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Maslowich
2 years ago
9

6.37 LAB*: Program: Authoring assistant(1) Prompt the user to enter a string of their choosing. Store the text in a string. Outp

ut the string. (1 pt)Ex:Enter a sample text:we'll continue our quest in space. there will be more shuttle flights and more shuttle crews and, yes, more volunteers, more civilians, more teachers in space. nothing ends here; our hopes and our journeys continue!You entered: we'll continue our quest in space. there will be more shuttle flights and more shuttle crews and, yes, more volunteers, more civilians, more teachers in space. nothing ends here; our hopes and our journeys continue!
Engineering
1 answer:
irina [24]2 years ago
6 0

Answer:

import java.util.Scanner;

public class Lab_Program

{

 //declare the variable for scanner

static Scanner in=new Scanner(System.in);

//main method

public static void main(String[] args)

 {

//declare the required variable

String str;

//prompt the user to enter a sample text

System.out.println("Enter a sample text: ");

//read the string

str=in.nextLine();

 //display the string

 System.out.println("You entered: "+str);

}

}

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6.28 A six-lane freeway (three lanes in each direction) in rolling terrain has 10-ft lanes and obstructions 4 ft from the right
dimulka [17.4K]

Answer:

Assume Base free flow speed (BFFS) = 70 mph

Lane width = 10 ft

Reduction in speed corresponding to lane width, fLW = 6.6 mph

Lateral Clearance = 4 ft

Reduction in speed corresponding to lateral clearance, fLC = 0.8 mph

Interchanges/Ramps = 9/ 6 miles = 1.5 /mile

Reduction in speed corresponding to Interchanges/ramps, fID = 5 mph

No. of lanes = 3

Reduction in speed corresponding to number of lanes, fN = 3 mph

Free Flow Speed (FFS) = BFFS – fLW – fLC – fN – fID = 70 – 6.6 – 0.8 – 3 – 5 = 54.6 mph

Peak Flow, V = 2000 veh/hr

Peak 15-min flow = 600 veh

Peak-hour factor = 2000/ (4*600) = 0.83

Trucks and Buses = 12 %

RVs = 6 %

Rolling Terrain

fHV = 1/ (1 + 0.12 (2.5-1) + 0.06 (2.0-1)) = 1/1.24 = 0.806

fP = 1.0

Peak Flow Rate, Vp = V / (PHV*n*fHV*fP) = 2000/ (0.83*3*0.806*1.0) = 996.54 ~ 997 veh/hr/ln

Vp < (3400 – 30 FFS)

S = FFS

S = 54.6 mph

Density = Vp/S = (997) / (54.6) = 18.26 veh/mi/ln

7 0
2 years ago
Tech A says that as moisture levels increase in brake fluid, the boiling point of the brake fluid decreases. Tech B says that fi
dezoksy [38]

Answer: Both Technician A and B are correct.

Explanation:

Technicians A and B are both right about their diagnosis. The Society of Automotive Engineers performed extensive research on vehicle brake fluids and found that there is typically a 2% moisture content in the brake fluid after a year of operating a vehicle. And as the moisture content of the brake fluid rises, the boiling point of the brake fluid decreases as well.

4 0
2 years ago
3/63 A 2‐kg sphere S is being moved in a vertical plane by a robotic arm. When the angle θ is 30°, the angular velocity of the a
miss Akunina [59]

Answer:

Ps=19.62N

Explanation:

The detailed explanation of answer is given in attached files.

5 0
2 years ago
A steel rotating-beam test specimen has an ultimate strength Sut of 1600 MPa. Estimate the life (N) of the specimen if it is tes
ziro4ka [17]

Answer:

the life (N) of the specimen is 46400 cycles

Explanation:

given data

ultimate strength Su = 1600 MPa

stress amplitude σa = 900 MPa

to find out

life (N) of the specimen

solution

we first calculate the endurance limit of specimen Se i.e

Se = 0.5× Su   .............1

Se = 0.5 × 1600

Se = 800 Mpa

and we know

Se for steel is 700 Mpa for Su ≥ 1400 Mpa

so we take endurance limit Se is = 700 Mpa

and strength of friction f  = 0.77 for 232 ksi

because for Se 0.5 Su at 10^{6} cycle = (1600 × 0.145 ksi ) = 232

so here coefficient value (a) will be

a = \frac{(f*Su)^2}{Se}    

a = \frac{(0.77*1600)^2}{700}  

a = 2168.3 Mpa

so

coefficient value (b) will be

a = -\frac{1}{3}log\frac{(f*Su)}{Se}

b =  -\frac{1}{3}log\frac{(0.77*1600)}{700}

b = -0.0818

so no of cycle N is

N =  (\frac{ \sigma a}{a})^{1/b}

put here value

N =  (\frac{ 900}{2168.3})^{1/-0.0818}

N = 46400

the life (N) of the specimen is 46400 cycles

5 0
2 years ago
. A piston-cylinder device whose piston is resting on top of a set of stops initially contains 0.5 kg of helium gas at 100 kPa a
Delvig [45]

Answer:

Qin = 1857 kJ

Explanation:

Given

m = 0.5 Kg

T₁ = 25°C = (25 + 273) K = 298 K

P₁ = 100 kPa

P₂ = 500 kPa

First, the temperature when the piston starts rising is determined from the ideal gas equations at the initial state and at that state:

T₂ = T₁*P₂/P₁

⇒  T₂ = 298 K*(500 kPa/100 kPa) = 1490 K

Until the piston starts rising no work is done so the heat transfer is the change in internal energy

Qin = ΔU = m*cv*(T₂-T₁)

⇒   Qin = 0.5*3.1156*(1490 - 298) kJ = 1857 kJ

8 0
2 years ago
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