Answer:
it will take 36.12 ms to reduce the capacitor's charge to 10 μC
Explanation:
Qi= C×V
then:
Vi = Q/C = 30μ/20μ = 1.5 volts
and:
Vf = Q/C = 10μ/20μ = 0.5 volts
then:
v = v₀e^(–t/τ)
v₀ is the initial voltage on the cap
v is the voltage after time t
R is resistance in ohms,
C is capacitance in farads
t is time in seconds
RC = τ = time constant
τ = 20µ x 1.5k = 30 ms
v = v₀e^(t/τ)
0.5 = 1.5e^(t/30ms)
e^(t/30ms) = 10/3
t/30ms = 1.20397
t = (30ms)(1.20397) = 36.12 ms
Therefore, it will take 36.12 ms to reduce the capacitor's charge to 10 μC.
In the writing of ionic chemical formulas the value of each ion's charge is crossed over in the crossover rule.
Rules for naming Ionic compounds
- Frist Rule
The cation (element with a negative charge) is written first in the name then the anion(element with a positive charge) is written second in the name.
- Second rule
When the formula unit contains two or more of the same polyatomic ion, that ion is written in parentheses with the subscript written outside the parentheses.
Example: Sodium carbonate is written as Na₂CO₃ not Na₂(CO)₃
- Third rule
If the cation is a metal ion with a fixed charge then the name of the cation will remain the same as the (neutral) element from which it is derived (Example: Na+ will be sodium).
If the cation is a metal ion with a variable charge, the charge on the cation is indicated using a Roman numeral, in parentheses, immediately following the name of the cation (example: Fe³⁺ = iron(III)).
- Fourth rule
If the anion is a monatomic ion, the anion is named by adding the suffix <em>-ide</em> to the root of the element name (example: F = Fluoride).
The oxidation state of each ion is also important, thus in the crossover rule, the value of each ion's charge is crossed over.
Learn more about chemical formulas here:
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Answer:
I would live in the Atlantic ocean on a lux liner
Explanation:
:)
