control loop

Explanation:

Feedforward basically configured and used mainly to avoid errors in a control system entering or disrupting a control loop. Although Feedforward control seems to be a very attractive idea, it imposes a high responsibility on both the system developer and the operator to examine and consider mathematically the effect of disruptions on the process concerned.

example of feedforward is

Shower

which consist of following control points

Hear toilet flush (measurement)

Customize water to compensate

feedback refers to that point when water turns hot before the configuration changes

5 0

4 years ago

The Acme threading tool forms an inc luded angle of how many degrees? A. 30 B. 55 C. 29 D. 60

Schach [20]

Answer:

Explanation:

ACME THREAD ANGLE : it is the angle measured between the thread axis and thread the thread flanks, with the help of thread angle we calculate the shape of screw thread it is the mean of v thread and square thread the acme thread angle is denoted by β every thread has its own charactersistics which depends on the pitch and diameter of the thread

4 0

3 years ago

If the total length of a measurement is 10 1/2" what is half of this length?

Aneli [31]

Answer:

13.335 CM (1 ft, 1.335 cm)

I am 80% sure this is the answer, but i am not too keen on math so if i am wrong let me know and i will try my best to fix it!

I hope this helped! Have a good day :]

5 0

2 years ago