Answer:
The first part can be solved via conservation of energy.

For the second part,
the free body diagram of the car should be as follows:
- weight in the downwards direction
- normal force of the track to the car in the downwards direction
The total force should be equal to the centripetal force by Newton's Second Law.

where
because we are looking for the case where the car loses contact.

Now we know the minimum velocity that the car should have. Using the energy conservation found in the first part, we can calculate the minimum height.

Explanation:
The point that might confuse you in this question is the direction of the normal force at the top of the loop.
We usually use the normal force opposite to the weight. However, normal force is the force that the road exerts on us. Imagine that the car goes through the loop very very fast. Its tires will feel a great amount of normal force, if its velocity is quite high. By the same logic, if its velocity is too low, it might not feel a normal force at all, which means losing contact with the track.
An independent variable is the variable that is changed or controlled in a scientific experiment to test the effects on the dependent variable. A dependent variable is the variable being tested and measured in a scientific experiment.
Answer:
The answer to your question: d.
Explanation:
a. The rate of change of momentum of an object is equal to the net force applied to the object.
This is the second a law of motion, so this answer is incorrect.
b. In the absence of a net force acting on it, an object moves with constant velocity.
This is the first Newton law of motion, so this option is not correct.
c. For any force, there always is an equal and opposite reaction force.
This is the third law of motion, so this is not the right option.
d. What goes up must come down.
Newton said this sentence, but is not part of the law of motion.
Answer: rp/re= me/mp= 544 * 10^-6.
Explanation: To calculate this problem we have to consider the circular movement by the electron and proton inside a magnetic field.
Then the dynamic equation for the circular movement is given by:
Fcentripetal= m*ω^2.r
q*v*B=m*ω^2.r
we write this for each particle then we have the following:
q*v*B=me* ω^2*re
q*v*B=mp* ω^2*rp
rp/re=me/mp=9.1*10^-31/1.67*10^-27=544*10^-6