One form of Ohm's Law says . . . . . Resistance = Voltage / Current .
R = V / I
R = (12 v) / (0.025 A)
R = (12 / 0.025) (V/I)
<em>R = 480 Ohms</em>
I don't know if the current in the bulb is steady, because I don't know what a car's "accumulator" is. (Floogle isn't sure either.)
If you're referring to the car's battery, then the current is quite steady, because the battery is a purely DC storage container.
If you're referring to the car's "alternator" ... the thing that generates electrical energy in a car to keep the battery charged ... then the current is pulsating DC, because that's the form of the alternator's output.
The answer is b
300,000 km
Answer:
(a) The energy of the photon is 1.632 x
J.
(b) The wavelength of the photon is 1.2 x
m.
(c) The frequency of the photon is 2.47 x
Hz.
Explanation:
Let;
= -13.60 ev
= -3.40 ev
(a) Energy of the emitted photon can be determined as;
-
= -3.40 - (-13.60)
= -3.40 + 13.60
= 10.20 eV
= 10.20(1.6 x
)
-
= 1.632 x
Joules
The energy of the emitted photon is 10.20 eV (or 1.632 x
Joules).
(b) The wavelength, λ, can be determined as;
E = (hc)/ λ
where: E is the energy of the photon, h is the Planck's constant (6.6 x
Js), c is the speed of light (3 x
m/s) and λ is the wavelength.
10.20(1.6 x
) = (6.6 x
* 3 x
)/ λ
λ = 
= 1.213 x 
Wavelength of the photon is 1.2 x
m.
(c) The frequency can be determined by;
E = hf
where f is the frequency of the photon.
1.632 x
= 6.6 x
x f
f = 
= 2.47 x
Hz
Frequency of the emitted photon is 2.47 x
Hz.