Answer:
See the answer below
Explanation:
The optimal conditions for high biodiversity seem to be a <u>warm temperature</u> and <u>wet climates</u>. 
<em>The tropical areas of the world have the highest biodiversity and are characterized by an average annual temperature of above 18 </em> <em> and annual precipitation of 262 cm. The areas are referred to as the world's biodiversity hotspots. </em>
<em> and annual precipitation of 262 cm. The areas are referred to as the world's biodiversity hotspots. </em>
Consequently, it follows logically that the optimal conditions for high biodiversity would be a warm temperature of above 18  and wet environment with annual precipitation of not less than 262 cm.
 and wet environment with annual precipitation of not less than 262 cm.
The variation in temperature and precipitation across biomes can thus be said to be responsible for the variation in the level of biodiversity in them. 
 
        
             
        
        
        
Use the law of universal gravitation, which says the force of gravitation between two bodies of mass <em>m</em>₁ and <em>m</em>₂ a distance <em>r</em> apart is
<em>F</em> = <em>G m</em>₁ <em>m</em>₂ / <em>r</em>²
where <em>G</em> = 6.67 x 10⁻¹¹ N m²/kg².
The Earth has a radius of about 6371 km = 6.371 x 10⁶ m (large enough for a pineapple on the surface of the earth to have an effective distance from the center of the Earth to be equal to this radius), and a mass of about 5.97 x 10²⁴ kg, so the force of gravitation between the pineapple and the Earth is
<em>F</em> = (6.67 x 10⁻¹¹ N m²/kg²) (1 kg) (5.97 x 10²⁴ kg) / (6.371 x 10⁶ m)²
<em>F</em> ≈ 9.81 N
Notice that this is roughly equal to the weight of the pineapple on Earth, (1 kg)<em>g</em>, where <em>g</em> = 9.80 m/s² is the magnitude of the acceleration due to gravity, so that [force of gravity] = [weight] on any given planet.
This means that on this new planet with twice the radius of Earth, the pineapple would have a weight of
<em>F</em> = <em>G m</em>₁ <em>m</em>₂ / (2<em>r</em>)² = 1/4 <em>G m</em>₁ <em>m</em>₂ / <em>r</em>²
i.e. 1/4 of the weight on Earth, which would be about 2.45 N.
 
        
             
        
        
        
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)
=>       
  
    
 
        
             
        
        
        
250kg
                                         would have momentum that is being caried by the impact of the trow                                                                                                                    
 
        
             
        
        
        
Answer:middle
Explanation:
Because it will make the seasaw balanced