53cm. = 0.52m. Radius = 0.26m. Perimeter = 1.634 metres. This is covered 85 x per min. (60/85) x 1.634 = 1.1534m/sec. tangential V. 30g = 0.030kg. Force = mv^2/

**r**= 0.1535N.2 Answers · Science & Mathematics · 17/10/2013

speed = ω r = (64rev/min * 1min/60s * 2π rads/rev) * 0.23m = 1.3 m/s

1 Answers · Science & Mathematics · 03/11/2013

...period by w = 2π/T with T = 2 min = 120 s and radius r is r = L /2π being L = 3000 m the track length. Then Fc = m (2π/T)² L /2π = 2π m...

1 Answers · Science & Mathematics · 23/10/2010

42cm.diameter = 0.21m. radius, and 30g = 0.030kg. (60/60s) = 1rps. (1 x 2pi) = 6.2832 rad/sec. V = (6.2832 x 0.21m. radius) = 1.3195 m/sec. Force = (m*v^2)/

**r**= (0.03 x 1.3195^2)/0.21 = 0.2487N.2 Answers · Science & Mathematics · 12/11/2017

r = L / 2 π r = 1000 / 2 * 3.14159 = 159.155 m v = 76000 / 3600 = 21.111m a = v^2 / r = 445.679 / 159.155 = 2.8 m/s2 Goodbye

1 Answers · Science & Mathematics · 03/02/2013

... the track (i.e. the angle between the line from the object to the center of the track and the vertical), and let

**R**be the radius of the track . The object leaves the track because the the...1 Answers · Science & Mathematics · 13/09/2009

...origin 'O')) with the circular track marked. Mark the front of the train at ( r ,0), 'F'. Mark the rear...shift. If the train is length L , it subtends an angle α = L ...

2 Answers · Science & Mathematics · 31/03/2013

... provided by the wall and has magnitude Fc = m v^2 / r Therefore the friction force is F_fric = μk m v^2 / r ...

1 Answers · Science & Mathematics · 26/04/2016

ω = 60/60 * 2pi rad/s; r = 0.22 m; m = 0.028 kg Fx = mω^2r Fy = mg F = sqrt(Fx^2 + Fy^2)

2 Answers · Science & Mathematics · 21/10/2007

I'm guessing the track is horizontal and the L -shape is the track 's cross-sectional shape. So the setup is like this: |_......._| The centripetal force = mv²/ r and is provided by the inwards normal reaction of the track (F...

2 Answers · Science & Mathematics · 11/11/2014