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NeX [460]
3 years ago
8

Why is it important to know the accuracy and precision of a measuring device? Do you think that the dial caliper manufacturer’s

claim that the ""accuracy"" of the instrument is ±.001 is appropriate? Why or why not? Do you think that either of the dial calipers needs to be adjusted to accurately display measurements? Explain.
Engineering
1 answer:
Leto [7]3 years ago
3 0

Answer:

Accuracy and precision allow us to know how much we can rely on a measuring device readings. ±.001 as a "accuracy" claim is vague because there is no unit next to the figure and the claim fits better to the definition of precision.

Explanation:

Accuracy and Precision: the golden couple.

Accuracy and precision are key elements to define if a measuring device is reliable or not for a specific task. Accuracy determines how close are the readings from the ideal/calculated values. On the other hand, precision refers to repeatability, that is to say how constant the readings of a device are when measuring the same element at different times. One of those two key concepts may not fulfill the criteria for measuring tool to be used on certain engineering projects where lack of accuracy (disntant values from real ones) or precision (not constant readings) may lead to malfunctons and severe delays on the project development.

±.001 what unit?

The manufacturer says that is an accuracy indicator, nevertheless there is now unit stated so this is not useful to see how accurate the device is. Additionally, That notation is more used to refer to device tolerances, that is to say the range of possible values the instrument may show when reading and element. It means it tells us more about the device precision during measurments than actual accuracy. I would recommend the following to the dial calipers manufacturers to better explain its measurement specifications:

  1. Use  ±.001 as  a reference for precision. It is important to add the respective unit for that figure.
  2. Condcut test to define the actual accuracy value an present it using one of the common used units for that:  Error percentage or ppm.

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Answer:

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Explanation:

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8 0
3 years ago
Write an application named EnterUppercaseLetters that asks the user to type an uppercase letter from the keyboard. If the charac
ozzi

Answer:

The solution code is given below

  1. using System;
  2. using System.Linq;
  3.      
  4. public class Program
  5. {
  6. public static void Main()
  7. {
  8.  string[] letters = {"A", "B", "C", "D", "E", "F", "G", "H", "I", "J", "K", "L", "M", "N", "O", "P", "Q", "R", "S", "T","U", "V", "W", "X", "Y", "Z"};
  9.  
  10.  Console.WriteLine("Please input a letter: ");
  11.  string input_letter = Console.ReadLine();
  12.  
  13.  if(letters.Contains(input_letter)){
  14.   Console.WriteLine("OK");
  15.  }else{
  16.   Console.WriteLine("Error. Not uppercase letter.");
  17.  }  
  18. }
  19. }

Explanation:

Firstly we import the necessary libraries, System and Linq (Line 1-2).

Next we create a string array to hold all uppercase letters (Line 8).

Next, we prompt user to input a letter using the ReadLine() method (Line 10 - 11)

At last, we define if and else conditions to check if the letters contains the input letter. If so, print ok else print an error message. (Line 13-17)

6 0
3 years ago
For the following circuit, V"#$=120∠30ºV.Redraw the circuit in your solution.a.(4) Calculate the total input impedance seen by t
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Answer:

Check the explanation

Explanation:

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8 0
3 years ago
Your driver license will be _____ if you race another driver on a public road, commit a felony using a motor vehicle, or are fou
Georgia [21]

Hello there,

In the problems given in the question, the driver's license is confiscated and suspended.

So our answer is: A)

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6 0
4 years ago
Read 2 more answers
A household refrigerator that has a power input of 450 W and a COP of 1.5 is to cool five large watermelons, 10 kg each, to 8°C.
Andreyy89

Answer:

6222.22 sec

Explanation:

Given data the power input to the refrigerator is 450 W

The COP of refrigerator is 1.5

Temperature T_1=8^{\circ}C

T_2=28^{\circ}C

mass of watermelon =10 kg

specific heat =4.2 KJ/kg°C

The amount of heat removed from 5 watermelon

Q=mc_pdt=5\times 10\times 4.2\times (28-8)=4200 KJ

We know that COP=\frac{Q_1}{W}

1.5=\frac{Q_1}{450}

Q_1=675 W=0.675 KW

so time required to cool the watermelon is

t=\frac{Q_1}{Q_2}=\frac{4200}{0.675}=6222.22 sec  

4 0
3 years ago
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