At t1 = 2.00 s, the acceleration of a particle in counterclockwise circular motion is 6.00 i + 4.00 j m/s2 . it moves at constant speed. at time t2 = 5.00 s, its acceleration is 4.00 i - 6.00 j m/s2 . what is the radius of the path taken by the particle if t2-t1 is less than one period?

The particle moves with constant speed in a circular path, so its acceleration vector always points toward the circle's center.

At time , the acceleration vector has direction such that

which indicates the particle is situated at a point on the lower left half of the circle, while at time the acceleration has direction such that

which indicates the particle lies on the upper left half of the circle.

Notice that . That is, the measure of the major arc between the particle's positions at and is 270 degrees, which means that is the time it takes for the particle to traverse 3/4 of the circular path, or 3/4 its period.

Recall that

where is the radius of the circle and is the period. We have

and the magnitude of the particle's acceleration toward the center of the circle is

So we find that the path has a radius of

answer: d. a cme increases the amount of energy in the solar wind

explanation:

coronal mass ejections, also called cmes, are large clouds of plasma and magnetic field that erupt from the sun. only when the cloud is aimed at earth will the cme hit earth and therefore cause impacts. high-speed solar wind streams come from areas on the sun known as coronal holes.