A dentist's drill starts from rest. After 3.00 s of constant angular acceleratio

Physics
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A dentist's drill starts from rest. After 3.00 s of constant angular acceleration it turns at a rate of 2.0  104 rev/min.

(a) Find the drill's angular acceleration.
 rad/s2

(b) Determine the angle (in radians) through which the drill rotates during this period.
 rad
Jun 8th, 2015

α = (ω - ωo)/t 
α = (2 e4 rev/min)(2π rad/rev)(min / 60 s) / (3s) 
α = 698.13 rad/s^2 

(b) 
θ = αt^2 / 2 
θ = (698.13rad/s^2)(3 s)^2 / 2 
θ = 3141.59 rad = 3.142 * 10^3 rad

Please let me know if you have any questions and best me if you are satisfactory.


Please let me know if you need any clarification. I'm always happy to answer your questions.
Jun 8th, 2015

(b) is incorrect 


Jun 8th, 2015

Do you have the options?

3141.59 rad 

It can be written in different forms

Jun 8th, 2015

3.142 * 10^3 or 31.42 * 10^2 rad


Jun 8th, 2015

i got it thanks


Jun 8th, 2015

Can you please rate me?

Jun 8th, 2015

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Jun 8th, 2015
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