Answer:
t₂ = t₁ / 5
Explanation:
Rotational kinematics using: ωf = ωi + αt
Starting from rest and speeding up:
ω₁ = 0 + αt₁ .. Eq1
Starting from ω₁ and slowing to a stop:
0 = ω₁ - 5αt₂
Substituting for ω₁ from Eq 1
0 = αt₁ - 5αt₂
5αt₂ = αt₁
5t₂ = t₁
t₂ = t₁ / 5
C.) Mitochondria is the answer...
Answer:
Explanation:
if you are pulling in the positive direction and Upward is also positive.
F = -0.96i + 0.58(9.8)j
F = -0.96i + 5.7j
F = √(-0.96² + 5.7²) = 5.8 N
A car moves along an x axis through a distance of 900 m, starting at rest (at x = 0) and ending at rest (at x = 900 m). Through the first 1/4 of that distance, its acceleration is +6.25 m/s2. Through the next 3/4 of that distance, its acceleration is -2.08 m/s2. What are (a) its travel time through the 900 m and (b) its maximum speed?
<span>Solve for the time at the 1/4 mark. That's 225 m. How? d = (1/2)at^2 ( initial velocity zero). Thus 225 = (1/2) 6.25 t^2. t^2 = ( 225 * 2 ) / 6.25. t = 8.5 sec. </span>
<span>At the other end t^2 = (675 * 2) / 2.08 -- we reversed the sign and ran time backwards. t = 25.5 sec. </span>
<span>So total time is 8.5 + 25.5 or 34 sec. </span>
<span>Since zero initial velocity: v^2 = 2 a d. Here, v^2 = 2 * 6.25 * 225. v = 53 m/s. That's the fastest speed since braking then occurs.</span>