Answer:
15.66 rad/s
Explanation:
The vertical motion and horizontal motion are independent of each other.
t = √ ( 2 s/ g) where t = time for the ball to reach the ground and s is the height of the cliff = 18.0 m
t = √ ( 36 / 9.81 ) = 1.916 secs
horizontal distance travel = ut where u is the horizontal velocity of the stone = 30 × r (radius)
tangential velocity V = angular velocity ( ω) × radius
distance traveled = ω × r × t = 30 × r
radius cancelled on both side
ω = 30 / 1.9156 = 15.66 rad/s
If you were to sit a hot cup of water out side it would frezze faster
I believe it’s stay in motion if it’s not acted on by an unbalanced force
Answer:
Explanation:
Given that,
Initial angular velocity is 0
ωo=0rad/s
It has angular velocity of 11rev/sec
ωi=11rev/sec
1rev=2πrad
Then, wi=11rev/sec ×2πrad
wi=22πrad/sec
And after 30 revolution
θ=30revolution
θ=30×2πrad
θ=60πrad
Final angular velocity is
ωf=18rev/sec
ωf=18×2πrad/sec
ωf=36πrad/sec
a. Angular acceleration(α)
Then, angular acceleration is given as
wf²=wi²+2αθ
(36π)²=(22π)²+2α×60π
(36π)²-(22π)²=120πα
Then, 120πα = 8014.119
α=8014.119/120π
α=21.26 rad/s²
Let. convert to revolution /sec²
α=21.26/2π
α=3.38rev/sec
b. Time Taken to complete 30revolution
θ=60πrad
∆θ= ½(wf+wi)•t
60π=½(36π+22π)t
60π×2=58πt
Then, t=120π/58π
t=2.07seconds
c. Time to reach 11rev/sec
wf=wo+αt
22π=0+21.26t
22π=21.26t
Then, t=22π/21.26
t=3.251seconds
d. Number of revolution to get to 11rev/s
∆θ= ½(wf+wo)•t
∆θ= ½(0+11)•3.251
∆θ= ½(11)•3.251
∆θ= 17.88rev.
