The greater the cross-sectional area of an object, the greater the amount of air resistance it encounters since it collides with more air molecules. ... It will have to accelerate for a longer period of time before there is enough upward air resistance to balance the downward force of gravity.
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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Explanation:
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The water pressure on the first floor must be 455 PSI in order to push the water to the 13th floor at the given pressure.
The given parameters;
- <em>Pressure on the 13 th floor, P₁ = 35 PSI</em>
- <em>Distance between each floor, d = 10 ft</em>
The vertical pressure of the water is calculated as follows;
The vertical height of the first floor from the 13th floor = 130 ft
The vertical height of the 13 ft floor = 10 ft
Thus, the water pressure on the first floor must be 455 PSI in order to push the water to the 13th floor at the given pressure.
Learn more about vertical height and pressure here: brainly.com/question/15691554
