True. Being an electrical engineer, you learn very quickly that current will take All paths of resistance. But, the higher the resistance the lower the voltage. So, if a high resistance is shorted and the current flows through the short, there will be some small voltage across it, so some small amount of current will still flow through the high resistance.
The one that's completely submerged is displacing more water, so the buoyant force on it is greater.
Answer:
correct answer is C
Explanation:
The photoelectric effect was correctly described by Einstein, where he assumes that the light ray is formed by photons that are articulated and behaves like an elastic shock, the energy of this particular is described by the Planck equation.
K = h f + Ф
where k is the kinetic energy of the electrons, f the frequency of the photons and Ф the work function of the material.
In this experiment, red light removes electrons, it is assumed that each photon spreads an electron if we have another light with more energy and 10% more intense, that is, with 10% more shapes and each arcane an electron the number of electrons removed of; material is increased by 10%.
The change in wavelength and consequently the frequency
c = λ f
f = c /λ
therefore, the wavelength of the voilet λ = 400 num has a higher frequency and therefore more energy, so that the turned-on turns have more kinetic energy.
With these approaches we examine the final answers where the correct answer is C
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The potential difference across the parallel plate capacitor is 2.26 millivolts
<h3>Capacitance of a parallel plate capacitor</h3>
The capacitance of the parallel plate capacitor is given by C = ε₀A/d where
- ε₀ = permittivity of free space = 8.854 × 10⁻¹² F/m,
- A = area of plates and
- d = distance between plates = 4.0 mm = 4.0 × 10⁻³ m.
<h3>Charge on plates</h3>
Also, the surface charge on the capacitor Q = σA where
- σ = charge density = 5.0 pC/m² = 5.0 × 10⁻¹² C/m² and
- a = area of plates.
<h3>
The potential difference across the parallel plate capacitor</h3>
The potential difference across the parallel plate capacitor is V = Q/C
= σA ÷ ε₀A/d
= σd/ε₀
Substituting the values of the variables into the equation, we have
V = σd/ε₀
V = 5.0 × 10⁻¹² C/m² × 4.0 × 10⁻³ m/8.854 × 10⁻¹² F/m
V = 20.0 C/m × 10⁻³/8.854 F/m
V = 2.26 × 10⁻³ Volts
V = 2.26 millivolts
So, the potential difference across the parallel plate capacitor is 2.26 millivolts
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