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timofeeve [1]
2 years ago
11

What is the most basic level of knowledge that contains the most specific concepts

Physics
1 answer:
sdas [7]2 years ago
4 0

\huge \bold \red { \underline {levels \: of \: concept}}

It is the most general is the superordinate concept, followed by the basic concept, and the most specific is the subordinate concept.

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Describe how a battery works
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In a series RLC ac circuit, a second resistor is connected in series with the resistor previously in the circuit. As a result of
AnnyKZ [126]

Answer:

* The first thing we observe is that the frequency response does not change

* The current that circulates in the circuit decreases due to the new resistance at the resonance point,

          Z = R + R₂

Explanation:

The impedance of a series circuit is

          Z₀² = R² + (X_L-X_C) ²

when we place another resistor in series the initial resistance impedance changes to

          Z² = (R + R₂) ² + (X_L - X_C) ²

           

let's analyze this expression

* The first thing we observe is that the frequency response does not change

* The current that circulates in the circuit decreases due to the new resistance at the resonance point,

          Z = R + R₂

8 0
3 years ago
A parallel-plate capacitor with plates of area 360 cm2 is charged to a potential difference V and is then disconnected from the
Softa [21]

Answer:

Q=3.9825\times 10^{-9} C

Explanation:

We are given that a parallel- plate capacitor is charged to a potential difference V and then disconnected from the voltage source.

1 m =100 cm

Surface area =S=\frac{360}{10000}=0.036 m^2

\Delta d=0.8 cm=0.008 m

\Delta V=100 V

We have to find the charge Q on the positive plates of the capacitor.

V=Initial voltage between plates

d=Initial distance between plates

Initial Capacitance of capacitor

C=\frac{\epsilon_0 S}{d}

Capacitance of capacitor after moving plates

C_1=\frac{\epsilon_0 S}{(d+\Delta d)}

V=\frac{Q}{C}

Potential difference between plates after moving

V=\frac{Q}{C_1}

V+\Delta V=\frac{Q}{C_1}

\frac{Qd}{\epsilon_0S}+100=\frac{Q(d+\Delta d)}{\epsilon_0S}

\frac{Q(d+\Delta d)}{\epsilon_0 S}-\frac{Qd}{\epsilon_0S}=100

\frac{Q\Delta d}{\epsilon_0 S}=100

\epsilon_0=8.85\times 10^{-12}

Q=\frac{100\times 8.85\times 10^{-12}\times 0.036}{0.008}

Q=3.9825\times 10^{-9} C

Hence, the charge on positive plate of capacitor=Q=3.9825\times 10^{-9} C

6 0
3 years ago
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