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3 years ago

How often do very active sunspot and solar flare cycles take place?

Physics

2 answers:

Alchen [17]3 years ago

5 0

Answer:

Every 11 years The solar cycle is the cycle that the Sun's magnetic field goes through approximately every 11 years.

Explanation:

Our Sun is a Large ball of electrically-charged hot gas. This charged gas moves, generating a powerful magnetic field. The Sun's magnetic field goes through a cycle, called the solar cycle.

Every 11 years or so, the Sun's magnetic field completely flips. This means that the Sun's north and south poles switch places. Then it takes about another 11 years for the Sun’s north and south poles to flip back again.

The solar cycle affects activity on the surface of the Sun, such as sunspots which are caused by the Sun's magnetic fields. As the magnetic fields change, so does the amount of activity on the Sun's surface.

Hope this helps!

Brain-List?

alexira [117]3 years ago

3 0

11 Years

Usually sunspot and solar flare cycles vary. But the average duration occurs every 11 years, sometimes 9-14 years as well.

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Earth rotates on is axis approximately once every 24 hours, causing the day to night cycle and?

katovenus [111]

Yes this is true this causes day and night

6 0

3 years ago

An green hoop with mass mh = 2.8 kg and radius rh = 0.13 m hangs from a string that goes over a blue solid disk pulley with mass

Otrada [13]

The only force on the system is the mass of the hoop F net = 2.8kg*9.81m/s^2 = 27.468 N The mass equal of the rolling sphere is found by: the sphere rotates around the contact point with the table.
So by applying the theorem of parallel axes, the moment of inertia of the sphere is computed by:I = 2/5*mR^2 for rotation about the center of mass + mR^2 for the distance of the axis of rotation from the center of mass of the sphere.
I = 7/5*mR^2 M = 7/5*m
Therefore, linear acceleration is computed by:F/m = 27.468 / (2.8 + 1/2*2 + 7/5*4) = 27.468/9.4 = 2.922 m/s^2

7 0

4 years ago

Rain drops fall on a tile surface at a density of 4638 drops/ft2. There are 17 tiles/ft2. How many drops fall on each tile? Answ

Vinil7 [7]

Answer: 272.82 drop/tile

Explanation:

Given that the Rain drops fall on a tile surface at a density of 4638 drops/ft2. There are 17 tiles/ft2. How many drops fall on each tile?

Tiles/ft^2 × drop/tiles = drop/ft^2

Tiles will cancel out. Leaving the answer to be drop/ ft^2

Substitutes all the magnitude of the above units.

17 × drop/tiles = 4638

Make drop/tiles the subject of formula

Drop/tiles = 4638/17

Drop/tiles = 272.82

Therefore, 272.82 drop/tile drops fall on each tile?

8 0

3 years ago

Which plane divides the body into right and left halves

Andru [333]

Median plane - Sagittal plane through the midline of the body; divides the body or any of its parts into right and left halves.

4 0

2 years ago

Drops of rain fall perpendicular to the roof of a parked car during a rainstorm. The drops strike the roof with a speed of 15 m/

IrinaK [193]

Answer: (a) 1.065 N (b) 2.13 N

Explanation:

<h2>(a) average force exerted by the rain on the roof</h2><h2 />

According Newton's 2nd Law of Motion the force $F$ is defined as <u>the variation of linear momentum</u> $p$ <u>in time:</u>

$F=\frac{dp}{dt}$ (1)

Where the linear momentum is:

$p=mV$ (2) Being $m$ the mass and $V$ the velocity.

In the case of the rain drops, which initial velocity is $V_{i}=15m/s$ and final velocity is $V_{f}=0$ (we are told the drops come to rest after striking the roof). The momentum of the drops $p_{drops}$ is:

$p_{drops}=mV_{i}+mV_{f}$ (3)

If $V_{f}=0$ , then:

$p_{drops}=mV_{i}$ (4)

Now the force $F_{drops}$ exerted by the drops is:

$F_{drops}=\frac{dp_{drops}}{dt}=\frac{d}{dt}mV_{i}$ (5)

$F_{drops}=\frac{dm}{dt}V_{i}+m\frac{dV_{i}}{dt}$ (6)

At this point we know the mass of rain per second (mass rate) $\frac{dm}{dt}=0.071 kg/s$ and we also know the initial velocity does not change with time, because that is the velocity at that exact moment (instantaneous velocity). Therefore is a constant, and the derivation of a constant is zero.

This means (6) must be rewritten as:

$F_{drops}=\frac{dm}{dt}V_{i}$ (7)

$F_{drops}=(0.071 kg/s)(15m/s)$ (8)

$F_{drops}=1.065kg.m/s^{2}=1.065N$ (9) This is the force exerted by the rain drops on the roof of the car.

<h2>(b) average force exerted by hailstones on the roof </h2><h2 />

Now let's assume that instead of rain drops, hailstones fall on the roof of the car, and let's also assume these hailstones bounce back up off after striking the roof (this means they do not come to rest as the rain drops).

In addition, we know the hailstones fall with the same velocity as the rain drops and have the same mass rate.

So, in this case the linear momentum $p_{hailstones}$ is: