Answer: C) The two particles will move away from each other
Explanation:
When two electrically charged bodies come closer, appears a force that attracts or repels them, depending on the sign of the charges of this two bodies.
This is stated by Coulomb's Law:
"The electrostatic force  between two point charges
 between two point charges  and
 and  is proportional to the product of the charges and inversely proportional to the square of the distance
 is proportional to the product of the charges and inversely proportional to the square of the distance  that separates them, and has the direction of the line that joins them"
 that separates them, and has the direction of the line that joins them"  
Mathematically this law is written as:  
 
  
Where  is a proportionality constant.
 is a proportionality constant.  
Now, if  and
 and  have the same sign charge (both positive or both negative), a repulsive force will act on these charges.
 have the same sign charge (both positive or both negative), a repulsive force will act on these charges.
 
        
             
        
        
        
A. because I had this question yesterday.
        
             
        
        
        
Answer:
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<h2><u><em>
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Explanation:
3) For a displacement time graph, straight line denotes constant speed. For a velocity time graph, the graph parallel to time axis denotes constant speed. Hence, the correct option is a) and d).
 
        
             
        
        
        
Using the equation v(average)=x traveled/time
v = 100/2.5
You get 40 kilometers per hour
Hope this helped!
        
             
        
        
        
Answer:
μk = 0.26885
Explanation:
Conceptual analysis
We apply Newton's second law:
∑Fx = m*a (Formula 1)
∑F : algebraic sum of the forces in Newton (N)
m : mass in kilograms (kg)
a : acceleration in meters over second square (m/s²)
Data:
a= -0.9  m/s²,  
g = 9.81 m/s² : acceleration due to gravity
W= 75 N :  Block weight
W= m*g   
m =  W/g = 75/9.8= 7.65 kg :  Block mass
Friction force : Ff
Ff= μk*N
μk: coefficient of kinetic friction
N : Normal force (N)  
Problem development
We apply the formula (1)
∑Fy = m*ay    , ay=0
N-W-25 = 0
N = 75
+25
N= 100N
∑Fx = m*ax     
20-Ff= m*ax     
20-μk*100
= 7.65*(-0.90 ) 
20+7.65*(0.90) = μk*100
μk = ( 20+7.65*(0.90)) / (100)
μk = 0.26885