LakalakalalamLlalaa hahahaha
m = mass of the circular hoop
r = radius of the hoop
I = moment of inertia of the hoop
moment of inertia of the hoop about the center of hoop is given as
I = m r²
k = distance of the point of suspension from center of mass = r
using parallel axis theorem
I' = moment of inertia of hoop about the point of suspension
I' = I + m k²
I' = m r² + m k²
I' = m r² + m r²
I' = 2 m r²
Time period of oscillation for the hoop is given as
T = 2π sqrt(I'/mgk)
T = 2π sqrt(2 m r²/(mgr))
T = 2π sqrt(2 r/g)
since 2r = diameter = d
T = 2π sqrt(d/g)
Answer:
the maximum height of Joe's ball will be 4 times Bill's ball.
Explanation:
let bill's velocity be v then Joe's velocity is 2v.
and initial velocities of both bill and Joe are 0
for Bill


for Joe


thus we can write that
h'=4h
the maximum height of Joe's ball will be 4 times Bill's ball.
Answer:
0 is your answereeeeerrrrr
Since the mass is 5 grams the acceleration is: 4000 m/s^2
But if the mass is 5 Kilograms the acceleration is: 4 m/s^2