C. The higher the altitude the less gravity affects you
Solution :
Given :
Mass attached to the spring = 4 kg
Mass dropped = 6 kg
Force constant = 100 N/m
Initial amplitude = 2 m
Therefore,
a). 

= 10 m/s
Final velocity, v at equilibrium position, v = 5 m/s
Now, 
A' = amplitude = 1.4142 m
b). 
m' = 2m
Hence, 
c). 

Therefore, factor 
Thus, the energy will change half times as the result of the collision.
Answer:
Matter
Pure substances Mixture
Element compound Homogenous Heterogenous
Answer:


Explanation:
In order to calculate the equivalent spring constant we need to use the next formula:

Replacing the data provided:


Finally, to calculate the frequency of oscillation we use this:

Replacing m and k:

To find the ratio of planetary speeds Va/Vb we need the orbital velocity formula:
V=√({G*M}/R), where G is the gravitational constant, M is the mass of the distant star and R is the distance of the planet from the star it is orbiting.
So Va/Vb=[√( {G*M}/Ra) ] / [√( {G*M}/Rb) ], in our case Ra = 7.8*Rb
Va/Vb=[ √( {G*M}/{7.8*Rb} ) ] / [√( {G*M}/Rb )], we put everything under one square root by the rule: (√a) / (√b) = √(a/b)
Va/Vb=√ [ { (G*M)/(7.8*Rb) } / { (G*M)/(Rb) } ], when we cancel out G, M and Rb we get:
Va/Vb=√(1/7.8)/(1/1)=√(1/7.8)=0.358 so the ratio of Va/Vb = 0.358.