What is The Difference between Accuracy and Precision?

2 answers:

7 0

In other words, accuracy describes the difference between the measurement and the part's actual value, while precision describes the variation you see when you measure the same part repeatedly with the same device

GrogVix [38]3 years ago

7 0

Answer:

Accuracy describes the difference between measurement and the parts actually value, while precision describes the variation you see when you measure the same part repeatedly with the same device.

You might be interested in

After the collapse of a nebular cloud, atoms begin gravitating together to form a condensed center. What happens next in the sta

ELEN [110]

Well, first of all, I don't think "After the collapse of a nebular cloud ..."
is the first time that "atoms begin gravitating together". Seems to me like
that's what was going on all the time, and it's what caused the nebular cloud
to collapse in the first place.

In any case, once the pressure and temperature at the center get high enough,
you get "ignition" of nuclear fusion, and that's when you first have a "star".

8 0

4 years ago

Read 2 more answers

A negative charge of 20 x 10-6C and another charge of 15 x 10-6C are separated by as distance of 0.7 m.

denpristay [2]

Answer:

Approximately $5.5\; \rm N$ , assuming that the volume of these two charged objects is negligible.

Explanation:

Assume that the dimensions of these two charged objects is much smaller than the distance between them. Hence, Coulomb's Law would give a good estimate of the electrostatic force between these two objects regardless of their exact shapes.

Let $q_1$ and $q_2$ denote the magnitude of two point charges (where the volume of both charged object is negligible.) In this question, $q_1 = 20 \times 10^{-6}\; \rm C$ and $q_2 = 15 \times 10^{-6}\; \rm C$ .

Let $r$ denote the distance between these two point charges. In this question, $r = 0.7\; \rm m$ .

Let $k$ denote the Coulomb constant. In standard units, $k \approx 8.98755\times 10^{9}\; \rm kg \cdot m^{3}\cdot s^{-2}\cdot C^{-2}$ .

By Coulomb's Law, the magnitude of electrostatic force (electric force) between these two point charges would be:

$\begin{aligned}F &= \frac{k \cdot q_1 \cdot q_2}{r^{2}}\end{aligned}$ .

Substitute in the values and evaluate:

$\begin{aligned}F &= \frac{k \cdot q_1 \cdot q_2}{r^{2}}\\ &\approx 8.98755 \times 10^{9}\; \rm kg \cdot m^{3}\cdot s^{-2}\cdot C^{-2} \\ &\quad \times 20\times 10^{-6}\; \rm C\\ &\quad \times 15\times 10^{-6}\; \rm C \\ &\quad \times \frac{1}{{(0.7\; \rm m)}^{2}}\\ &\approx 5.5\; \rm N \end{aligned}$ .