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.
The correct answer is: [D]:
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"The amount of pressure exerted by a solid is solely dependent on its mass."
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Ummm I’m not sure let me do the work
Answer:
the moment of inertia with the arms extended is Io and when the arms are lowered the moment
I₀/I > 1 ⇒ w > w₀
Explanation:
The angular momentum is conserved if the external torques in the system are zero, this is achieved because the friction with the ice is very small,
L₀ = L_f
I₀ w₀ = I w
w =
w₀
where we see that the angular velocity changes according to the relation of the angular moments, if we approximate the body as a cylinder with two point charges, weight of the arms
I₀ = I_cylinder + 2 m r²
where r is the distance from the center of mass of the arms to the axis of rotation, the moment of inertia of the cylinder does not change, therefore changing the distance of the arms changes the moment of inertia.
If we say that the moment of inertia with the arms extended is Io and when the arms are lowered the moment will be
I <I₀
I₀/I > 1 ⇒ w > w₀
therefore the angular velocity (rotations) must increase
in this way the skater can adjust his spin speed to the musician.
