Answer:
Yes. Towards the center. 8210 N.
Explanation:
Let's first investigate the free-body diagram of the car. The weight of the car has two components: x-direction: towards the center of the curve and y-direction: towards the ground. Note that the ground is not perpendicular to the surface of the Earth is inclined 16 degrees.
In order to find whether the car slides off the road, we should use Newton's Second Law in the direction of x: F = ma.
The net force is equal to 
Note that 95 km/h is equal to 26.3 m/s.
This is the centripetal force and equal to the x-component of the applied force.

As can be seen from above, the two forces are not equal to each other. This means that a friction force is needed towards the center of the curve.
The amount of the friction force should be 
Qualitatively, on a banked curve, a car is thrown off the road if it is moving fast. However, if the road has enough friction, then the car stays on the road and move safely. Since the car intends to slide off the road, then the static friction between the tires and the road must be towards the center in order to keep the car in the road.
So impulse is a change in momentum.
Mass*(final velocity - initial velocity)
I dont think you will be able to find the average force with the given info because you need to know the time it takes for the car to slow down.
Answer:
20.7 s
Explanation:
The equation to calculate the velocity for a uniform acceleration a, time t and initial velocity v₀:
v = a*t + v₀
Solve for t:
t = (v - v₀)/a
Answer:
(a) 0.942 m
(b) 18.84 m/s
(c) 2366.3 m/s²
(d) 0.05 s
Explanation:
(a) In one revolution, it travels through one circumference, 2πr = 2 × 3.14 × 0.15 m = 0.942 m.
(b) Its frequency, f, is 1200 rev/min =
rev/s = 20 rev/s.
Its angular frequency, ω = 2πf = 2π × 20 = 40π
The speed is given by
v = ωr = 40π × 0.15 = 6π = 18.84 m/s
(c) Its acceleration is given by, a = ω²r = (40π)² × 0.15 = 2366.3 m/s²
(d) The period is the inverse of the frequency because it is the time taken to complete one revolution.

T = 1/20 = 0.05 s