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3 years ago

5

Which feature of the sun appears in cycles of about 11 years?

1 answer:

goldfiish [28.3K]3 years ago

8 0

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.

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If the action force is a player kicking a soccer ball, then what is the reaction force?

LiRa [457]

The force acting on his feet.

4 0

3 years ago

A 100-kg spacecraft is in a circular orbit about Earth at a height h = 2RE .

maria [59]

To solve this problem it is necessary to apply the concepts related to the conservation of the Gravitational Force and the centripetal force by equilibrium,

$F_g = F_c$

$\frac{GmM}{r^2} = \frac{mv^2}{r}$

Where,

m = Mass of spacecraft

M = Mass of Earth

r = Radius (Orbit)

G = Gravitational Universal Music

v = Velocity

Re-arrange to find the velocity

$\frac{GM}{r^2} = \frac{v^2}{r}$

$\frac{GM}{r} = v^2$

$v = \sqrt{\frac{GM}{r}}$

PART A ) The radius of the spacecraft's orbit is 2 times the radius of the earth, that is, considering the center of the earth, the spacecraft is 3 times at that distance. Replacing then,

$v = \sqrt{\frac{(6.67*10^{-11})(5.97*10^{24})}{3*(6.371*10^6)}}$

$v = 4564.42m/s$

From the speed it is possible to use find the formula, so

$T = \frac{2\pi r}{v}$

$T = \frac{2\pi (6.371*10^6)}{4564.42}$

$T = 8770.05s\approx 146min\approx 2.4hour$

Therefore the orbital period of the spacecraft is 2 hours and 24 minutes.

PART B) To find the kinetic energy we simply apply the definition of kinetic energy on the ship, which is

$KE = \frac{1}{2} mv^2$

$KE = \frac{1}{2} (100)(4564.42)^2$

$KE = 1.0416*10^9J$

Therefore the kinetic energy of the Spacecraft is 1.04 Gigajules.

8 0

3 years ago

How do you find the average speed of an object?

Aleonysh [2.5K]

Answer:

Average speed is total distance divided by total time.

v = d / t

5 0

3 years ago

Read 2 more answers

A 56.6-kg crate rests on a level floor at a shipping dock. The coefficients of static and kinetic friction are 0.517 and 0.260,

kifflom [539]

Answer:

Explanation:

Magnitude of frictional force = μ mg

μ is either static or kinetic friction.

To start the crate moving , static friction is calculated .

a ) To start crate moving , force required = μ mg where μ is coefficient of static friction .

force required =.517 x 56.6 x 9.8 = 286.76 N .

b ) to slide the crate across the dock at a constant speed , force required

= μ mg where μ is coefficient of kinetic friction , where μ is kinetic friction

= .26 x 56.6 x 9.8 = 144.21 N .

3 0

2 years ago

Skater 1 has a mass of 105.0 kg and a velocity of 2.0 m/s to the left. He

mart [117]

The final velocity of skater 1 is 3.7 m/s to the right. The right option is O A. 3.7 m/s to the right.

<h3>What is velocity?</h3>

Velocity can be defined as the ratio of the displacement and time of a body.

To calculate the final velocity of Skater 1 we use the formula below.

Formula:

mu+MU = mv+MV............ Equation 1

Where:

m = mass of the first skater
M = mass of the second skater
u = initial velocity of the first skater
U = initial velocity of the second skater
v = final velocity of the first skater
V = final velocity of the second skater.

make v the subject of the equation.

v = (mu+MU-MV)/m................ Equation 2

Note: Let left direction represent negative and right direction represent positive.

From the question,

Given:

m = 105 kg
u = -2 m/s
M = 71 kg
U = 5 m/s
V = -3.4 m/s.

Substitute these values into equation 2

v = [(105×(-2))+(71×5)-(71×(-3.4))]/105
v = (-210+355+241.4)/105
v = 386.4/105
v = 3.68 m/s
v ≈ 3.7 m/s

Hence, the final velocity of skater 1 is 3.7 m/s to the right. The right option is O A. 3.7 m/s to the right.

Learn more about velocity here: brainly.com/question/25749514

7 0

2 years ago