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2 years ago

If it is safe to do so you should shut off your engine if it’s going to be idling for longer than?

Engineering

1 answer:

Lesechka [4]2 years ago

8 0

Answer: turn your ignition off it you are waiting more than 10 seconds

Explanation: idling for just 10 seconds waste more gas than restarting the engine and idling increases overall engine by causing the car to operate longer

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Wattage is:

Ksju [112]

Answer:

c.Both A and B.

Explanation:

the wattage is c and d

7 0

3 years ago

A rigid tank contains 2 kg of N2 and 4 kg of Co2 at temperature of 25 C and 1 MPa. Find the partial pressure of each gas respect

lions [1.4K]

Answer: Partial pressures are 0.6 MPa for nitrogen gas and 0.4 MPa for carbon dioxide.

Explanation: Dalton's Law of Partial Pressure states when there is a mixture of gases the total pressure is the sum of the pressure of each individual gas:

$P_{total} = P_{1}+P_{2}+...$

The proportion of each individual gas in the total pressure is expressed in terms of mole fraction:

$X_{i}$ = moles of a gas / total number moles of gas

The rigid tank has total pressure of 1MPa.

Nitrogen gas:

molar mass = 14g/mol

mass in the tank = 2000g

number of moles in the tank: $n=\frac{2000}{14}$ = 142.85mols

Carbon Dioxide:

molar mass = 44g/mol

mass in the tank = 4000g

number of moles in the tank: $n=\frac{4000}{44}$ = 90.91mols

Total number of moles: 142.85 + 90.91 = 233.76 mols

To calculate partial pressure:

$P_{i}=P_{total}.X_{i}$

For Nitrogen gas:

$P_{N_{2}}=1.\frac{142.85}{233.76}$

$P_{N_{2}}$ = 0.6

For Carbon Dioxide:

$P_{total}=P_{N_{2}}+P_{CO_{2}}$

$P_{CO_{2}} = P_{total}-P_{N_{2}}$

$P_{CO_{2}}=1-0.6$

$P_{CO_{2}}=$ 0.4

Partial pressures for N₂ and CO₂ in a rigid tank are 0.6MPa and 0.4MPa, respectively.

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

t is desired to produce aligned carbon fiber-epoxy matric composite having a longitudinal tensile strength of 800 MPa. Given (a)

Nesterboy [21]

Answer:

The value of critical length = 3.46 mm

The value of volume of fraction of fibers = 0.43

Explanation:

Given data

$\sigma_{T}$ = 800 M pa

D = 0.017 mm

L = 2.3 mm

$\sigma_{f}$ = 5500 M pa

$\sigma_{m}$ = 18 M pa

$\sigma_c$ = 13.5 M pa

(a) Critical fiber length is given by

$L_{c} = \sigma_{f} (\frac{D}{2 \sigma_{c} } )$

Put all the values in above equation we get

$L_{c} =5500 (\frac{0.017}{(2) (13.5)} )$

$L_{c} = 3.46$ mm

This is the value of critical length.

(b).Since this critical length is greater than fiber length Than the volume fraction of fibers is given by

$V_{f} = \frac{\sigma_T - \sigma_m}{\frac{L\sigma_c}{D} - \sigma_m }$

Put all the values in above formula we get

$V_{f} = \frac{800-18}{\frac{(2.3)(13.5)}{0.017} - 18 }$

$V_{f}$ = 0.43

This is the value of volume of fraction of fibers.

3 0

4 years ago

Consider a cubic block whose sides are 6 cm long and a cylindrical block whose height and diameter are also 6 cm. both blocks ar

bija089 [108]

Answer:

Check the explanation

Explanation:

Kindly check the attached image below to see the step by step explanation to the question above.

5 0

3 years ago