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Biologists have studied the running ability of the northern quoll, a marsupial indigenous to Australia. In one set of experiments, they studied the maximum speed that quolls could run around a curved path without slipping. One quoll was running at 2.4 m/sm/s around a curve with a radius of 1.4 mm when it started to slip. What was the coefficient of static friction between the quoll's feet and the ground in this trial?

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Answer:

Step-by-step explanation:

velocity of quoll v = 2.4 m/s

radius of circular path R = 1.4 m .

The centripetal force in its circular motion will be provided by friction force which will act towards the center

centripetal acceleration = v² / R

= 2.4 x 2.4 / 1.4

= 4.11 m/s²

force = mass x acceleration

friction force = m x 4.11 , where m is mass of quoll

Normal force N = mg

coefficient of friction = friction force / normal force

= m x 4.11 / m x 9.8

= .42 Ans

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