A 16-Ω loudspeaker, an 8.0-Ω loudspeaker, and a 4.0-Ω loudspeaker are connected in parallel across the terminals of an amplifier
1 answer:
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Answer:
66.98 db
Explanation:
We know that
L_T= Total signal level in db
n= number of sources
L_S= signal level from signal source.
= 66.98 db
Answer: X
Explanation:
This situation can be illustrated as a car in circular motion (image attached).
In circular motion the acceleration vector is always directed toward the center of the circumference (that's why it's called centripetal acceleration).
So, in this case the arrow labeled X is the only that points toward the center, hence it represents the car's centripetal acceleration
Answer:
Connect C₁ to C₃ in parallel; then connect C₂ to C₁ and C₂ in series. The voltage drop across C₁ the 2.0-μF capacitor will be approximately 2.76 volts.
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Explanation:
Consider four possible cases.
<h3>Case A: 12.0 V.</h3>In case all three capacitors are connected in parallel, the capacitor will be connected directed to the battery. The voltage drop will be at its maximum: 12 volts.
In case the capacitor is connected in parallel with the
capacitor, and the two capacitors in parallel is connected to the
capacitor in series.
The effective capacitance of two capacitors in parallel is the sum of their capacitance: 2.0 + 1.5 = 3.5 μF.
The reciprocal of the effective capacitance of two capacitors in series is the sum of the reciprocals of the capacitances. In other words, for the three capacitors combined,
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What will be the voltage across the 2.0 μF capacitor?
The charge stored in two capacitors in series is the same as the charge in each capacitor.
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Voltage is the same across two capacitors in parallel.As a result,
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Similarly,
- the effective capacitance of the two capacitors in parallel is 5.0 μF;
- the effective capacitance of the three capacitors, combined:
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Charge stored:
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Voltage:
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Connect all three capacitors in series.
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For each of the three capacitors:
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For the capacitor:
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