The coefficient of static friction is 0.222
Explanation:
In order for the car to remain in circular motion, the frictional force must be able to provide the necessary centripetal force. Therefore, the car will start skidding when the two forces are equal:

where the term on the left is the frictional force, while the term on the right is the centripetal force, and where
is the coefficient of static friction
m is the mass of the car
g is the acceleration of gravity
v is the speed of the car
r is the radius of the track
In this problem, we have:
r = 564 m
v = 35 m/s

And re-arranging the equation for
, we can find the coefficient of static friction:

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C is the awnser because c is single replacement
Hello
The bullet is moving by uniformly accelerated motion.
The initial velocity is

, the final velocity is

, and the total time of the motion is

, so the acceleration is given by
where the negative sign means that is a deceleration.
Therefore we can calculate the total distance covered by the bullet in its motion using

So, the bullet penetrates the sandbag 1.8 meters.