Answer:
3.83×10¯⁴ N
Explanation:
From the question given above, the following data were obtained:
Charge 1 (q₁) = +2.4x10¯⁸ C
Charge 2 (q₂) = +1.8x10¯⁶ C
Distance apart (r) = 1.008 m
Electrical constant (K) = 9×10⁹ Nm²/C²
Force (F) =?
The magnitude of the electrical force acting between the two charges can be obtained as follow:
F = Kq₁q₂ / r²
F = 9×10⁹ × 2.4x10¯⁸ × 1.8x10¯⁶ / (1.008)²
F = 0.0003888 / 1.016064
F = 3.83×10¯⁴ N
Thus the magnitude of the electrical force acting between the two charges is 3.83×10¯⁴ N
♥️♥️♥️♥️♥️♥️♥️♥️♥️♥️♥️♥️♥️♥️
The force of friction is equal to the product of the vertical force applied by the surface to the object in the coefficient of friction.
♥️♥️♥️♥️♥️♥️♥️♥️♥️♥️♥️♥️♥️♥️
In this question ,
surface vertical force = Weight of the object
Thus ;
svf = ( mass ) × ( gravity acceleration )
_________________________________
If gravity acceleration is 10 :
svf = 10 × 10 = 100 N
So ;
frictional force = 100 × 0.20
frictional force = 20 N
##############################
If gravity acceleration is 9.8 :
svf = 10 × 9.8 = 98 N
So ;
frictional force = 98 × 0.20
frictional force = 19.6 N
_________________________________
♥️♥️♥️♥️♥️♥️♥️♥️♥️♥️♥️♥️♥️♥️
No it does not . That is the answer
