Answer: 39.8 μC
Explanation:
The magnitude of the electric field generated by a capacitor is given by:
d is the distance between the plates.
For a capacitor, charge Q = CV where C is the capacitance and V is the voltage.
where A is the area of the plate and ε₀ is the absolute permittivity.
substituting, we get
It is given that the magnitude of the electric field that can exist in the capacitor before air breaks down is, E = 3 × 10⁶ N/C.
radius of the plates of the capacitor, r = 69 cm = 0.69 m
Area of the plates, A = πr² = 1.5 m²
Thus, the maximum charge that can be placed on disks without a spark is:
Q = E×ε₀×A
⇒ Q = 3 × 10⁶ N/C × 8.85 × 10⁻¹² F/m × 1.5 m² = 39.8 × 10⁻⁶ C = 39.8 μC.
Examples of strong acids are hydrochloric acid (HCl), perchloric acid (HClO4), nitric acid (HNO3) and sulfuric acid (H2SO4). ... For example, hydrogen chloride is a strong acid in aqueous solution, but is a weak acid when dissolved in glacial acetic acid.
Answer:
E. The gas absorbs 50 of energy as heat and does 50」ot work
Explanation:
This is following the law of thermodynamics that energy is neither created nor destroyed
<em><u>One</u></em>
Givens
- delta B = 0.20 T/s
- A = 0.07 m^2
- R = 3.5 ohms
Formula
Φ = ΔB*A
e = Φ
Solution (first part)
e = 0.2 * 0.07
e = 0.014 emf
Solution (second part)
i = e/R
i = 0.014 / 3.5
i = 4 * 10^-3
i = 4 ma
Answer
A
<em><u>Two</u></em>
Givens
N = 200 turns
Φ = 30 degrees
Delta B = 0.45 T/s
phi = 30 degrees
r = 0.06 meters
Formula
e = -N * delta B * A * Cos(phi)
Solution
e = -200 * 0.45 (pi r^2) * Cos(30)
e = - 200 * 0.45 * (3.14 * 0.06^2) * cos(30)
e = 0.881 emf
Answer
A
