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1 year ago

Characteristics, such as field name and data type, are_____ that define a field.

Physics

1 answer:

Aleonysh [2.5K]1 year ago

3 0

Characteristics, such as field name and data type are properties that define a field.

<h3>What are the properties of a given discipline or research field?</h3>

The properties of a given discipline or research field are those that characterize the use of a particular type of information within the range of the attained field.

For example, genomics is a field characterized to make use of sequencing data, which refers to the linear order of nucleotides in the DNA sequence to provide useful information about the gene content and the presence of regulatory non-coding sequences.

Therefore, with this data, we can see that the properties of a given discipline or research field are mainly associated with the type of information that such discipline uses to make predictions in a given research and they are also related to the outcomes we can expect from such data.

Learn more about the properties of a scientific research field here:

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Inthisexperiment,aspringforcewasusedtokeep moving object travelingin a circularpath.The size of a spring force should be proport

poizon [28]

Answer:

By calculation, it can be shown that;

K = $\frac{F_{angular}}{2\times x\times c}$

Whereby for constant K, as ${F_{angular}}$ increases, x also increases.

Explanation:

The experiment set up consisted of the use of a spring force to maintain the object in circular path.

The energy in the spring is given by

$\frac{1}{2}\cdot k\cdot x^2$ .

Rotational kinetic energy = $\frac{1}{2}$ ·I·ω²

Inertia, I = $\frac{1}{2}$ ·m·r²

ω = $\frac{v}{r}$

Substituting gives

Rotational kinetic energy = $\frac{1}{2}$ ·

= $\frac{1}{4}$ ·m· v²

Equating both equations gives

K = $\frac{2\times m\times v^2}{4\times x^{2} }$ = $\frac{1\times m\times v^2}{2\times x^{2} }$

Within the proportionality limit, x ∝ r

therefore we can write x = c·r which gives

$\frac{ m\times v^2}{2\times x\times c\times r }$ = $\frac{v^{2} }{r} \times\frac{m}{2\times x\times c}$

Since $\frac{v^{2} }{r}$ = angular acceleration, α, then

m× $\frac{v^{2} }{r}$ = Angular force

Therefore K = $\frac{F_{angular}}{2\times x\times c}$

Therefore as Force, F increases, x also increases and the size of a spring force should be proportional to the amount of stretch in the spring.

3 0

3 years ago

A planet has two

lozanna [386]

Kepler's third law hypothesizes that for all the small bodies in orbit around the
same central body, the ratio of (orbital period squared) / (orbital radius cubed)
is the same number.

<u>Moon #1:</u> (1.262 days)² / (2.346 x 10^4 km)³

<u>Moon #2:</u> (orbital period)² / (9.378 x 10^3 km)³

If Kepler knew what he was talking about ... and Newton showed that he did ...
then these two fractions are equal, and may be written as a proportion.

Cross multiply the proportion:

(orbital period)² x (2.346 x 10^4)³ = (1.262 days)² x (9.378 x 10^3)³

Divide each side by (2.346 x 10^4)³:

(Orbital period)² = (1.262 days)² x (9.378 x 10^3 km)³ / (2.346 x 10^4 km)³

= 0.1017 day²

Orbital period = <u>0.319 Earth day</u> = about 7.6 hours.

7 0

3 years ago

Part 1: A rope has one end tied to a vertical support. You hold the other end so that the rope is horizontal. If you move the en

Blizzard [7]

Answer:

c. 2 m/s

Explanation:

The speed of a wave is given by:

$v=f \lambda$