All categories

4 years ago

When the polarity of the Sun's magnetic field is taken into account, the solar activity cycle lasts ____.

Physics

1 answer:

Bas_tet [7]4 years ago

8 0

Answer:

10.6 Years

Explanation:

The Sun is a very volatile body. It shows a lot of features like solar flares, sun spots, prominence etc. These features are result of the magnetic field and differential rotation of the Sun. The regarding the no. of sun spots visible, solar flares, intensity of flares etc. have been collected for years and it can be seen that these activities follow a cycle with maxima and minima.

This cycle is of 10.6 years. The change in the solar activity also affects some terrestrial phenomenon such as Aurora.

You might be interested in

A billiard ball is moving in the x-direction at 30.0 cm/s and strikes another billiard ball moving in the y-direction at 40.0 cm

PSYCHO15rus [73]

Explanation:

Given that,

Initial speed of the billiard ball 1, u = 30i cm/s

Initial speed of another billiard ball 2, u' = 40j cm/s

After the collision,

Final speed of first ball, v = 50 cm/s

Final speed of second ball, v' = 0 (as it stops)

Let us consider that both balls have same mass i.e. m

Initial kinetic energy of the system is :

$K_i=\dfrac{1}{2}mu^2+\dfrac{1}{2}mu'^2\\\\K_i=\dfrac{1}{2}m(u^2+u'^2)\\\\K_i=\dfrac{1}{2}m((30)^2+(40)^2)\\\\K_i=1250m\ J$

Final kinetic energy of the system is :

$K_f=\dfrac{1}{2}mv^2+\dfrac{1}{2}mv'^2\\\\K_f=\dfrac{1}{2}m(v^2+v'^2)\\\\K_f=\dfrac{1}{2}m((50)^2+(0)^2)\\\\K_f=1250m\ J$

The change in kinetic energy of the system is equal to the difference of final and initial kinetic energy as :

$\Delta K=K_f-K_i\\\\\Delta K=1250m-1250m\\\\\Delta K=0$

So, the change in kinetic energy of the system as a result of the collision is equal to 0.

7 0

4 years ago

Healthy pregnant women should get at least 150 minutes of moderate-intensity aerobic activity, spread over a 7-day period, each

kirill115 [55]

Healthy pregnant women should get at least 150 minutes of moderate-intensity aerobic activity, spread over a 7-day period, each week - True

<u>Explanation:</u>

Women who are healthy but moderately active must be engaged with aerobic activities moderately. they should be engaged with this activities at least 2 and a half hours in all 7 days of a week. They must be engaged with this activity during the pregnancy time and also in the period after their delivery.

During the time of pregnancy all women must be healthy and physically active. The ways for maintaining a very good health in the pregnancy time will be guided by the health care providers. Women should be highly active during and after pregnancy time. This helps in having a baby that is also very healthy.

3 0

3 years ago

If an 800.-kg sports car slows to 13.0 m/s to check out an accident scene and the 1200.-kg pick-up truck behind him continues tr

Alborosie

Answer: 20.2 m/s

Explanation:

From the question above, we have the following data;

M1 = 800kg

M2 = 1200kg

V1 = 13m/s

V2 = 25m/s

U (common velocity) =?

M1V1 + M2V2 = (M1 + M2). U

(800*13) + (1200*25) = (800+1200) * U

10400 + 30000 = 2000u

40400 = 2000u

U = 40400 / 2000

U = 20.2 m/s

5 0

3 years ago

Read 2 more answers

A proton initially at rest is accelerated by a uniform electric field. The proton moves 4.76 cm in 1.10 x 10^-6 s. Find the volt

iVinArrow [24]

Answer:

The voltage drop through which the proton moves is 39.1 V.

Explanation:

Given that,

Distance = 4.76 cm

Time $t=1.10\times10^{-6}\ s$

We need to calculate the acceleration

Using equation of motion

$s = ut+\dfrac{1}{2}at^2$

Where, s = distance

a = acceleration

t = time

Put the value in the equation

$4.76\times10^{-2}=\dfrac{1}{2}\times a \times(1.10\times10^{-6})^2$

$a=\dfrac{2\times 4.76\times10^{-2}}{(1.10\times10^{-6})^2}$

$a=7.87\times10^{10}\ m/s^2$

We need to calculate the voltage drop

Using formula of electric field

$F=qE$

$F = q\dfrac{V}{d}$ ....(I)

Using newton's second law

$F = ma$ ....(II)

Put the value of F in equation (I) from equation (II)

$ma=\dfrac{qV}{d}$

$V=\dfrac{mad}{q}$

Where, q = charge

a = acceleration

d = distance

m= mass of proton

Put the value into the formula

$V=\dfrac{1.67\times10^{-27}\times7.87\times10^{10}\times4.76\times10^{-2}}{1.6\times10^{-19}}$

$V=39.1\ V$

Hence, The voltage drop through which the proton moves is 39.1 V.

8 0

3 years ago

Describe how the particles of a substance behave at its boiling point.

fgiga [73]

Particles begin to move out of their ordered arrangement due to energy. They are unable to escape because of external pressure pushing down on the liquid.

4 0

3 years ago