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
I'm pretty sure that there should be an options to choose. Anyway, I've seen this question before and I know that this is an example of <span>the phi phenomenon.</span>
C. Radiosonde is the answer
the above mentioned is not correct
Answer:
Explanation:
We know that acceleration is change in velocity over time.
v is the final velocity and u is the initial velocity.
Solve for v.
Multiply both sides by t.
Add u to both sides.
For the answer to the question above
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--Going past 5--
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Very (VERY) basic CRM for small businesses
I hope my answer helped you.
