Answer:
B. w=12.68rad/s
C. α=3.52rad/s^2
Explanation:
B)
We can solve this problem by taking into account that (as in the uniformly accelerated motion)
( 1 )
where w0 is the initial angular speed, α is the angular acceleration, s is the arc length and r is the radius.
In this case s=3.7m, r=16.2cm=0.162m, t=3.6s and w0=0. Hence, by using the equations (1) we have
![\theta=\frac{3.7m}{0.162m}=22.83rad](https://tex.z-dn.net/?f=%5Ctheta%3D%5Cfrac%7B3.7m%7D%7B0.162m%7D%3D22.83rad)
![22.83rad=\frac{1}{2}\alpha (3.6s)^2\\\\\alpha=2\frac{(22.83rad)}{3.6^2s}=3.52\frac{rad}{s^2}](https://tex.z-dn.net/?f=22.83rad%3D%5Cfrac%7B1%7D%7B2%7D%5Calpha%20%283.6s%29%5E2%5C%5C%5C%5C%5Calpha%3D2%5Cfrac%7B%2822.83rad%29%7D%7B3.6%5E2s%7D%3D3.52%5Cfrac%7Brad%7D%7Bs%5E2%7D)
to calculate the angular speed w we can use![\alpha=\frac{\omega _{f}-\omega _{i}}{t _{f}-t _{i}}\\\\\omega_{f}=\alpha t_{f}=(3.52\frac{rad}{s^2})(3.6)=12.68\frac{rad}{s}](https://tex.z-dn.net/?f=%5Calpha%3D%5Cfrac%7B%5Comega%20_%7Bf%7D-%5Comega%20_%7Bi%7D%7D%7Bt%20_%7Bf%7D-t%20_%7Bi%7D%7D%5C%5C%5C%5C%5Comega_%7Bf%7D%3D%5Calpha%20t_%7Bf%7D%3D%283.52%5Cfrac%7Brad%7D%7Bs%5E2%7D%29%283.6%29%3D12.68%5Cfrac%7Brad%7D%7Bs%7D)
Thus, wf=12.68rad/s
C) We can use our result in B)
![\alpha=3.52\frac{rad}{s^2}](https://tex.z-dn.net/?f=%5Calpha%3D3.52%5Cfrac%7Brad%7D%7Bs%5E2%7D)
I hope this is useful for you
regards
Your answer would be A. Halogens
Average <u>speed</u> = (distance covered) / (time to cover the distance) =
(5m) / (15 sec) =
(5/15) (m/s) = <em>1/3 m/s</em> .
Average <u>velocity</u> =
(displacement) / (time spent traveling) in the direction of the displacement
Average velocity = (5m) / (15 sec) left =
(5/15) / (m/sec) left =
<em>1/3 m/s left</em>.
A number without a direction is a speed, not a velocity.
He should confront her about it and if after that point she continues report it to the chess team