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:
The main reason is that very young calves are more noticeable to predators when mixed with older calves from the previous year
Explanation:
Answer:
1.A compound is when two substances are put together and there is a reaction creating a new substance Ex. H20. A mixture is when two substances are put together, there is no reaction and they can be separated easier than compounds. Ex. salad dressing.
2.Atomic bonds
3.Compound
4.Element
5.Mixture
Explanation:
Answer:
3000 J
Explanation:
Kinetic energy is:
KE = ½ mv²
If m = 15 kg and v = -20 m/s:
KE = ½ (15 kg) (-20 m/s)²
KE = 3000 J
Answer:
e. 1.2 x 10²³
Explanation:
According to the problem, The current equation is given by:
Here time is in seconds.
Consider at t=0 s the current starts to flow due to battery and the current stops when the time t tends to infinite.
The relation between current and number of charge carriers is:
Here the limits of integration is from 0 to infinite. So,
q = 1.90 x 10⁴ C
Consider N be the total number of charge carriers. So,
q = N e
Here e is electronic charge and its value is 1.69 x 10⁻¹⁹ C.
N = q/e
Substitute the suitable values in the above equation.
N = 1.2 x 10²³