If a poems has a regular rhythm throughout the poem, it has: PLEASE HELP MEH I WILL GIVE YOU BRAINLEIST!!

A. tone
B. imagery
C. irony
D. meter

Answers

Answer 1

Answer:

D, meter.

Explanation:

Rhythm is associated with meter, which identifies units of stressed and unstressed syllables.

Answer 2

If a poem has a regular rhythm throughout the poem, it has a meter. Option D is correct.

What is meter in poem?

Meter, which distinguishes between stressed and unstressed syllables, is related to rhythm. The fundamental rhythmic framework of a stanza or a line of poetry is known as meter.

The number of feet in the poem serves as a measure of the poem's meter, which is the rhythm of the language.

Many traditional poem forms call for a certain verse meter or a group of meters that alternate in a specified pattern. Prosody refers to both the study of meters and other types of versification, as well as their practical application.

Therefore, option D is correct.

To learn more about the poem, refer to:

https://brainly.com/question/12155529

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Related Questions

the importance of reading a circuit diagram to interpret a wiring diagram?

Answers

Answer:

The ability to read electrical schematics is a really useful skill to have. To start developing your schematic reading abilities, it's important to memorize the most common schematic symbols. ... You should also be able to get a rough idea of how the circuit works, just by looking at the schematic.

Explanation:

The quantity of bricks required increases with the surface area of the wall, but the thickness of a masonry wall does not affect the total quantity of bricks used in the wall

True or False

Answers

Answer:

false

Explanation:

False is the answer:)!

How do you describe sound? (SELECT ALL THAT APPLY.) PLEASE HELP AND SELECT ALL THAT APPLY PLEASE!! A. Sound waves have to have a medium to travel through. B. The volume of a sound is known as amplitude. C. Loud sounds have high amplitude and vibrate with more energy than soft sounds. D. Sound waves are compression waves that cause energy transfer in air molecules.

Answers

Answer:

Sound waves are compression waves that cause energy transfer in air molecules

Sound waves have to have a medium to travel through

Loud sounds have high amplitude and vibrate with more energy than soft sounds

Explanation:

Sound waves is a form of energy composed of compression and rare factions. Sound waves are compression waves that cause energy transfer in air molecules.

Sound is an example of a mechanical wave hence it requires a material medium for propagation.

The amplitude of a sound wave determines its loudness or volume. A larger amplitude implies that we will have a louder sound, and a smaller amplitude means that we will have a softer sound.

An open-circuit wind tunnel draws in air from the atmosphere through a well-contoured nozzle. In the test section, where the flow is straight and nearly uniform, a static pressure tap is drilled into the tunnel wall. A manometer connected to the tap shows that static pressure within the tunnel is 45 mm of water below atmospheric. Assume that the air is incompressible, and at 25 C, 100 kPa absolute. Calculate the air speed in the wind-tunnel test section

Answers

Answer:

Air speed in the wind-tunnel [tex]v_{2}[/tex] = 27.5 m/s

Explanation:

Given data :

Manometer reading ; p1 - p2 = 45 mm of water

Pressure at section ( I ) p1 = 100 kPa ( abs )

temperature ( T1 ) = 25°C

Pw ( density of water ) = 999 kg/m3

g = 9.81 m/s^2

next we apply Bernoulli equation at section 1 and section 2

p1 - p2 = [tex]\frac{PairV^{2} _{2} }{2}[/tex]     ----------  ( 1 )

considering  ideal gas equation

Pair ( density of air ) = [tex]\frac{P}{RT}[/tex] ------------------- ( 2 )

R ( constant ) = 287 NM/kg.k

T = 25 + 273.15 = 298.15 k

P1 = 100 kN/m^2 = 100 * 10^3 or N/m^2

substitute values into equation ( 2 )

= 100 * 10^3 / (287 * 298.15)

= 1.17 kg/m^3

Also note ; p1 - p2 = PwgΔh  ------- ( 3 )

finally calculate the Air speed in the wind-tunnel test section by equating equation ( 1 ) and ( 3 )

[tex]\frac{PairV^{2} _{2} }{2}[/tex]   =  PwgΔh  

[tex]V^{2} _{2}[/tex] = [tex]\frac{2*999* 9.81* 0.045}{1.17}[/tex]  =  753.86

[tex]v_{2}[/tex] = 27.5 m/s

trevor moves a magnetic toy train away from a magnet that cannot move. what happens to the potential energy in the system of magnets during the movement?

Answers

Answer:a

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