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

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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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