Answer:
2.15 m/s²
Step-by-step explanation:
We'll begin by calculating the force of attraction between two charges. This can be obtained as follow:
Charge of 1st object (q₁) = +11.5 μC = +11.5×10¯⁶ C
Charge of 2nd object (q₂) = –7.55 μC = –7.55×10¯⁶ C
Electrical constant (K) = 9×10⁹ Nm²/C²
Distance apart (r) = 0.925 m
Force (F) =?
F = Kq₁q₂ / r²
F = 9×10⁹ × 11.5×10¯⁶ × 7.55×10¯⁶/ 0.925²
F = 0.781425 / 0.855625
F = 0.91 N
Finally, we shall determine the acceleration of the object. This can be obtained as follow:
Mass of object (m) = 0.423 Kg
Force (F) = 0.91 N
Acceleration (a) =?
F = ma
0.91 = 0.423 × a
Divide both side by 0.423
a = 0.91 / 0.423
a = 2.15 m/s²
Thus, the magnitude of the object's acceleration is 2.15 m/s²