Answer:
Magnitude of the net force on q₁-
Fn₁=1403 N 
Magnitude of the net force on q₂+ 
Fn₂= 810 N
Magnitude of the net force on q₃+ 
Fn₃= 810 N
Explanation:
Look at the attached graphic:
The charges of the same sign exert forces of repulsion and the charges of   opposite sign exert forces of attraction.
Each of the charges experiences 2 forces and these forces are equal and we calculate them with Coulomb's law: 
F= (k*q*q)/(d)²
F= (9*10⁹*3*10⁻⁶*3*10⁻⁶)(0.01)² =810N
Magnitude of the net force on q₁-
Fn₁x= 0
Fn₁y= 2*F*sin60 = 2*810*sin60° = 1403 N
Fn₁=1403 N 
Magnitude of the net force on q₃+ 
Fn₃x= 810- 810 cos 60° = 405 N
Fn₃y= 810*sin 60° = 701.5 N

Fn₃ = 810 N
Magnitude of the net force on q₂+ 
Fn₂ = Fn₃ = 810 N
 
        
             
        
        
        
Answer:
C
Explanation:
the formula is a + b = ab
 
        
             
        
        
        
Answer:
KE_2 = 3.48J
Explanation:
Conservation of Energy
E_1 = E_2
PE_1+KE_1 = PE_2+KE_2
m*g*h+(1/2)m*v² = m*g*h+(1/2)m*v²
(0.0780kg)*(9.81m/s²)*(5.36m)+(.5)*(0.0780kg)*(4.84m/s)² = (0.0780kg)*(9.81m/s²)*(2m)+KE_2
4.10J+0.914J = 1.53J + KE_2
5.01J = 1.53J + KE_2
KE_2 = 3.48J
 
        
             
        
        
        
Answer: cleft grafting, inlay grafting, four-flap grafting, and whip grafting.
Explanation:
I hope I helped , Have a great day or night . Xoxoxo.
 
        
             
        
        
        
Answer:
The error in tapping is ±0.02828 ft.
Explanation:
Given that,
Distance = 200 ft
Standard deviation = ±0.04 ft
Length = 100 ft
We need to calculate the number of observation
Using formula of number of observation

Put the value into the formula


We need to calculate the error in tapping
Using formula of error


Put the value into the formula


Hence, The error in tapping is ±0.02828 ft.