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Two 2.0 kg masses are 1.1 m apart (center to center) on a frictionless table. Each has + 9.6 μC of charge. PART A
What is the magnitude of the electric force on one of the masses?
Express your answer to two significant figures and include the appropriate units.

PART B
What is the initial acceleration of the mass if it is released and allowed to move?
Express your answer to two significant figures and include the appropriate units.

Respuesta :

Answer:

A) Force = 0.69 N

B) Acceleration = 0.34 m/s^2

Explanation:

The electric force is given by the expression:

[tex]F_e= K *\frac{q_1*q_2}{r^2}[/tex]

K is the Coulomb constant equal to 9*10^9 N*m^2/C^2, q1 and 12 is the charge of the particles, and r is the distance:

[tex]F_e = 9*10^9 Nm^2/C^2 * \frac{(9.6*10^{-6}C)^2}{(1.1m)^2} = 0.69 N[/tex]

Part B.

For the acceleration, you need Newton's second Law:

F = m*a

Then,

[tex]a = \frac{F}{m} = \frac{0.69 N}{2 kg} = 0.34 m/s^2[/tex]