Your answer is correct the last procedure of a vehicle starting is selecting the appropriate gear for the right situation. (D)
Answer:
The solution code is written in Java.
- public class Main {
-
- public static void main(String[] args) {
-
- Scanner inNum = new Scanner(System.in);
- System.out.print("Enter number of toss: ");
- int num = inNum.nextInt();
-
- for(int i=0; i < num; i++){
- System.out.println(toss());
- }
- }
-
- public static String toss(){
- String option[] = {"heads", "tails"};
- Random rand = new Random();
- return option[rand.nextInt(2)];
- }
- }
Explanation:
Firstly, we create a function <em>toss()</em> with no parameter but will return a string (Line 14). Within the function body, create an option array with two elements, "heads" and "tails" (Line 15). Next create a Random object (Line 16) and use <em>nextInt()</em> method to get random value either 0 or 1. Please note we need to pass the value of 2 into <em>nextInx() </em>method to ensure the random value generated is either 0 or 1. We use this generate random value as an index of <em>option </em>array and return either "heads" or "tails" as output (Line 17).
In the main program, we create Scanner object and use it to prompt user to input an number for how many times to toss the coin (Line 6 - 7). Next, we use the input num to control how many times a for loop should run (Line 9). In each round of the loop, call the function <em>toss() </em>and print the output to terminal (Line 10).
Answer:
design hour volumes will be 4000 to 6000
Explanation:
given data
AADT = 150000 veh/day
solution
we get here design hour volumes that is express as
design hour volumes = AADT × k × D ..............1
here k is factor and its range is 8 to 12 % for urban
and D is directional distribution i.e traffic equal divided by the direction
so here design hour volumes will be 4000 to 6000
Answer:
a) Please see attached copy below
b) 0.39KJ
c) 20.9‰
Explanation:
The three process of an air-standard cycle are described.
Assumptions
1. The air-standard assumptions are applicable.
2. Kinetic and potential energy negligible.
3. Air in an ideal gas with a constant specific heats.
Properties:
The properties of air are gotten from the steam table.
b) T₁=290K ⇒ u₁=206.91 kj/kg, h₁=290.16 kj/kg.
P₂V₂/T₂=P₁V₁/T₁⇒ T₂=P₂T₁/P₁ = 380/95(290K)= 1160K
T₃=T₂(P₃/P₂)⁽k₋1⁾/k =(1160K)(95/380)⁽⁰°⁴/₁.₄⁾ =780.6K
Qin=m(u₂₋u₁)=mCv(T₂-T₁)
=0.003kg×(0.718kj/kg.k)(1160-290)K= 1.87KJ
Qout=m(h₃₋h₁)=mCp(T₃₋T₁)
=0.003KG×(1.005kj/kg.k(780.6-290)K= 1.48KJ
Wnet, out= Qin-Qout = (1.87-1.48)KJ =0.39KJ
c)ηth= Wnet/W₍in₎ =0.39KJ/1.87KJ = 20.9‰
Answer and Explanation:
Flame Front Generator: It is a ignition system which is very useful in flaring system .In this system the air and gases are mixed together and make a combustible air gas mixture. There is a flame front region where the combustion reaction takes place , it is the region where gases as like hydrogen and air mixed with each other and form combustible gases.