Answer:
29.16 J
Explanation:
From Hook's law,
W = 1/2(ke²)..................... Equation 1
Where W = work done, k = Spring constant, e = extension.
Given: W = 9 J, e = 0.5 m.
Substitute into equation 1
9 = 1/2(k×0.5²)
Solve for k
k = 18/0.5²
k = 72 N/m.
The work done required to stretch the spring by additional 0.4 m is 
W = 1/2(72)(0.4+0.5)²
W = 36(0.9²)
W = 29.16 J.
 
        
             
        
        
        
A bond with elements from B.
        
                    
             
        
        
        
Answer:
Who are the people for you then I can help you format the essay
Explanation:
 
        
                    
             
        
        
        
Complete Question 
Planet D has a semi-major axis = 60 AU and an orbital period of 18.164 days. A piece of rocky debris in space has a semi major axis of 45.0 AU.  What is its orbital period?
Answer:
The value  is   
  
Explanation:
From the question we are told that 
    The semi - major axis of the rocky debris  
    The semi - major axis of  Planet D is  
     The orbital  period of planet D is  
Generally from Kepler third law 
           
Here T is the  orbital period  while a is the semi major axis 
So  
         
=>     ![T_R  = T_D *  [\frac{a_R}{a_D} ]^{\frac{3}{2} }](https://tex.z-dn.net/?f=T_R%20%20%3D%20T_D%20%2A%20%20%5B%5Cfrac%7Ba_R%7D%7Ba_D%7D%20%5D%5E%7B%5Cfrac%7B3%7D%7B2%7D%20%7D) 
  
=>     ![T_R  = 18.164  *  [\frac{ 45}{60} ]^{\frac{3}{2} }](https://tex.z-dn.net/?f=T_R%20%20%3D%2018.164%20%20%2A%20%20%5B%5Cfrac%7B%2045%7D%7B60%7D%20%5D%5E%7B%5Cfrac%7B3%7D%7B2%7D%20%7D)
=>      