Answer:
(1) En to n-1 = 0.55 eV
(2) En-1 to n-2 = 0.389 eV
(3) ninit =4
(4) L =483.676 ×10^-11 nm
(5) λlongest= 1773.33 nm
Explanation:
The comprehensive details regarding the answer are provided in the attached files.
<span>You are presented with a circuit that includes a 6.0-v battery, a 4.0-ohm resistor, a 0.60 microfarad capacitor, an ammeter, and a switch all connected in series. Your task is to determine the current reading once the switch is closed. Ohm's law should be used, which states V = IR where V signifies voltage, I indicates current, and R represents resistance.</span>
V = IR
I = V/R
I = 6 volts / 4 ohms
I = 1.5A
Upon closing the switch, the cathode side plate starts accumulating electrons if it was previously empty. As this process continues, the current diminishes. Eventually, when the capacitor reaches its maximum electron retention, the current will cease. An increased capacitance means a greater capacity for electron storage.
Answer:
Explanation:
The beacon is rotating at an angular speed of
so we have
We know that
At this point we have
So we can conclude with
The charge on the plastic cube is determined as follows.
Answer:
The convergence of light rays redirects them toward the focal point, resulting in a magnifying effect.
Explanation:
