28. PART 1: Caleb is investigating the effect of friction on the motion of an
object. He uses the following supplies for the investigation:
1. a wooden block
2. three ramps, each ramp is the same height and length but
has a different surface (smooth wood, carpet, rough
sandpaper)
3. a stopwatch
Caleb tests his hypothesis that the smooth wood ramp surface
causes less friction than the carpet and that the carpet causes less
friction than the rough sandpaper.
A. Describe a procedure using these supplies that Caleb may
follow to test his hypothesis.

Is a  vector quantity  that refers to "the rate at which an object changes its position." imagine a person moving rapidly - one step forward and one step back - always returning to the original starting position. while this might result in a frenzy of activity, it would result in a zero velocity. because the person always returns to the original position, the motion would never result in a change in position. since velocity is defined as the rate at which the position changes, this motion results in zero velocity. if a person in motion wishes to maximize their velocity, then that person must make every effort to maximize the amount that they are displaced from their original position. every step must go into moving that person further from where he or she started. for certain, the person should never change directions and begin to return to the starting position.

Answerthe energy required is 1.60 × 10⁻¹⁸ j, to three significant figures.explanationwe define electric potential as the ratio electric potential energy to the charge,    then the following gives us the potential difference,      the formula  may be on your formula sheet.the change in electric potential energy would be the energy that would needed to move it from point a to point b. solving for that, we get    using q = 8.00  × 10⁻¹⁹ c and  δv = 2.00 v, we get    the energy required is 1.60 × 10⁻¹⁸ j, to three significant figures.

If the heat engine is a perfect carnot cycle engine, thenefficiency = (t2-t1)/t2   where t1=temperature of cold reservoir ( ° k) = 135            t2=temperature of hot reservoir ( ° k)   = 325thereforeefficiency = (t2-t1)/t2 = (325-135)/325 = 0.5846 = 58%  note: this is the theoretical maximum efficiency of a perfect heat engine.   real-life engines do not achieve this value.