2. When performing an alignment, what action should be taken immediately after putting a vehicle on the rack?

The path taken to move items from their origin to their destination is known as which of the following?

mrs_skeptik [129]

Answer:d

Explanation:

7 0

3 years ago

WIL MARK U AS BRAINLIEST HELP PLZ

ss7ja [257]

Answer:

The 1st one:Your natural ability

4 0

3 years ago

1 import java.util.Scanner; 3 public class EqualityAndRelational { 4 public static void main (String args) args) { int userBonus

Anastaziya [24]

Missing Part of the Question

Complete the expression so that userPoints is assigned with 0 if userBonus is greater than 20 (second branch). Otherwise, userPoints is assigned with 10 (first branch

import java.util.Scanner;

public class EqualityAndRelational {

public static void main (String args) args) {

int userBonus; int userPoints;

userPoints=0;

Scanner scnr = new Scanner(System.in);

userBonus = scnr.nextInt();

// Program will be tested with values : 15, 20, 25, 30, 35. 12 13 14 15 16 17 18 19

( Your solution goes here)

{

userPoints= 10 ;

}

else {

userPoints = 0;

}

Answer;

Replace

( Your solution goes here)

With

if(userBonus>20).

The full program becomes

import java.util.Scanner;

public class EqualityAndRelational {

public static void main (String args) args) {

int userBonus; int userPoints;

userPoints=0;

Scanner scnr = new Scanner(System.in);

userBonus = scnr.nextInt();

// Program will be tested with values : 15, 20, 25, 30, 35. 12 13 14 15 16 17 18 19

if(userBonus>20)

{

userPoints= 10 ;

}

else {

userPoints = 0;

}

7 0

4 years ago

A liquid with a specific gravity of 2.6 and a viscosity of 2.0 cP flows through a smooth pipe of unknown diameter, resulting in

sasho [114]

.......,.,.,.,.,.,,.,.,,.,.,.,.,.,.,.,,.,.,,.,,,.,,.,,.,.,.,.,,............,,,,,,’mmdjidvdhxxkf jkkk he in d

6 0

3 years ago

Assume the average fuel flow rate at the peak torque speed (1500 rpm) is 15kg/hr for a sixcylinder four-stroke diesel engine und

sveticcg [70]

Answer:

Q = 8.845 DEGREE

Explanation:

given data:

combine Mass for 6 cylinder (M) =15 Kg/hr

mass of each cylinder (m) = 15/6 = 2.5 Kg/hr = 0.000694 Kg/ sec

Engine speed (N)= 1500rpm

Diameter of one nozzle hole ( d) = 200 micrometer = 0.0002 m

Discharge Coefficient (Cd) = 0.75

Pressure difference = 100 MPa

Density of fuel = 800 kg/m^3

velocity of fuel is $v = cd\sqrt{\frac{2*P}{p}}$

$v = 0.75 \sqrt{\frac{2\times 100\times 10^6}{800}} = 375 m/sec$

injected fuel volume (V) =Area of given Orifices × Fuel velocity × time of single injection × no of injection/sec

we know that p = m/ V

So $V = \frac{0.000694}{800} =8.68\times10^{-7} m3/sec$

putting these value in volume equation and solve for Discharge $8.68\times 10^{-7} = (\frac{(3.14}{4})\times 6\times( .0002\times .0002) \times 375 \times \frac{(Q}{360}) \times \frac{30}{750} \times \frac{(750}{60)}$

Q = 8.845 DEGREE

4 0

3 years ago