Why is it important to consider the experimental error in all the empirical results presented?

Physics

2 answers:

irinina [24]3 years ago

6 0

Answer:

It is very important because scientists, especially the ones with empirical experiments and results, are prone to error and the empirical data is in need to be under strict observation done not only by many scientists but also by expermiented ones. This guards everybody to change the parameters suddenly which can affect the real results of an experiment

Explanation:

brilliants [131]3 years ago

3 0

Answer:

Explanation:

Suppose you want to know how many miles a tire gets before it must be replaced.

Suppose further that you live in a place where the roads are not very good. You would be quite concerned about how you were going to take road conditions into account.

Is rain different from snow? Will the tire get more miles in rain or snow?

Does the tire respond differently after the rain has stopped? It likely will, but how differently.

What about getting stuck especially if an alligator is the only thing looking at you being stuck and he has lunch written all over his face. Is the tire able to get you out?

All of these factors will be very important because no two seasons are alike and one July (say) of one year is not the same as another July in a different year. There has to be a way of determining what the numbers mean.

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A parallel plate capacitor with air between the plates has a potential difference of 71.0 V. Determine the potential difference

Gwar [14]

Answer:

15.8 V

Explanation:

The relationship between capacitance and potential difference across a capacitor is:

$q=CV$

where

q is the charge stored on the capacitor

C is the capacitance

V is the potential difference

Here we call C and V the initial capacitance and potential difference across the capacitor, so that the initial charge stored is q.

Later, a dielectric material is inserted between the two plates, so the capacitance changes according to

$C'=kC$

where k is the dielectric constant of the material. As a result, the potential difference will change (V'). Since the charge stored by the capacitor remains constant,

$q=C'V'$