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3 years ago

Which element of art creates edges of shapes and visual cues for the viewer in terms of both motion and space?

2 answers:

7 0

The element of art that creates edges of shapes and visual cues for the viewer in terms of both motion and space is line. The correct option among all the options that are given in the question is the third option or option "c". The lines have the ability to create a sense of motion when drawn in spaces.

GaryK [48]3 years ago

5 0

The correct answer for the question that is being presented above is this one: "c. line." The <span>element of art creates edges of shapes and visual cues for the viewer in terms of both motion and space is the line. Line is one of the basic elements of art.</span>

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Provided following are four different ranges of stellar masses. Rank the stellar mass ranges based on how many stars in each ran

elena-s [515]

Highest to lowest number:

-less than 1 solar mass

-between 1 and 10 solar masses

-between 10 and 30 solar masses

-between 30 and 60 solar masses

<h3>What is Stellar masses ?</h3>

Stellar mass is a phrase that is used by astronomers to describe the mass of a star.

It is usually enumerated in terms of the Sun's mass as a proportion of a solar mass ( M ☉). Hence, the bright star Sirius has around 2.02 M ☉.

Stellar masses are not fixed, although they change for single stars only on long periods.

Learn more about Stellar masses here:

brainly.com/question/1128503

#SPJ4

3 0

2 years ago

A guitar player tunes the fundamental frequency of a guitar string to 560 Hz. (a) What will be the fundamental frequency if she

lawyer [7]

Answer:

(a) if she increases the tension in the string is increased by 15%, the fundamental frequency will be increased to 740.6 Hz

(b) If she decrease the length of the the string by one-third the fundamental frequency will be increased to 840 Hz

Explanation:

(a) The fundamental, f₁, frequency is given as follows;

$f_1 = \dfrac{\sqrt{\dfrac{T}{\mu}} }{2 \cdot L}$

Where;

T = The tension in the string

μ = The linear density of the string

L = The length of the string

f₁ = The fundamental frequency = 560 Hz

If the tension in the string is increased by 15%, we will have;

$f_{(1 \, new)} = \dfrac{\sqrt{\dfrac{T\times 1.15}{\mu}} }{2 \cdot L} = 1.3225 \times \dfrac{\sqrt{\dfrac{T}{\mu}} }{2 \cdot L} = 1.3225 \times f_1$

$f_{(1 \, new)} = 1.3225 \times f_1 = (1 + 0.3225) \times f_1$

$f_{(1 \, new)} = 1.3225 \times f_1 =\dfrac{132.25}{100} \times 560 \ Hz = 740.6 \ Hz$

Therefore, if the tension in the string is increased by 15%, the fundamental frequency will be increased by a fraction of 0.3225 or 32.25% to 740.6 Hz

(b) When the string length is decreased by one-third, we have;

The new length of the string, $L_{new}$ = 2/3·L

The value of the fundamental frequency will then be given as follows;

$f_{(1 \, new)} = \dfrac{\sqrt{\dfrac{T}{\mu}} }{2 \times \dfrac{2 \times L}{3} } =\dfrac{3}{2} \times \dfrac{\sqrt{\dfrac{T}{\mu}} }{2 \cdot L} = \dfrac{3}{2} \times 560 \ Hz = 840 \ Hz$

When the string length is reduced by one-third, the fundamental frequency increases to one-half or 50% to 840 Hz.

6 0

3 years ago

A light wave traveling through medium 1 strikes a boundary with medium 2 at at a 45 degree angle. the light then enters the seco

Alona [7]

Here light ray strikes to interface at an angle of 45 degree and then refracts into other medium such that it will bend towards boundary.

So here the angle of incidence will be less than the angle of refraction as light moves towards the boundary after refraction which mean it will bend away from the normal

here it can be said that medium 2 will be rarer then medium 1

So here the possible options are

1. Water

Air

2. Diamond

Air

So in above two options medium 1 is denser and medium 2 is rarer

8 0

3 years ago

In science and physics what is the standard unit of measure for speed?

attashe74 [19]

meters per second or m/s

6 0

3 years ago

Find an expression for the square of the orbital period.

jenyasd209 [6]

Answer:

T²= 4π²R³/GM

Explanation:

First we know that

Fg= Fc

Because centripetal force must equal gravitational force

GMm/R² = Mv²/R

But velocity is 2πR/T

So by substitution we have

GMm/R²= M (2πR/T)/T

We have

T²= 4π²R³/GM as period

8 0

3 years ago