<span>this  may help you
As far as the field goes, the two charges opposite each other cancel! 
So E = kQ / d² = k * Q / (d/√2)² = 2*k*Q / d² ◄ 
and since k = 8.99e9N·m²/C², 
E = 1.789e10N·m²/C² * Q / d² </span>
        
                    
             
        
        
        
Answer:
See answers below
Explanation:
a. 
F = mg,
15.5 N = m(9.8 m/s²)
m = 1.58 kg
b. 
Fnet = Applied force - resistance,
Fnet = 18 N - 4.30 N,
Fnet = 13.70 N
Fnet = ma
13.70 N = (1.58 kg)a
a = 8.67 m/s²
For the free body diagram, draw a box with an upward arrow labeled 15.5 N, a downward label labeled 15.5 N, a right label labeled 18 N, and a left label labeled 4.30 N.
 
        
             
        
        
        
Answer:
Q = 40.1 degrees 
Explanation:
Given:
- The weight of the timber W = 670 N
- Water surface level from pivot y = 2.1 m 
- The specific density of water Y = 9810 N / m^3
- Dimension of timber = (0.15 x 0.15 x 0.0036) m
Find:
- The angle of inclination Q that the timber makes with the horizontal.
Solution:
- Calculate the Flamboyant Force F_b acting upwards at a distance x along the timber, which is unknown:
                                    F_b = Y * V_timber
                                    F_b = 9810*0.15*0.15*x 
                                    F_b = 226.7*x N
- Take static equilibrium conditions for the timber, and take moments about the pivot:
                                    (M)_p = 0
                                    W*0.5*3.6*cos(Q) - x/2 * F_b*cos(Q) = 0
- Plug values in:
                                    670*0.5*3.6 - x^2 * 0.5*226.7 = 0
                                    x^2 = 1206 / 113.35
                                    x = 3.26 m
- Now use the value of x and vertical height y to compute the angle of inclination to be:
                                    sin(Q) = y / x
                                    sin(Q) = 2.1 / 3.26
                                    Q = sin^-1 (0.6441718)
                                    Q = 40.1 degrees 
 
        
             
        
        
        
On sources it says it would  just be the super giant star 
        
             
        
        
        
Answer:
the less shielding of electrons