F=ma
11.6=3.8*a
a=11.6/3.8
a=3.05m/s
Gamma rays because it has more penetrating power and frequency but shorter wavelength.
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
Learn more about potential difference across parallel plate capacitor here:
brainly.com/question/12993474
Answer:
Explanation:
Let the angle between the first polariser and the second polariser axis is θ.
By using of law of Malus
(a)
Let the intensity of light coming out from the first polariser is I'
.... (1)
Now the angle between the transmission axis of the second and the third polariser is 90 - θ. Let the intensity of light coming out from the third polariser is I''.
By the law of Malus
So,
(b)
Now differentiate with respect to θ.
Answer:
Uh No thanks but make me brainiest!
Explanation:
