Answer: B
Explanation: look at the chart, easy
 
        
             
        
        
        
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)
=>       
  
    
 
        
             
        
        
        
Inductive reactance (Z) = ω L  =  2Πf L = (2Π) (12,000) (L)
I = V / Z
4 A = 16v / (24,000Π L)
Multiply each side by (24,000 Π L):
96,000 Π L = 16v
Divide each side by  (96,000 Π) :
L = 16 / 96,000Π  =  5.305 x 10⁻⁵ Henry
L = 53.05 microHenry
        
             
        
        
        
It would take at less 10 minte i guess this the right awnser
        
             
        
        
        
Hi There! :)
An equilibrium constant is not changed by a change in pressurea. True
b. False
False! :P