Answer:
Question: A car (assumed to be a Ford Taurus) is traveling around a turn that is banked at 7 degrees. The turn has a radius of 29 m. The car has a mass of 1300 kg. The coefficient of static friction between the tires and the road is 0.68.
1. What is the "ideal speed?" That is, what speed would allow the car to make the turn without requiring friction?
2. What is the maximum speed the car can go around the turn without sliding?
Answer: Green Light
Explanation:
The color green has a higher frequency than the color red and the higher the frequency the more energy. If you looks at a rainbow you'll see red on one end and violet on the other. Red has the least energy Violet has the most.
Answer:
I believe <u>chemicals</u>, though I might be wrong.
Explanation:
To solve this problem it is necessary to apply the concepts given in the kinematic equations of movement description.
From the perspective of angular movement, we find the relationship with the tangential movement of velocity through

Where,
Angular velocity
v = Lineal Velocity
R = Radius
At the same time we know that the acceleration is given as the change of speed in a fraction of the time, that is

Where
Angular acceleration
Angular velocity
t = Time
Our values are




Replacing at the previous equation we have that the angular velocity is



Therefore the angular speed of a point on the outer edge of the tires is 66.67rad/s
At the same time the angular acceleration would be



Therefore the angular acceleration of a point on the outer edge of the tires is 
Use vector analysis and calculate resultant vector using Pythagoras theorem:
5^2 + 5^2 = 50
Square root of 50 = approx 7.1 km NE
Therefore the answer is D