<span>Subrahmanyan Chandrasekhar
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Below are the choices that can be found from other source:
a. frequency
b. the Doppler effect
c. the resonance effect
<span>d. intensity
The answer is B.
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Answer:
Option B. 3.0×10¯¹¹ F.
Explanation:
The following data were obtained from the question:
Potential difference (V) = 100 V.
Charge (Q) = 3.0×10¯⁹ C.
Capacitance (C) =..?
The capacitance, C of a capacitor is simply defined as the ratio of charge, Q on either plates to the potential difference, V between them. Mathematically, it is expressed as:
Capacitance (C) = Charge (Q) / Potential difference (V)
C = Q/V
With the above formula, we can obtain the capacitance of the parallel plate capacitor as follow:
Potential difference (V) = 100 V.
Charge (Q) = 3.0×10¯⁹ C.
Capacitance (C) =..?
C = Q/V
C = 3.0×10¯⁹ / 100
C = 3.0×10¯¹¹ F.
Therefore, the capacitance of the parallel plate capacitor is 3.0×10¯¹¹ F.
Because the electrons collide with the particles inside the conductor so are therefore slowed down seen as current is the rate of flow of electrons
Answer:
The answer to the questions is;
In terms of standing waves, the listener moves from a location with high amplitude to one with lower amplitude or vibration (anti-node to node)
The distance 4.1 cm is equivalent to λ/4
Explanation:
For standing waves we have is a stationary wave comprising of two opposite direction moving waves that have equal amplitude and frequency, resulting in the superimposition of the waves. As such certain points are fixed along the wave path that is the peaks amplitude of the wave oscillation is constant at a particular point. A node occurring at a point and an anti-node occurring at another fixed point
When the listener moves 4.1 cm he or she has left the anti-node to the node hence the faintness of the sound
The distance from the node to the anti-node is 1/4 wavelength, or 1/4×λ
Therefore 4.1 cm is λ/4
