A car is traveling at a constant speed along the road ABCDE shown in the drawing. Section Ab and DE are straight. Rank the acceleration in each of the four sections according to magnitude (smallest first). -AB & DE tie, CD, BC
-all tie
-CD, BC, AB & DE tie
-AB & DE tie, BC , CD
-BC, CD, AB & DE tie

AB = DE <CD <BC

Explanation:

This is an exercise in kinetics, the accelerations defined as the change in velocity over the time interval, therefore the accelerations of a vector.

Because the acceleration is a vector, it has two parts, the modulus that the numerical value of the magnitude and the direction, a change in any of them implies the existence of a relationship.

Let's apply these reasoning to our problem.

AB Path  

this path is straight and as they indicate that the constant speed the acceleration is zero

DE path

This path is straight and since the velocity is constant the zero steps

BC path

This path is a curve and the velocity modulus is constant, but its directional changes therefore there is an acceleration called centripetal, given by the expression

          = v² / r

where r is the radius of the curve and the direction of acceleration is towards the center of the curve

CD path

This path is a curve and it also has centripetal acceleration, as can be seen in the drawing, the radius of the curve is greater than in section BC, therefore the acceleration is less

              >

In  summary lower accelerations are

 AB = DE <CD <BC

part 1: which substance has a higher solubility?

a

smallest amount left undissolved atall temps in table shown most has dissolved high sol'by