A bicycle tire is spinning clockwise at 2.50 rad/s. During a time period Dt 5 1.25 s, the tire is stopped and spun in the oppo- site (counterclockwise) direction, also at 2.50 rad/s. Calculate (a) the change in the tire’s angular velocity Dv and (b) the tire’s average angular acceleration aav .

a) The change in the tire’s angular velocity is 5 rads/sec

b) The tire’s average angular acceleration is 4 rad/s²

Explanation:

initial angular speed , ω (initial) = - 2.5 rad/s

final speed , ω (final) = 2.5 rad/s

time = 1.25 s

a) The change in tire's angular speed

ω = ω (final) - ω (initial)

ω = 2.5 - ( -2.5) rad/s

ω = 2.5 + 2.5 (rad/s)

ω = 5 rad/s

b) The average angular acceleration

a = ω ÷ t

a = 5 rad/s ÷ 1.25 s

a = 4 rad/s²

potential energy is directly proportional to height.

more height = more potential energy.

greatest increase in height = greatest increase in potential energy

where are the points with the greatest increase in height ?

has to be c-e or d-e .

scientific laws are the result of testing theories and proving them to be facts.