Hi! Let me help you!
a = (Vf - Vi)/t ; where distance d = [2(t)]/(4+t), t = 5secs, and Vi = 0
a = [(2t)/(4+t)]/t <---- working equation
a = {[2(5)]/9}/5 <---- cancel 5
a = 2/9 ft/s^2 <---- Answer
Try 40. it seems correct. i’m sorry if i’m wrong.
Answer:
r = 0.02 m
Explanation:
from the question we have :
speed = 1 rps = 1x 60 = 60 rpm
coefficient of friction (μ) = 0.1
acceleration due to gravity (g) = 9.8 m/s^{2}
maximum distance without falling off (r) = ?
to get how far from the center of the disk the coin can be placed without having to slip off we equate the formula for the centrifugal force with the frictional force on the turntable force
mv^2 / r = m x g x μ
v^2 / r = g x μ .......equation 1
where
velocity (v) = angular speed (rads/seconds) x radius
angular speed (rads/seconds) = (\frac{2π}{60} ) x rpm
angular speed (rads/seconds) = (\frac{2 x π}{60} ) x 60 = 6.28 rads/ seconds
now
velocity = 6.28 x r = 6.28 r
now substituting the value of velocity into equation 1
v^2 / r = g x μ
(6.28r)^2 / r = 9.8 x 0.1
39.5 x r = 0.98
r = 0.02 m
