Answer:
The electric potential is approximately 5.8 V
The resulting direction of the electric field will lie on the line that joins the charges but since it is calculated in the midpoint and the charges are the same we can directly say that its magnitude is zero
Explanation:
The two protons can be considered as point charges. Therefore, the electric potential is given by the point charge potential:
 (1)
 (1)
where  is the charge of the particle,
 is the charge of the particle,  the electric permittivity of the vacuum (I assuming the two protons are in a vacuum) and
 the electric permittivity of the vacuum (I assuming the two protons are in a vacuum) and  is the distance from the point charge to the point where the potential is being measured. Because the electric potential is an scalar, we can simply add the contribution of the two potentials in the midpoint between the protons. Thus:
 is the distance from the point charge to the point where the potential is being measured. Because the electric potential is an scalar, we can simply add the contribution of the two potentials in the midpoint between the protons. Thus:

Substituting the values  ,
,  and
 and  we obtain:
 we obtain:

The resulting direction of the electric field will lie on the line that joins the charges but since it is calculated in the midpoint and the charges are the same we can directly say that its magnitude is zero. 
 
        
             
        
        
        
Answer:a supposition or proposed explanation made on the basis of limited evidence as a starting point for further investigation.
Hope This helps!!
 
        
             
        
        
        
Noise does not affect the digital signal making it more reliable
        
                    
             
        
        
        
The solution would be like
this for this specific problem:
<span>5.5 g = g + v^2/r </span><span>
<span>4.5 g =
v^2/r </span>
<span>v^2 = 4.5
g * r </span>
<span>v = sqrt
( 4.5 *9.81m/s^2 * 350 m) </span>
v = 124
m/s</span>
So the pilot will black out for this dive at 124
m/s. I am hoping that these answers have satisfied your query and it
will be able to help you in your endeavors, and if you would like, feel free to
ask another question.