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2 years ago

Artistic works in the animal style often contain ________ and images of abstracted animals.

1 answer:

5 0

The answer to the above question is "Interlace".

Artistic works in the animal style often contain INTERLACE and images of abstracted animals. In the medieval art, this artistic work contains decorative elements. Some elements are knotted while others are braided.

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If you subtract vector 3.7 cm at 45° North of East from vector 4.5 cm at 57° West of North using a scale drawing, what is the re

Gelneren [198K]

The resultant vector is 5.2 cm at a direction of 12⁰ west of north.

<h3>Resultant of the two vectors</h3>

The resultant of the two vectors is calculated as follows;

R = a² + b² - 2ab cos(θ)

where;

θ is the angle between the two vectors = 45° + (90 - 57) = 78⁰
a is the first vector
b is the second vector

R² = (3.7)² + (4.5)² - (2 x 3.7 x 4.5) cos(78)

R² = 27.02

R = 5.2 cm

<h3>Direction of the vector</h3>

θ = 90 - 78⁰

θ = 12⁰

Thus, the resultant vector is 5.2 cm at a direction of 12⁰ west of north.

Learn more about resultant vector here: brainly.com/question/28047791

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3 0

1 year ago

Problem: The frequency of an FM radio station is 89.3 MHz. Calculate its period. Part B: From the Library, select the general eq

vekshin1

Answer:

Time period, $T=1.11\times 10^{-8}\ s$

Explanation:

We have,

The frequency of an FM radio station is 89.3 MHz.

It is required to find the period of the wave.

The reciprocal of frequency is called time period of a wave. It can be given by :

$T=\dfrac{1}{f}\\\\T=\dfrac{1}{89.3\times 10^{6}\ Hz}\\\\T=1.11\times 10^{-8}\ s$

So, the period of the wave is $1.11\times 10^{-8}\ s$ .

5 0

2 years ago

3. What is a manometer used to determine?

Gemiola [76]

Answer:

It's used to indicate pressure

3 0

2 years ago

Read 2 more answers

Which of these objects has kinetic energy?

Damm [24]

Answer:

A ball moving through the air.

Explanation:

The ball has momentum which is a form of kinetic energy.

I don't know if that is correct, but I hope it helps!!!!

3 0

3 years ago

Read 2 more answers

A 1.0-cm-tall object is 13 cm in front of a converging lens that has a 40 cm focal length.

kicyunya [14]

A) Image position: -19.3 cm

B) Image height: 1.5 cm, upright

Explanation:

In order to calculate the image position, we can use the lens equation:

$\frac{1}{p}+\frac{1}{q}=\frac{1}{f}$

where

p is the distance of the object from the lens

q is the distance of the image from the lens

f is the focal length

In this problem, we have:

p = 13 cm (object distance)

f = 40 cm (focal length, positive for a converging lens)

So the image distance is

$\frac{1}{q}=\frac{1}{f}-\frac{1}{p}=\frac{1}{40}-\frac{1}{13}=-0.0519\\q=\frac{1}{-0.0519}=-19.3 cm$

The negative sign means that the image is virtual.

In order to calculate the image height, we use the magnification equation:

$\frac{y'}{y}=-\frac{q}{p}$

where

y' is the image height

y is the object height

In this problem, we have:

y = 1.0 cm (object height)

p = 13 cm

q = -19.3 cm

Therefore, the image heigth is

$y'=-\frac{qy}{p}=-\frac{(-19.3)(1.0)}{13}=1.5 cm$

And the positive sign means the image is upright.

6 0

3 years ago