Question

At takeoff a commercial jet has a 75.0 m/s speed. Its tires have a diameter of 0.400 m. (a) At how many rpm are the tires rotating? 3581 (b) What is the centripetal acceleration at the edge of the tire? 28125m/s (c) With what force must a determined 10 15 kg bacterium cling to the rim? rpm (d) Take the ratio of this force to the bacterium's weight. 2870(force from part (e)/ bacterium's weight)

Answer 1

Solution)

Part C)

Here, Force is the centripetal force= mv^2/r

= 10^-15*(75)^2/0.2 =2.81*10^-11 N (Ans)

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