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motikmotik
3 years ago
11

How are sunspots related to prominences and solar flares?

Physics
1 answer:
solniwko [45]3 years ago
7 0
<h2><u> Answer:</u></h2><h2><u /></h2>

The Sun is a star that experiences changes in its activity in a cycle of about 11 years, known as the solar cycle. During that time, the amount of sunspots on its surface may vary.

But first it is necessary to define what a sunspot is.

A <u>sunspot </u>is a region of the Sun that has a lower temperature than its surroundings, and with an intense magnetic activity. It consists of a dark central region, called <em>"umbra"</em>, surrounded by a lighter region<em>"penumbra"</em>.

Now, a solar protuberance or <u>prominence</u> is a large gaseous structure located on the surface of the Sun, often in a loop shape, which emerges from the surface of the Sun, the photosphere, and extends to reach the solar Corona. It is caused by disturbances in the Sun's magnetic field (hence it is<u> related to sunspots</u>) and, although most of the gas expelled returns to the surface, sometimes this can release particles that can reach the Earth.

In this context, a <u>solar flare</u> is a sudden, rapid and intense variation of brightness that is observed on the surface of the Sun due to an explosion of hot gases. It releases a large amount of energy in the form of electromagnetic radiation, energetic particles and mass in motion. These phenomena take place in active regions of the Sun, areas of strong magnetic field, and especially in sunspots with great magnetic complexity.

These eruptions are also known as Coronal Mass Ejections (CME), since they occur in the Sun's Corona layer; and can produce X-rays and gamma rays. However, <u>they are classified according to the peak of X-ray flow</u> by the letters A, B, C, M or X, where each letter (or class)  represents an increase in energy production 10 times greater than the previous one.

In summary:

<h2> When a sunspot is observed it is because in that region there was a coronal mass ejection, also known as a solar flare. This in turn is related to solar prominences. </h2>

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A mass of 0.250 kg is attached to a spring and undergoes simple harmonic oscillations with a period of 0.640 s. What is the forc
Paha777 [63]

The force constant of the spring is approximately 24.038 newtons per meter.

As we are talking about Simple Harmonic Motion. In this exercise we need to determine the Spring Constant (k), in newtons per meter, from the equation of the Period (T), in seconds, which is described below:

T = 2\pi\cdot \sqrt{\frac{m}{k} } (1)

Where m is the mass of the moving element, in kilograms.

If we know that T = 0.640\,s and m = 0.250\,kg, then the spring constant of the spring is:

0.640 = 2\pi\cdot \sqrt{\frac{0.250}{k} }

\sqrt{\frac{0.250}{k} } \approx 0.102

\frac{0.250}{k} \approx 0.0104

k \approx 24.038\,\frac{N}{m}

The force constant of the spring is approximately 24.038 newtons per meter.

Please see this question related to Simple Harmonic Motion for further details: brainly.com/question/17315536

6 0
3 years ago
Read 2 more answers
An element's atomic number is the​
Zepler [3.9K]

the atomic number of a chemical element (also known as its proton number) is the number of protons found in the nucleus of an atom of that element, and therefore identical to the charge number of the nucleus.

Hope this helped

8 0
3 years ago
What is the relationship in Newton's second law between force, mass and acceleration?
Anuta_ua [19.1K]

Answer:

Newton's second law of motion states that the acceleration of a system is directly proportional to and in the same direction as the net external force acting on the system, and inversely proportional to its mass.  If the only force acting on an object is due to gravity, the object is in free fall.

5 0
3 years ago
A plant is thrown straight
olchik [2.2K]

The time taken for the plant to hit the ground from a distance of 7.01m and at a velocity of 8.84m/s is 1.59s.

<h3>How to calculate time?</h3>

The time taken for a motion to occur can be calculated using the following formula:

v² = u² - 2as

Where;

  • v = final velocity
  • u = initial velocity
  • s = distance
  • a = acceleration

8.84² = 0² + 2 × a × 7.01

78.15 = 14.02a

a = 5.57m/s²

V = u + at

8.84 = 0 + 5.57t

t = 1.59s

Therefore, the time taken for the plant to hit the ground from a distance of 7.01m and at a velocity of 8.84m/s is 1.59s.

Learn more about time at: brainly.com/question/13170991

#SPJ1

4 0
2 years ago
You drive 4 km at 30 km/h and then another 4 km at 50 km/h. What is your average speed for the whole 8-km trip?
Setler79 [48]

Answer:

option A

Explanation:

given,

drive 4 km at 30 km/h and

another 4 km at 50 km/h

average speed for the whole 8 Km trip = ?

time for the trip of first 4 km

 distance = speed x time

t_1 = \dfrac{d}{s}

t_1 = \dfrac{4}{30}

     t₁= 0.1333 hr

time from another 4 km trip

t_2 = \dfrac{d}{s}

t_2= \dfrac{4}{50}

     t₂ = 0.08 hr

average speed

     s= \dfrac{d}{t}  

     s= \dfrac{8}{0.1333 + 0.08}  

     s= \dfrac{8}{0.2133}  

            s = 37.50 Km/hr

Less than 40 Km/h

the correct answer is option A

6 0
3 years ago
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