The tub within a washer goes into its spin cycle, starting from rest and reaching an angular speed of 13 π rad/s in 6.6 s. at this point, the lid is opened, and a safety switch turns off the washer. the tub slows to rest in 12.2 s. through how many revolutions does the tub turn? assume constant angular acceleration while the machine is starting and stopping. answer in units of rev.

θ  = 21.45 revolutions

Explanation:

Angular acceleration of the tub

We apply the equations of circular motion uniformly accelerated

ωf= ω₀ + α*t  Formula (1)

Where:  

α : Angular acceleration (rad/s²)  

ω₀ : Initial angular speed ( rad/s)  

ωf : Final angular speed ( rad

t :  time interval (s)

Data

ω₀ = 0

ωf = 13π rad/s

t= 6.6 s

We replace data in the formula (1) :

ωf= ω₀ + α*t

13π = 0 + α* (6.6)

α = (13π) / (6.6)

α = 6.18 rad/s²

Revolutions made by the tub

We apply the equations of circular motion uniformly accelerated

ωf²= ω₀²+ 2α*θ    Formula (2)

Where:  

θ : Angle that the body has rotated in a given time interval (rad)

We replace data in the formula (2):  

(ωf)²= ω₀²+ 2α*θ

(13π)²= (0)²+ 2(6.18 )*θ

θ = (13π)²/ (12.376)

θ = 134.77 rad

1 rev = 2π rad  

θ = 134.77 rad* (1 rev/2π rad)

θ  = 21.45 revolutions

u=13

v=30

s/t=u+1\2at

=13+431/2

=13+265.5

=278.5

answer: strength i guess

explanation: