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.
 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. 
 
        
             
        
        
        
Can cause children into getting cyber bullying, being a thief, sexual behavior, anxiety, and depression
        
             
        
        
        
Answer:
Your hands get warm by fire because chemical energy gets converted into heat energy. When the chemical bonds in the wood are released in the air which then mixes with oxygen and emit heat. This is the reason why it is always hot when you go near something that is burning or is up in flames.
Explanation:
 
        
                    
             
        
        
        
Answer:
   k = 6,547 N / m
Explanation:
This laboratory experiment is a simple harmonic motion experiment, where the angular velocity of the oscillation is
          w = √ (k / m)
angular velocity and rel period are  related
          w = 2π / T
substitution
          T = 2π √(m / K)
in Experimental measurements give us the following data
   m (g)     A (cm)    t (s)   T (s)
   100        6.5         7.8    0.78
   150        5.5          9.8   0.98
    200      6.0        10.9    1.09
    250       3.5        12.4    1.24
we look for the period that is the time it takes to give a series of oscillations, the results are in the last column
         T = t / 10
To find the spring constant we linearize the equation
         T² = (4π²/K)    m
therefore we see that if we make a graph of T² against the mass, we obtain a line, whose slope is
          m ’= 4π² / k
where m’ is the slope
            k = 4π² / m'
the equation of the line of the attached graph is
        T² = 0.00603 m + 0.0183
therefore the slope
        m ’= 0.00603  s²/g
     we calculate
          k = 4 π² / 0.00603
           k = 6547 g / s²
we reduce the mass to the SI system
          k = 6547 g / s² (1kg / 1000 g)
          k = 6,547 kg / s² =
          k = 6,547 N / m
let's reduce the uniqueness
          [N / m] = [(kg m / s²) m] = [kg / s²]