Answer:
F = K Q1 Q2 / R^2       where K = 9 * 10E9  (1 / 4 pi ∈0)
F = 9.00E9 * (4.6E-16)^2 / .01 = 1.90E-19 N
 
        
             
        
        
        
Answer:

Explanation:
let the ladder is of mass "m" and standing at an angle with the ground
So here by horizontal force balance we will have

by vertical force balance we have

now by torque balance about contact point on ground we will have

so we will have

now from first equation we have


 
        
             
        
        
        
Answer:
ωB = 300 rad/s
ωC = 600 rad/s
Explanation:
The linear velocity of the belt is the same at pulley A as it is at pulley D.
vA = vD
ωA rA = ωD rD
ωD = (rA / rD) ωA
Pulley B has the same angular velocity as pulley D.
ωB = ωD
The linear velocity of the belt is the same at pulley B as it is at pulley C.
vB = vC
ωB rB = ωC rC
ωC = (rB / rC) ωB
Given:
ω₀A = 40 rad/s
αA = 20 rad/s²
t = 3 s
Find: ωA
ω = αt + ω₀
ωA = (20 rad/s²) (3 s) + 40 rad/s
ωA = 100 rad/s
ωD = (rA / rD) ωA = (75 mm / 25 mm) (100 rad/s) = 300 rad/s
ωB = ωD = 300 rad/s
ωC = (rB / rC) ωB = (100 mm / 50 mm) (300 rad/s) = 600 rad/s