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.
Answer: a dark (absorption) line
Explanation:
This is as a result of absorption of electromagnetic a specific wavelength. The pattern followed by such lines is characteristic of specific atoms in the path of radiation.
Answer:
What inferences can you make about the melting points of the different substances and the motion of their particles, based on the data? (ignore that needed it here so i could see it better.)
Explanation:
Butter has a lower melting point than the cheese and the wax. The motion of the cheese were a little separated while the butter articles have more space in between. The wax had the closest particles.
I dont know if that makes sense?
The net force on the box is:
50 + 30 - 65
= 15 Newtons
This can be used in conjunction with
F = ma
to calculate the acceleration of the box if its mass is known.
Answer:
1.3 rev/s
Explanation:
= Moment of inertia when arms and one leg of the skater is out = 3.5 kgm²
= Moment of inertia when arms and legs of the skater are in = 0.80 kgm²
= Angular speed of skater when arms and legs of the skater are in = 5.5 rev/s
= Angular speed of skater when arms and legs of the skater are out = ?
Using conservation of angular momentum
=
(0.80) (5.5) = (3.5)
= 1.3 rev/s