The coefficient of static friction be in order to keep the quarter from sliding outward when the turntable is in motion will be 0.31.
It is a type of opposition force acting on the surface of the body that tries to oppose the motion of the body. its unit is Newton (N).
Mathematically it is defined as the product of the coefficient of friction and normal reaction.
On resolving the given force and acceleration in the different components and balancing the equation gets. The friction force balances centripetal force.
The given data in the problem will be;
m is the mass =6.25 g
r is the radius =25 cm=0.25 m
t is the time = 0.18 sec
μ is the coefficient of static friction =?
[tex]\rm F_c= mw^2r \\\\ f= \mu mg \\\\ F_C=f\\\\ mw^2r= \mu mg \\\\ \rm \mu = \frac{\omega^2r}{g} \\\\ \rm \mu = (\frac{2\pi}{1.8} )^2\frac{0.25}{9.81} \\\\ \rm \mu = 0.31[/tex]
Hence the coefficient of static friction will be 0.31.
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