Answer :
Answer:
[tex]\mu=0.74[/tex]
Explanation:
It is given that,
Speed of one quoll around a curve, v = 3.2 m/s (maximum speed)
Radius of the curve, r = 1.4 m
On the curve, the centripetal force is balanced by the frictional force such that the coefficient of frictional is given by :
[tex]\mu=\dfrac{v^2}{rg}[/tex]
[tex]\mu=\dfrac{(3.2\ m/s)^2}{1.4\ m\times 9.8\ m/s^2}[/tex]
[tex]\mu=0.74[/tex]
So, the coefficient of static friction between the quoll's feet and the ground in this trial is 0.74. Hence, this is the required solution.
Answer:
[tex]\mu = 0.746[/tex]
Explanation:
As we know that the force required to move in circle at uniform speed is known as centripetal force
So here we know that centripetal force is given as
[tex]F = \frac{mv^2}{R}[/tex]
so here the force is due to friction force
so it is given as
[tex]\mu mg = \frac{mv^2}{R}[/tex]
[tex]\mu = \frac{v^2}{Rg}[/tex]
now we have
[tex]\mu = \frac{3.2^2}{1.4 \times 9.8}[/tex]
[tex]\mu = 0.746[/tex]