A125-kg merry-go-round in the shape of a uniform, solid, horizontal disk of radius 1.50 m is set in motion by wrapping a rope about the rim of the disk and pulling on the rope. what constant force would have to be exerted on the rope to bring the merry-go-round from rest to an angular speed of 0.700 rev/s in 2.00 s? (state the magnitude of the force.)

201.6 N

Explanation:

m = mass of disk shaped merry-go-round = 125 kg

r = radius of the disk = 1.50 m

w₀ = Initial angular speed = 0 rad/s

w = final angular speed = 0.700 rev/s = (0.700) (2π) rad/s = 4.296 rad/s

t = time interval = 2 s

α = Angular acceleration

Using the equation

w = w₀ + α t

4.296 = 0 + 2α

α = 2.15 rad/s²

I = moment of inertia of merry-go-round

Moment of inertia of merry-go-round is given as  

I = (0.5) m r² = (0.5) (125) (1.50)² = 140.625 kgm²

F = constant force applied

Torque equation for the merry-go-round is given as

r F = I α

(1.50) F = (140.625) (2.15)

F = 201.6 N

are there options? cause you ain’t. real specific

answer: at 24 february, 2014 the population gets twice of its initial population.

explanation:

since we have given that

the population for alpha city, t years after january 1, 2004 is given as

first we find out the initial population, i.e. t=0,

so, our quadratic equation becomes,

according to question, we have said that if the population has twice its initial population, then it becomes

so, time taken to reach this above mentioned population is given by

now, we know that time can't be negative so, we take t=9.149 years.

hence, at 24 february, 2014 the population gets twice of its initial population.