Ω₀ = the initial angular velocity (from rest)
t = 0.9 s, time for a revolution
θ = 2π rad, the angular distance traveled
Let
α = the angular acceleration
ω = the final angular velocity
The angular rotation obeys the equation
(1/2)*(α rad/s²)*(0.9 s)² = (2π rad)
α = 15.514 rad/s²
The final angular velocity is
ω = (15.514 rad/s²)*(0.9 s) = 13.963 rad/s
If the thrower's arm is r meters long, the tangential velocity of release will be
v = 13.963r m/s
Answer: 13.963 rad/s
Answer:F=1.7802
Explanation:
Since we've been given the mass to be .18kg,we are asked to find the Force of which the formulae is
F=ma where f-force,m-mass and a-acceleration due to gravity
So we can just substitute
F-?.m=.18 and a9.89
F=.18×9.89
F=1.7802N
Answer:what are the answer options?
Explanation
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Explanation:
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