The energy is 3.06 electronvolts, E = 3.06eV
1eV = 1.6 * 10^-19 J
3.06 eV = 3.06* 1.6 * 10^-19 J = 4.896 * 10^-19 J
1. All the relevant resistors are in series, so the total (or equivalent) resistance is the sum of the resistances of the resistors: 20 Ω + 80 Ω + 50 Ω = 150 Ω [choice A].
2. The ammeter will read the current flowing through this circuit. We can find the ammeter reading using Ohm's law in terms of the electromotive force provided by the battery: I = ℰ/R = (30 V)(150 Ω) = 0.20 A [choice C].
3. The voltmeter will measure the potential drop across the 50 Ω resistor, i.e., the voltage at that resistor. We know from question 2 that the current flowing through the resistor is 0.20 A. So, from Ohm's law, V = IR = (0.20 A)(50 Ω) = 10. V, which will be the voltmeter reading [choice F].
4. Trick question? If the circuit becomes open, then no current will flow. Moreover, even if the voltmeter were kept as element of the circuit, voltmeters generally have a very high resistance (an ideal voltmeter has infinite resistance), so the current moving through the circuit will be negligible if not nil. In any case, the ammeter reading would be 0 A [choice B].
The answer is Monocline. And I checked it, it's correct.
The formula for the pendulum experiment is based on the assumption that the amplitude is small so that the angle is approximately equal to the Sine of the angle.
Answer:
a. λ = 647.2 nm
b. I₀ 9.36 x 10⁻⁵
Explanation:
Given:
β = 56.0 rad , θ = 3.09 ° , γ = 0.170 mm = 0.170 x 10⁻³ m
a.
The wavelength of the radiation can be find using
β = 2 π / γ * sin θ
λ = [ 2π * γ * sin θ ] / β
λ = [ 2π * 0.107 x 10⁻³m * sin (3.09°) ] / 56.0 rad
λ = 647.14 x 10⁻⁹ m ⇒ λ = 647.2 nm
b.
The intensity of the central maximum I₀
I = I₀ (4 / β² ) * sin ( β / 2)²
I = I₀ (4 / 56.0²) * [ sin (56.0 /2) ]²
I = I₀ 9.36 x 10⁻⁵
