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asambeis [7]
3 years ago
12

What is the main similarity between ""fog"" and frost’s poem mending wall?

Physics
2 answers:
Anettt [7]3 years ago
7 0

Answer:

They both use iambic pentameter

Explanation:

<em>The main similarity between "fog" and frost's poem "mending wall" is that they both use iambic pentameter </em>

<em> Iambic pentameter is a line of verse with five metrical feet, each comprising of one short or unstressed syllable which is followed by one long or stressed) syllable, Currently, this technique is certainly the usual technique that is used in a poetry.</em>

muminat3 years ago
5 0

Answer:

Both use everyday language.

Explanation:

Language is the system through which man or animals communicate their ideas and feelings, either through speech, writing or other conventional signs, being able to use all the senses to communicate.

The term language is of Latin lingua origin.   The human being uses a complex language that expresses with sound sequences and graphic signs. Animals, meanwhile, communicate through sound and body signs, which even man has not been able to decipher, and which in many cases are far from simple.

Depending on the social context in which the language is produced, the speaker can use the formal language or technical language that is produced in situations that require the use of the standard language, for example, in classrooms or work meetings or language informal or everyday language that is used when there is intimacy between speakers, using colloquial expressions.

Everyday language is the use of informal, familiar language and is characterized by being a spontaneous, relaxed and expressive language. In everyday language, the speaker uses onomatopoeia, short sentences, repetitions, redundancies, among others. In turn, technical language is used by scientific and professional people.

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nirvana33 [79]

Answer:

t = 444.125 sec

Explanation:

Given data:

V = 24 volt

I  = 0.1 ampere

mass of water mw = 51 gm

cr = 4.18 J/gm degree K^-1

mass of resistor = 8 gm

cr = 3.7 J/gm degree K^-1

we know that power is given as

Power P = VI

But P =E/t

so equating both side we have

\frac{E}{t} = VI

solving for t

t = \frac{E}{VI}

t = \frac{m_w C_w \Delta T}{VI}

t = \frac{51 \times 4.18 \times (5 -0)}{24\times 0.1}

t = 444.125 sec

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3 years ago
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St. John’s

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10. John does 176 J of work lifting himself a distance of 0.40 m. How
ch4aika [34]
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Convert 252 cL into uL
OleMash [197]

Answer:

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In 0.601 s, a 13.1-kg block is pulled through a distance of 4.19 m on a frictionless horizontal surface, starting from rest. The
Naya [18.7K]

Answer:

0.615 m

Explanation:

We need to determine the force on the spring first. By Newton's second law of motion, force is the product of the mass and acceleration. The mass is given.

The acceleration is determined using the equation of motion.

Given parameters:

Initial velocity, <em>u</em> = 0.00 m/s

Distance, <em>s</em> = 4.19 m

Time, <em>t</em> = 0.601 s

We use the equation

s = ut+\frac{1}{2}at^2

With <em>u</em> = 0.00 m/s,

s = \frac{1}{2}at^2

a = \dfrac{2s}{t^2}

a = \dfrac{2\times4.19\text{ m}}{(0.601\text{ s})^2} = 23.2\text{ m/s}^2

The force is

F = (13.1\text{ kg})(23.2\text{ m/s}^2) = 303.92 \text{ N}

From Hooke's law, the extension, <em>e</em>, of a string is given by

e = \dfrac{F}{k}

where <em>k</em> is the spring constant.

Hence,

e = \dfrac{303.92\text{ N}}{494\text{ N/m}} = 0.615\text{ m}

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3 years ago
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