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
Co carbon monoxide
sorry friend i don't know other ones
Answer:
v= 1.71 m/s
Explanation:
Given that
Distance between two successive crests = 4.0 m
λ = 4 m
T= 7 sec
T is the time between 3 waves.
3 waves = 7 sec
1 wave = 7 /3 sec
So t= 7/3 s
We know that frequency f
f= 1/t= 3/7 Hz
Lets take speed of the wave is v
v= f λ
f=frequency
λ=wavelength
v= 3/7 x 4 = 12 /7
v= 1.71 m/s
No,because they may have more particles
