This question is incomplete, the complete question is;
A parallel-plate capacitor is made from two aluminum-foil sheets, each 3.0 cm wide and 5.00 m long. Between the sheets is a mica strip of the same width and length that is 0.0225 mm thick. What is the maximum charge?
(The dielectric constant of mica is 5.4, and its dielectric strength is 1.00×10⁸ V/m)
Answer: the maximum charge q is 716.85 μF
Explanation:
Given data;
with = 3.0 cm = 0.03
breathe = 5.0 m
Area = 0.03 × 5 = 0.15 m²
dielectric strength E = 1.00 × 10⁸
∈₀ = 8.85 × 10⁻¹²
constant K = 5.4
maximum charge = ?
the capacitor C = KA∈₀ / d
q = cv so c = q/v
now
q/v = KA∈₀ / d
q = vKA∈₀/d = EKA∈₀
we substitute
q = (1.00 × 10⁸) × 5.4 × 0.15 × 8.85 × 10⁻¹²
q = 716.85 × 10⁻⁶ F
q = 716.85 μF
the maximum charge q is 716.85 μF
Answer:
Wavelength = 3.74 m
Explanation:
In order to find wavelength in "metres", we must first convert megahertz to hertz.
1 MHz = 1 × 10⁶ Hz
80.3 Mhz = <em>x</em>
<em>x </em>= 80.3 × 1 × 10⁶ = 8.03 × 10⁷ Hz
The formula between wave speed, frequency and wavelength is:
v = fλ [where v is wave speed, f is frequency and λ is wavelength]
Reorganise the equation and make λ the subject.
λ = v ÷ f
λ = (3 × 10⁸) ÷ (8.03 × 10⁷)
λ = 3.74 m [rounded to 3 significant figures]
4000 grams kilo means thousand
Answer:
Part a)
V = 18.16 V
Part b)

Part c)
P = 672 Watt
Part d)
V = 5.84 V
Part e)

Explanation:
Part a)
When battery is in charging mode
then the potential difference at the terminal of the cell is more than its EMF and it is given as

here we have



now we have

Part b)
Rate of energy dissipation inside the battery is the energy across internal resistance
so it is given as



Part c)
Rate of energy conversion into EMF is given as



Now battery is giving current to other circuit so now it is discharging
now we have
Part d)



Part e)
now the rate of energy dissipation is given as



Answer:
(A) She needs to move the decimal point by 3 places