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Artyom0805 [142]
4 years ago
5

Does research lessen its value for its being cyclical? Enumerate Two Possible advantages of a cyclical research

Physics
1 answer:
Tanya [424]4 years ago
7 0

Answer:

Research usually is a cyclical process because it starts with a problem and ends with a problem.

This is bad or lessens the value of the research?

No, because there are problems that in nature are cyclical, (for example the related ones to action research).

And it also may be a good thing, solving one "problem" leads to another problem, but in the process of solving the first one we may win a lot of knowledge, and the same happens with the next one, and so on.

Two possible advantages of cyclical research are:

Knowing beforehand that the research will be cyclical, will allow us to estimate better the amount of time and money needed because we already know (more or less) where to aim.

It also may lead to a better end product, as we already know that we must focus in solving one thing and then we can focus in the next one.

Another possible advantage may be that we know that after the work, there will be a new thing to research, and it is fun, so if you are curious enough this may be a good thing (especially in scientific areas, physics, chemistry, etc).

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Xamine the graph. Select the statement that best describes the energy change in the particles of a substance during melting.
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Answer:

  • Option B) Absorbed energy results in the change in potential energy.

Explanation:

Please, find attached the graph that accompanies this question.

The<em> melting</em> proces is the change from solid phase to liquid phase. It is represented with the lower flat line with the symbol ΔHfus over it.

The line is flat because the temperature remains constant during this process. Thus, you know the option "C) As the temperature increases during melting, the kinetic energy also increases" is FALSE.

What happens during this process is:

  • Most of the energy received by the particles from heating, during the melting process, goes to overcome the intermolecular bonds between the particles. This results in increasing the distance between the particles, so the internal potential energy increases. This is what the option <em>"B) Absorbed energy results in the change in potential energy" correctly describes.</em> Hence, option B) is TRUE.

Althoug most of the heat energy received is transformed into potential energy, yet a small part of the heat energy increases a bit the kinetic energy of the particles, because the particles will vibrate faster around their relatively fixed positions. Hence, the option "<em>A) The kinetic energy of the particles remains unchanged</em>" is FALSE.

As for option D) it is not reasonable at all: none chemical or physical priciple can be used to state that <em>the kinetic energy decreases as the particles move farther apart</em>. Thus, this is FALSE.

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An ideal spring hangs from the ceiling. A 1.25-kg mass is hung from the spring. After all vibrations have died away, the spring
ch4aika [34]

The kinetic energy of the mass at the instant it passes back through its equilibrium position is about 1.20 J

\texttt{ }

<h3>Further explanation</h3>

Let's recall Elastic Potential Energy formula as follows:

\boxed{E_p = \frac{1}{2}k x^2}

where:

<em>Ep = elastic potential energy ( J )</em>

<em>k = spring constant ( N/m )</em>

<em>x = spring extension ( compression ) ( m )</em>

Let us now tackle the problem!

\texttt{ }

<u>Given:</u>

mass of object = m = 1.25 kg

initial extension = x = 0.0275 m

final extension = x' = 0.0735 - 0.0275 = 0.0460 m

<u>Asked:</u>

kinetic energy = Ek = ?

<u>Solution:</u>

<em>Firstly , we will calculate the spring constant by using </em><em>Hooke's Law</em><em> as follows:</em>

F = k x

mg = k x

k = mg \div x

k = 1.25(9.8) \div 0.0275

k = 445 \frac{5}{11} \texttt{ N/m}

\texttt{ }

<em>Next , we will use </em><em>Conservation of Energy</em><em> formula to solve this problem:</em>

Ep_1 + Ek_1 = Ep_2 + Ek_2

\frac{1}{2}k (x')^2 + mgh + 0 = \frac{1}{2}k x^2 + Ek

Ek = \frac{1}{2}k (x')^2 + mgh - \frac{1}{2}k x^2

Ek = \frac{1}{2}k ( (x')^2 - x^2 ) + mgh

Ek = \frac{1}{2}(445 \frac{5}{11}) ( 0.0460^2 - 0.0275^2 ) + 1.25(9.8)(0.0735)

\boxed {Ek \approx 1.20 \texttt{ J}}

\texttt{ }

<h3>Learn more</h3>
  • Kinetic Energy : brainly.com/question/692781
  • Acceleration : brainly.com/question/2283922
  • The Speed of Car : brainly.com/question/568302
  • Young Modulus : brainly.com/question/9202964
  • Simple Harmonic Motion : brainly.com/question/12069840

\texttt{ }

<h3>Answer details</h3>

Grade: High School

Subject: Physics

Chapter: Elasticity

8 0
3 years ago
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