B. Extra text to get to 20 characters.
The power that the light is able to utilize out of the supply is only 0.089 of the given.
Power utilized = (0.089)(22 W)
= 1.958 W
= 1.958 J/s
The energy required in this item is the product of the power utilized and the time. That is,
Energy = (1.958 J/s)(1 s) = 1.958 J
Thus, the light energy that the bulb is able to produce is approximately 1.958 J.
Answer:
D
Explanation:
The color you see is the color the object reflectes. The rest of the color are absorbed by that object.
Answer:
Explanation:
The application of Gauss's law is used in the derivation as shown with detailed step by step in the attached file.
The potential difference on this spherical capacitor is ΔV = Va - Vb = kQ/a - kQ/b = kQ(1/a - 1/b)
