1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
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.

You might be interested in
What is civil engineering​
kkurt [141]

Answer:

engineering that works on building structures.

Explanation:

7 0
4 years ago
Which of the following are examples of engineering controls? Select all that apply.
Neporo4naja [7]

The examples of engineering controls is Biohazard waste containers and Spill clean up kits.

What is engineering controls?

An engineering controls is a workplace process that protect workers by removing hazardous conditions or by placing a barrier between the worker and the hazard.

An example of engineering controls is installation of exhaust ventilation to remove airborne emissions to shield the worker.

Hence, the examples of engineering controls is Biohazard waste containers and Spill clean up kits.

Therefore, the Option C and D is correct.

8 0
2 years ago
PLEASE HELP, TEST MULTIPLE CHOICE QUESTIONS
Rina8888 [55]

Answer:

c

Explanation:

dbfjex vadamhqrmtwg

8 0
3 years ago
Read 2 more answers
What kind of microscope would be used to study a whole or opaque object?
djyliett [7]

Answer:

the compound light microscope

Explanation:

The stereomicroscope is to study section to study the entire objects in three dimensions at low magnification. A Compound light microscope is used for small or thinly sliced objects under higher magnification than stereomicroscope.

4 0
4 years ago
The input shaft to a gearbox rotates at 2300 rpm and transmits a power of 42.6 kW. The output shaft power is 34.84 kW at a rotat
Marrrta [24]

Answer:

Torque at input shaft will be 176.8695 N-m

Explanation:

We have given input power P_{IN}=42.6KW=42.6\times 10^3W

Angular speed = 2300 rpm

For converting rpm to rad/sec we have multiply with \frac{2\pi }{60}

So 2300rpm=\frac{2300\times 2\pi }{60}=240.855rad/sec

We have to find torque

We know that  power is given by P=\tau \omega, here \tau is torque and \omega is angular speed

So 42.6\times 10^3=\tau \times 240.855

\tau =176.8695N-m

So torque at input shaft will be 176.8695 N-m

4 0
4 years ago
Other questions:
  • The primary heat transfer mechanism that warms me while I stand next to a campfire is: a)- Conduction b)- Impeadance c)- Convect
    7·2 answers
  • All circuit conductors between the service equipment, the source of a separately derived system, or other power supply source an
    5·1 answer
  • 15. A cold-chamber die-casting machine operates automatically, supported by two industrial robots.The machine produces two zinc
    9·2 answers
  • 4. (3 pts) Sketch cylinder/cylinder head configurations to show the differences between PFI and GDI gasoline injection systems.
    5·2 answers
  • If gain of the critically damped system is increased, the system will behave as a) Under damped b) Over damped c) Critically dam
    13·1 answer
  • A well-established way of power generation involves the utilization of geothermal energy-the energy of hot water that exists nat
    9·1 answer
  • You want to plate a steel part having a surface area of 160 with a 0.002--thick layer of lead. The atomic mass of lead is 207.19
    15·1 answer
  • How can you drop two eggs the fewest amount of times without them breaking?
    15·1 answer
  • A pinion and gear pair is used to transmit a power of 5000 W. The teeth numbers of pinion
    8·1 answer
  • An engineer is tasked to design a combinational circuit with three inputs x, y, z and one output F to satisfy the following cond
    12·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!