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PHY 201
Your 'cq_1_25.1' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.
** CQ_1_25.1_labelMessages.txt **
Samantha Rogers
PHY 201
Seed 25.1
A steel ball of mass 110 grams moves with a speed of 30 cm / second around a circle of radius 20 cm.
• What are the magnitude and direction of the centripetal acceleration of the ball?
answer/question/discussion: ->->->->->->->->->->->-> :
We can find the acceleration by the equation v^2/r = a; therefore, (.3 m/s)^2/.2 m = 0.45 m/s^2
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• What is the magnitude and direction of the centripetal force required to keep it moving around this circle?
answer/question/discussion: ->->->->->->->->->->->-> :
We know the acceleration and mass of the steel ball; therefore, the F = ( .11 kg) * (.45 m/s^2) = 0.0495 N is the force of the centripetal motion.
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&#Very good responses. Let me know if you have questions. &#