Final answer:
The acceleration of the system where a 1.00 kg object A is connected to a 2.00 kg object B, and further connected to a 4.00 kg object C, is g/7, which means the correct answer is (D) None of the above.
Step-by-step explanation:
To determine the acceleration of each object in a system where a 1.00 kg object A is connected with a string to a 2.00 kg object B, which is connected with a second string over a pulley to a 4.00 kg object C, we must apply Newton's second law of motion. Since there is no friction or mass in the pulley and strings, we only consider the gravitational force acting on the objects. We can denote the acceleration of the system as a and the gravitational acceleration as g.
The net force on object C (downward) is the weight of object C minus the tension in the string, giving us the equation 4.00g - T = 4.00a. Similarly, the net force on objects A and B (to the right) is the tension minus their combined weight (due to the tension in the string connecting them), giving us the equation T - (1.00 + 2.00)g = (1.00 + 2.00)a. By combining these equations, we find that 4.00g - (1.00 + 2.00)g = (4.00 + 1.00 + 2.00)a, and thus a = g/7.
Therefore, the correct answer is (D) None of the above, since the acceleration of each object is g/7, which does not match any of the given options.