At takeoff, a commercial jet has a 60.0 m/s speed. its tires have a diameter of 0.850 m.(a) at how many rev/min are the tires rotating? (b) what is the centripetal acceleration at the edge of the tire? (c) with what force must a determined 1.00×10−15 kg bacterium cling to the rim? (d) take the ratio of this force to the bacterium's weight.

(a) Angular speed = 1344.8rev/min

(b) Centripetal acceleration = 8473.4m/s^2

(c) Force = 8.4734×10^-12N

(d) Ratio of force to bacterium's weight is 864.6

Explanation:

(a) Angular speed (w) = v/r

v = 60m/s, d = 0.850m, r = 0.850m/2 = 0.425m

w = 60/0.425 = 141.2rad/s = 141.2×9.524rev/min = 1344.8rev/min

(b) Centripetal acceleration = w^2r = 141.2^2 × 0.425 = 8473.4m/s^2

(c) Force = mass × centripetal acceleration = 1×10^-15 × 8473.4 = 8.4734×10^-12N

(d) Bacterium's weight = mass × acceleration due to gravity = 1×10^-15 × 9.8 = 9.8×10^-15N

Ratio of force to bacterium's weight = 8.4734×10^-12N/9.8×10^-15N = 864.6

explanation:

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a and c

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