Answer:
6.96 s
Explanation:
<u>Given:</u>
- u = initial speed of the automobile = 0 m/s
- a = constant acceleration of the automobile =
- v = constant speed of the truck = 8.7 m/s
<u>Assume:</u>
- t = time instant at which the automobile overtakes the truck.
At the moment the automobile and the truck both meat each other the distance travel by both vehicles must be the same.
Since t = 0 s is the initial condition. So, they both meet again at t = 6.96 s such that the automobile overtakes the truck.
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:
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The second one if it’s on edge
The most dangerous frequencies of electromagnetic energy are X-rays, gamma rays, ultraviolet light and microwaves. X-rays, gamma rays and UV light can damage living tissues, and microwaves can cook them. Hope this helps! =^-^=
