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MrRa [10]
4 years ago
9

Two charged bees land simultaneously on flowers that are separated by a finite distance. For a few moments, the charged bees res

t on the flowers. The charged bees both generate an electric field, and while the charged bees are resting on the flowers, the net electric field at some distance between them is zero. (a) Do the bees have the same or opposite signs of charge?
Same, the electric fields point in opposite directions and therefore cancel at some midpoint.
Same, the electric fields multiply together to equal zero.
Opposite, the electric fields point in the same direction summing to zero.
Opposite, the net electric field due to the two bees points in a direction perpendicular to the direction from one bee to the other.
Physics
1 answer:
damaskus [11]4 years ago
3 0

Answer:

Same, the electric fields point in opposite directions and therefore cancel at some midpoint.

Explanation:

The Electric field net at some point between them is zero, only if they point in opposite direction (they cancel to the each other). In order the electric fields  have opposite direction, at some point between the bees , the bees must have the same sign of electric charge

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   v  =   √ { 2*(KE) ] / m } ; 

Now, plug in the known values for "KE" ["kinetic energy"] and "m" ["mass"] ; 
        
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Explanation:
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The formula is:  KE = (½) * (m) * (v²) ;
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"Kinetic energy" = (½) * (mass) * (velocity , "squared")
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Note:  Velocity is similar to speed, in that velocity means "speed and direction";  however, if you "square" a negative number, you will get a "positive"; since:  a "negative" multiplied by a "negative" equals a "positive".
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So, we have the formula:
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KE = (½) * (m) * (v²) ;  to solve for "(v)" ; velocity, which is very similar to                                          the "speed"; 
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(KE) = (½) * (m) * (v²) ;  ↔  (½)*(m)* (v²) = (KE) ; 
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→ We have:  (½)*(m)* (v²) = (KE)  ; we isolate, "m" (mass) on one side of the equation:
______________________________________________________
   
→ We divide each side of the equation by: "[(½)* (m)]" ; 
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           →   [ (½)*(m)*(v²) ] /  [(½)* (m)]  = (KE) / [(½)* (m)]<span> ;
</span>______________________________________________________
 to get: 
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                           →   v²     =   (KE) / [(½)* (m)]
                     
                           →   v²     = 2 KE / m
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Take the "square root" of each side of the equation ;
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                          →  √ (v²)  =  √ { 2*(KE) ] / m }
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                          →     v  =   √ { 2*(KE) ] / m } ; 

Now, plug in the known values for "KE" ["kinetic energy"] and "m" ["mass"]; 
       
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