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