Answer:
59.4 meters
Explanation:
The correct question statement is :
A floor polisher has a rotating disk that has a 15-cm radius. The disk rotates at a constant angular velocity of 1.4 rev/s and is covered with a soft material that does the polishing. An operator holds the polisher in one place for 4.5 s, in order to buff an especially scuff ed area of the floor. How far (in meters) does a spot on the outer edge of the disk move during this time?
Solution:
We know for a circle of radius r and θ angle by an arc of length S at the center,
S=rθ
This gives
θ=S/r
also we know angular velocity
ω=θ/t where t is time
or
θ=ωt
and we know
1 revolution =2π radians
From this we have
angular velocity ω = 1.4 revolutions per sec = 1.4×2π radians /sec = 1.4×3.14×2×= 8.8 radians / sec
Putting values of ω and time t in
θ=ωt
we have
θ= 8.8 rad / sec × 4.5 sec
θ= 396 radians
We are given radius r = 15 cm = 15 ×0.01 m=0.15 m (because 1 m= 100 cm and hence, 1 cm = 0.01 m)
put this value of θ and r in
S=rθ
we have
S= 396 radians ×0.15 m=59.4 m
Answer:
The answer is true
Explanation:
could you brainliest again they said I plagarized when I didn't
143m/s if you just perhaps by what you know you'll figure it out
Answer:
3/7 ω
Explanation:
Initial momentum = final momentum
I(-ω) + (2I)(3ω) + (4I)(-ω/2) = (I + 2I + 4I) ωnet
-Iω + 6Iω - 2Iω = 7I ωnet
3Iω = 7I ωnet
ωnet = 3/7 ω
The final angular velocity will be 3/7 ω counterclockwise.
