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Inga [223]
3 years ago
14

What is the magnitude of δv12 if the bulb is removed from the socket (i.e. the circuit is not closed)?

Physics
2 answers:
zloy xaker [14]3 years ago
7 0
If the socket is open after removing the bulb, then the substitute value is
I = 0 A
ΔV₁₂ = IR
= (OA)(2Ω)
= 0 V
andreev551 [17]3 years ago
4 0

Since the circuit is incomplete or not closed, no current flows in the circuit. as per ohm's law , Voltage is directly proportional to current and is given as

V = Voltage = i R where i = current , R = resistance

as no current flows in the circuit, i = 0

the resistance R can not be zero. hence

V = 0 (R)

V = 0 Volts

so the magnitude of the Voltage is zero Volts

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Answer:

The new frequency (F₂ ) will be related to the old frequency by a factor of one (1)

Explanation:

Fundamental frequency = wave velocity/2L

where;

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Wave velocity = \sqrt{\frac{T}{\frac{M}{L}}}

Frequency (F₁) = \frac{\sqrt{\frac{T}{\frac{M}{L}}}}{2*L}

To obtain the new frequency with respect to the old frequency, we consider the conditions stated in the question.

Given:

L₂ =2L₁ = 2L

T₂ = 2T₁ = 2T

(M/L)₂ = 0.5(M/L)₁ = 0.5(M/L)

F₂ = \frac{\sqrt{\frac{2T}{0.5(\frac{M}{L})}}}{4*L} = \frac{\sqrt{4(\frac{T}{\frac{M}{L}}})}{4*L} = \frac{2}{2} [\frac{\sqrt{\frac{T}{\frac{M}{L}}}}{2*L}] = F_1

Therefore, the new frequency (F₂ ) will be related to the old frequency by a factor of one (1).

7 0
3 years ago
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Arisa [49]

Explanation:

Hope this helps,

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Explanation:

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