at athlete swings a ball, connected to the end of a chain, in a horizontal circl

Question

at athlete swings a ball, connected to the end of a chain, in a horizontal circle. The athlete is able to rotate the ball at the rate of 8.0/ rev/s when the length of the chain is 0.600 m. When he increases the length to 0.900 m, he is able to rotate the ball only 6.59 rev/s. Which rate of rotation gives the greater sped for the ball? What is the centripetal acceleration of the ball at 8.07 rev/s? m/s2 What is the centripetal acceleration at 6.59 rev/s? Your response differs significantly from the correct answer. Rework your solution from the beginning and check each step carefully. m/s2 a train slows down as it rounds a sharp horizontal turn, going from 80.0 km/h to 58.0 km/h in the 13.0 s that it takes to round the bend. The radius of the curve is 120 m. Compute the acceleration at the moment the train speed reaches 58.0 km/h. Assume the train continues to slow down at this time at the same rate. magnitude m/s2 direction degree backward (behind the radial line pointing inward)

Explanation / Answer

you asked two questions so here is the answer for the first question and the part you got wrong

a = v^2/r
but v=r

so v= 6.59rev/s * 2pi rad/rev *0.9=6.59*2*pi*0.9=37.26

a= v^2/r = 37.26^2/0.9=1542.6

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