Final answer:
To solve the physics problem concerning a capacitor lifting a mass, we use formulas for energy stored in a capacitor and equate it to gravitational potential energy to find the lifting height. For the second part, we reverse the process, finding the voltage needed for a specific lift height.
Step-by-step explanation:
To answer your questions: (a) Assuming 100% efficiency, the energy stored in the 0.310μF capacitor when charged by a 1.50 V battery can be calculated using the formula for the energy stored in a capacitor: E = ½ CV². Plugging in the values, we get E = 0.5 × 0.310× 10⁻¶ F × (1.50 V)² = 3.4875× 10⁻´ J. This energy can be used to lift a mass to a certain height by equating it to the gravitational potential energy (mgh), where m is the mass, g is the acceleration due to gravity, and h is the height. We can rearrange this to solve for h: h = E / (mg). Using m = 6.60g = 6.60× 10⁻3 kg and g = 9.81 m/s², we find the height h.
(b) To calculate the initial voltage required for the capacitor to lift a 6.60 g mass through a height of 1.00 cm, we first find the necessary energy using E = mgh (with the height converted to meters). Then, we solve the energy formula for the voltage: V = √(2E/C), where C is the capacitance.