Adentist's drill starts from rest. after 3.00 s of constant angular acceleration it turns at a rate of 2.90 ✕ 10^4 rev/min. (a) find the drill's angular acceleration. (b) determine the angle (in radians) through which the drill rotates during this period.

Initial angular velocity wi = 0

final angular velocity wf = 2.9 x 10⁴ rev/min

wf = 2.9 x 10⁴ (2*3.14 rad)/(60 sec)

wf = 3.035 x 10³ rad/sec

time t  = 3 sec

angular acceleration α

wf = wi + αt

3.035 x 10³ = 0 + α(3)

α = 1012 rad/s²

(b) angle covered is θ

θ = wi*t + (1/2)αt²

θ = 0*3  + (1/2)(1012)(3²)

θ = 4554 rad

low freq = 100 mhz

low wavelength =101.7m

length = 3x10^8/freq

what is the speed of a softball that travels 20m in 2secs

speed = 20/2 = 10 m/s