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Digiron [165]
3 years ago
13

Christopher has designed a fluid power system that repeatedly gets clogs. Which of the following objects should he choose to add

to his fluid power system to prevent these clogs in the future?
filter


pipe


valve


piston
Engineering
1 answer:
MA_775_DIABLO [31]3 years ago
3 0

Answer:

Valve

Explanation:

Its right

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When checking for a no-star concern, you notice that an engine has no spark Technician A says to turn on the ignition engine (en
lbvjy [14]

Answer:

Technician B

Explanation:

Technician B is correct in his argument. This is because according to what he said, as the computer pulses stimuli the coil will turn on and off, promoting an increase in the voltage that will cause the fluctuation. Technician A is incorrect because the procedure he indicated imposes that the voltage is checked at the negative terminal and not at the positive.

5 0
2 years ago
What is the rated power output in ( kw) of a 8 pole motor designed to an IEC 180L motor frame ?<br>​
ra1l [238]

Answer:

P=11 kW

Explanation:

Given that

Number of poles= 8

I.E.C. 180L motor frame

From data book , for 8 poles motor at 50 Hz

Speed = 730 rpm

Power factor = 0.75

Efficiency at 100 % load= 89.3 %

Efficiency at 50 % load= 89.1 %

Output power = 11 kW

Therefore the rated output power of 8 poles motor will be 11 kW. Thus the answer will be 11 kW.

P=11 kW

4 0
3 years ago
Explain the working of McLeod gauge for measurement of very low pressure.
AfilCa [17]

Explanation:

McLeod gauge:

  It is used to measure very low pressure of gas.It measure gas pressure by the help of mercury(Hg).it measure absolute pressure of gas.

McLeod gauge on the principle of Boyle's law.Boyle's law state that ,at constant temperature the pressure of gas is directly proportional to its  volume.From Boyle's law

P_1V_1=P_2V_2

A sample of gas is taken from vacuum then its pressure is measured by help  Hg.

4 0
3 years ago
(Gas Mileage) Drivers are concerned with the mileage their automobiles get. One driver has kept track of several trips by record
Effectus [21]

Answer:

import java.util.*;

public class Main {

   

   public static void main(String[] args) {

     

       double milesPerGallon = 0;

       int totalMiles = 0;

       int totalGallons = 0;

       double totalMPG = 0;

       

       Scanner input = new Scanner(System.in);

 

       while(true){

           System.out.print("Enter the miles driven: ");

           int miles = input.nextInt();

           if(miles <= 0)

               break;

           else{

               System.out.print("Enter the gallons used: ");

               int gallons = input.nextInt();

               totalMiles += miles;

               totalGallons += gallons;

               milesPerGallon = (double) miles/gallons;

               totalMPG = (double) totalMiles / totalGallons;

               System.out.printf("Miles per gallon for this trip is: %.1f\n", milesPerGallon);

               System.out.printf("Total miles per gallon is: %.1f\n", totalMPG);

           }

       }

   }  

}

Explanation:

Initialize the variables

Create a while loop that iterates until the specified condition is met inside the loop

Inside the loop, ask the user to enter the miles. If the miles is less than or equal to 0, stop the loop. Otherwise, for each trip do the following: Ask the user to enter the gallons. Add the miles and gallons to totalMiles and totalGallons respectively. Calculate the milesPerGallon (divide miles by gallons). Calculate the totalMPG (divide totalMiles by totalGallons). Print the miles per gallon and total miles per gallon.

6 0
3 years ago
Consider a cylindrical nickel wire 2.1 mm in diameter and 3.2 × 104 mm long. Calculate its elongation when a load of 280 N is ap
pshichka [43]

Answer:

Total elongation will be 0.012 m

Explanation:

We have given diameter of the cylinder = 2.1 mm

Length of wire L=3.2\times 10^4mm

So radius r=\frac{d}{2}=\frac{2.1}{2}=1.05mm=1.05\times 10^{-3}m

Load F = 280 N

Elastic modulus = 207 Gpa

Area of cross section A=\pi r^2=3.14\times (1.05\times 10^{-3})^2=3.461\times 10^{-6}m^2

We know that elongation in wire is given by \delta =\frac{FL}{AE}, here F is load, L is length, A is area and E is elastic modulus

So \delta =\frac{FL}{AE}=\frac{280\times 32}{3.461\times 10^{-6}\times 207\times 10^9}=0.012m

4 0
3 years ago
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