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2 years ago

11

Ls are used to analyze large unstructured data sets, such as e-mail, memos, survey responses, etc., to discover patterns and rel

ationships. question 10 options:

1 answer:

Kruka [31]2 years ago

6 0

The word that could fit to this is Text Mining. This word also referred to as text data mining, roughly equivalent to text analytics, a process of deriving high-quality information from text. This high quality information derived through the devising of patterns and trends through means such as statistical patter learning.

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An electron accelerated from rest through a voltage of 780 v enters a region of constant magnetic field. part a part complete if

maxonik [38]

The electron is accelerated through a potential difference of $\Delta V=780 V$ , so the kinetic energy gained by the electron is equal to its variation of electrical potential energy:
$\frac{1}{2}mv^2 = e \Delta V$
where
m is the electron mass
v is the final speed of the electron
e is the electron charge
$\Delta V$ is the potential difference

Re-arranging this equation, we can find the speed of the electron before entering the magnetic field:
$v= \sqrt{ \frac{2 e \Delta V}{m} } = \sqrt{ \frac{2(1.6 \cdot 10^{-19}C)(780 V)}{9.1 \cdot 10^{-31} kg} }=1.66 \cdot 10^7 m/s$

Now the electron enters the magnetic field. The Lorentz force provides the centripetal force that keeps the electron in circular orbit:
$evB=m \frac{v^2}{r}$
where B is the intensity of the magnetic field and r is the orbital radius. Since the radius is r=25 cm=0.25 m, we can re-arrange this equation to find B:
$B= \frac{mv}{er}= \frac{(9.1 \cdot 10^{-31}kg)(1.66 \cdot 10^7 m/s)}{(1.6 \cdot 10^{-19}C)(0.25 m)} =3.8 \cdot 10^{-4} T$

3 0

3 years ago

How did the continental Shelf form? Please help, thank you!! :)

shutvik [7]

The movements of the tectonic plates

6 0

2 years ago

Two astronauts are playing catch in a zero gravitational field. Astronaut 1 of mass m1 is initially moving to the right with spe

Ede4ka [16]

The final velocity ( $v_1_f$ ) of the first astronaut will be greater than the <em>final velocity</em> of the second astronaut ( $v_2_f$ ) to ensure that the total initial momentum of both astronauts is equal to the total final momentum of both astronauts <em>after throwing the ball</em>.

The given parameters;

Mass of the first astronaut, = m₁
Mass of the second astronaut, = m₂
Initial velocity of the first astronaut, = v₁
Initial velocity of the second astronaut, = v₂ > v₁
Mass of the ball, = m
Speed of the ball, = u
Final velocity of the first astronaut, = $v_f_1$
Final velocity of the second astronaut, = $v_f_2$

The final velocity of the first astronaut relative to the second astronaut after throwing the ball is determined by applying the principle of conservation of linear momentum.

$m_1v_1 + m_2v_2 = m_2v_2_f + m_1v_1_f$

if v₂ > v₁, then $v_1_f > v_2_f$ , to conserve the linear momentum.

Thus, the final velocity ( $v_1_f$ ) of the first astronaut will be greater than the <em>final velocity</em> of the second astronaut ( $v_2_f$ ) to ensure that the total initial momentum of both astronauts is equal to the total final momentum of both astronauts after throwing the ball.

Learn more here: brainly.com/question/24424291

5 0

2 years ago

Looking straight downward into a rain puddle whose surface is covered with a thin film of gasoline, you notice a swirling patter

Ivanshal [37]

Answer:

Explanation:

Point beneath you forms a beautiful iridescent green

refractive index of Gasoline $n=1.38$

Wavelength of Green light is $\lambda =540\ nm$

Here light first traverse from air(n=1) to gasoline , it reflects from front surface of gasoline(n=1.38) so it suffers a phase change. After this light reflect from rear surface of gasoline and there is a decrease in refractive index(n=1.38 to n=1.33), so there is no phase change occurs .

For constructive interference

$2t=(m+\frac{1}{2})\cdot \frac{\lambda }{n}$

here t= thickness of gasoline film

n=refractive index

for $m=0$

$t=\frac{\lambda }{4n}$

$t=\frac{540}{4\times 1.38}$

$t=97.82\approx 98\ nm$

4 0

2 years ago

A horizontal rectangular surface has dimensions 2.80 cm by 3.20 cm and is in a uniform magnetic field that is directed at an ang

koban [17]

Answer:

magnitude of the magnetic field 0.692 T

Explanation:

given data

rectangular dimensions = 2.80 cm by 3.20 cm

angle of 30.0°

produce a flux Ф = 3.10 × $10^{-4}$ Wb

solution

we take here rectangular side a and b as a = 2.80 cm and b = 3.20 cm

and here angle between magnitude field and area will be ∅ = 90 - 30

∅ = 60°

and flux is express as

flux Ф = $\int \vec{B}.d\vec{A}$ .................1

and Ф = BA cos∅ ............2

so B = $\frac{\phi }{Acos\theta }$

and we know

A = ab

so

B = $\frac{\phi }{abcos\theta }$ ..............3

put here value

B = $\frac{3.10\times 10^{-4} }{2.80 \times 10^{-2}\times 3.20 \times 10^{-2}\times cos60}$

solve we get

B = 0.692 T

8 0

2 years ago