Answer:
c. As we gain mass, the force of gravity on us increases
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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brainly.com/question/12993474
Answer: B.) 6
Explanation:
To answer this problem you get the number of students who attended Wednesday (18) and the number of students who attended Tuesday (12) and subtract 18 - 12 = 6
Answer = B.) 6
<span>In the physics lab, a cube slides down a frictionless incline as shown in the figure below, check the image for the complete solution:
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Answer:
Current, I = 0.0011 A
Explanation:
It is given that,
Diameter of rod, d = 2.56 cm
Radius of rod, r = 1.28 cm = 0.0128 m
The resistivity of the pure silicon, 
Length of rod, l = 20 cm = 0.2 m
Voltage, 
The resistivity of the rod is given by :
R = 893692.30 ohms
Current flowing in the rod is calculated using Ohm's law as :
V = I R
I = 0.0011 A
So, the current flowing in the rod is 0.0011 A. Hence, this is the required solution.
