U = 1794.005 × 10⁶ J. Explanation: Information provided indicates that the capacitance of the original capacitor is C = 1.27 F, and the potential difference applied to it is V = 59.9 kV, or 59.9 × 10³ V. The potential energy (U) for the capacitor is determined by the formula: U = (1/2) × C × V². Substituting the respective values, we find U = (1/2) × 1.27 × (59.9 × 10³)², resulting in U = 1794.005 × 10⁶ J.
The essential principle for this question is Ohm’s Law: V=IR, I=V/R, R=V/I. Therefore, the answer is (3) Resistance, as it is inversely related to Current (I=V/R).
In terms of light energy, a higher frequency corresponds to increased energy within the light.
We establish that frequency is essentially the inverse of wavelength:
frequency = 1 / wavelength
Calculating frequencies:
f UVA = 1/320 to 1/400
f UVA = 0.0031 to 0.0025
f UVB = 1/290 to 1/320
f UVB = 0.0034 to 0.0031
Since UVB occupies a higher frequency range, it consequently possesses greater energy than UVA.
