Chalecos no tienen  mangas. Vests don't have sleeves.<span />
        
             
        
        
        
The sun’s gravitational attraction and the planet’s inertia keeps planets moving is circular orbits.
Explanation:
The planets in the Solar System move around the Sun in a circular orbit. This motion can be explained as a combination of two effects:
1) The gravitational attraction of the Sun. The Sun exerts a force of gravitational attraction on every planet. This force is directed towards the Sun, and its magnitude is

where
G is the gravitational constant
M is the mass of the Sun
m is the mass of the planet 
r is the distance between the Sun and the planet
This force acts as centripetal force, continuously "pulling" the planet towards the centre of its circular orbit.
2) The inertia of the planet. In fact, according to Newton's first law, an object in motion at constant velocity will continue moving at its velocity, unless acted upon an external unbalanced force. Therefore, the planet tends to continue its motion in a straight line (tangential to the circular orbit), however it turns in a circle due to the presence of the gravitational attraction of the Sun.
Learn more about gravity:
brainly.com/question/1724648
brainly.com/question/12785992
#LearnwithBrainly
 
        
             
        
        
        
After the great 1906 San Francisco earthquake, geolophysicistHarry Fielding Reid examined the displacement of the ground surface along the San Andreas Fault. He concluded that the quake must have been the result of the elastic reboundof the strain energy in the rocks on either side of the fault.
strain energy is 0. 5x force x (compression) X (compression) 
There is a lot of force and a bit of compression when rocks squash up against other rocks causing earthquakes
        
             
        
        
        
Incomplete question.The Complete question is here
A flat uniform circular disk (radius = 2.00 m, mass = 1.00 ✕ 102 kg) is initially stationary. The disk is free to rotate in the horizontal plane about a friction less axis perpendicular to the center of the disk. A 40.0-kg person, standing 1.25 m from the axis, begins to run on the disk in a circular path and has a tangential speed of 2.00 m/s relative to the ground.
a.) Find the resulting angular speed of the disk (in rad/s) and describe the direction of the rotation.
b.) Determine the time it takes for a spot marking the starting point to pass again beneath the runner's feet.
Answer:
(a)ω = 1 rad/s
(b)t = 2.41 s
Explanation:
(a) initial angular momentum = final angular momentum  
0 = L for disk + L............... for runner
0 = Iω² - mv²r ...................they're opposite in direction
0 = (MR²/2)(ω²) - mv²r
................where is ω is angular speed which is required in part (a) of question
0 = [(1.00×10²kg)(2.00 m)² / 2](ω²) - (40.0 kg)(2.00 m/s)²(1.25 m)
0=200ω²-200
200=200ω²
ω = 1 rad/s
b.)
lets assume the "starting point" is a point marked on the disk.
The person's angular speed is  
v/r = (2.00 m/s) / (1.25 m) = 1.6 rad/s
As the person and the disk are moving in opposite directions, the person will run part of a revolution and the turning disk would complete the whole revolution.
(angle) + (angle disk turns) = 2π
(1.6 rad/s)(t) + ωt = 2π
t[1.6 rad/s + 1 rad/s] = 2π
t = 2.41 s