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:
For Xenon fluoride, the average bond energy is 132kj/mol
For tetraflouride,the average bond energy is 150.5kj/mol.
For hexaflouride, the average bond energy is 146.5 kj/mol
Explanation:
For xenon fluoride
105/2 = 52.5
For F-F
159/2 = 79.5
Average bond energy of Xe-F = 79.5 + 52.5 = 132kj/mole
For tetraflouride
284/4 = 71
For F-F
159/2 = 79.5
Average bond energy = 79.5 + 71 = 150.5kj/mol
For hexaflouride
402/6 = 67
F-F = 159/2 = 79.5
Average bond energy = 67 + 79.5 = 146.5kj/ mol
Answer:
power
Explanation:
it helps to do work without power we cant do any things
Answer:
B, It reacts with oxygen to form a new substance
Explanation:
B