Ask question

Ask question

All categories

2 years ago

11

What is the difference between accurate data and reproducible data?

2 answers:

Shtirlitz [24]2 years ago

5 0

Answer:

Accuracy is ‘the state of being correct’ and on the other hand precision is ‘the state of being exact.

Explanation:

Accuracy shows the nearness of the measurement with the actual measurement while precision shows the nearness of an individual measurement with other measurements.

Accuracy focuses on the errors caused by the problem in the instrument. while precision is concerned with errors, which occurs periodically with no recognizable pattern.

If the measurement is precise but inaccurate, the result will not be the expected one. And if the result is accurate but imprecise, then there are huge variations in the measurements. And finally, if the actual measurement is neither accurate nor precise, then the result would not have correctness and exactness at the same time.

vladimir1956 [14]2 years ago

4 0

Accuracy is a general concept while precision is more of a mathematical concept.

You might be interested in

In addition to a reference point you also need distance and __________ to describe location

Lana71 [14]

displacement

btw in not sure

6 0

2 years ago

An electric current in a metal consists of moving

Alik [6]

The answer would be (A) Protons

6 0

2 years ago

30pts<br><br> what type of energy is feeling the air around you?

miv72 [106K]

⚡️⚡️⚡️Kinetic energy ⚡️⚡️⚡️

3 0

2 years ago

Read 2 more answers

A charged wire of negligible thickness has length 2L units and has a linear charge density λ. Consider the electric field E-vect

Stels [109]

Answer:

$E=2K\lambda d\dfrac{L }{d^2\sqrt{L^2+d^2}}$

Explanation:

Given that

Length= 2L

Linear charge density=λ

Distance= d

K=1/(4πε)

The electric field at point P

$E=2K\int_{0}^{L}\dfrac{\lambda }{r^2}dx\ sin\theta$

$sin\theta =\dfrac{d}{\sqrt{d^2+x^2}}$

$r^2=d^2+x^2$

So

$E=2K\lambda d\int_{0}^{L}\dfrac{dx }{(x^2+d^2)^{\frac{3}{2}}}$

Now by integrating above equation

$E=2K\lambda d\dfrac{L }{d^2\sqrt{L^2+d^2}}$

4 0

2 years ago

A 1,500 kg car hits a boulder with a force of 500 N, and the collision takes 0.5

Anuta_ua [19.1K]

Answer: C.

250 kg-m/s

Explanation:

Given that the

Mass M = 1,500 kg

Force F = 500 N

Time t = 0.5 seconds

From Newton's second law of motion which state that the rate of change of momentum is proportional to the applied force.

F = mV/t

Ft = mV

Where ft = impulse: the product of time and applied force

Substitutes force and time into the formula

Ft = 500 × 0.5 = 250 Ns

7 0

3 years ago