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Olenka [21]
2 years ago
13

Cuál es la resistencia de un foco que está conectado a una fuente de 20 v por el cuál pasa una intensidad de 20 mA

Physics
1 answer:
LUCKY_DIMON [66]2 years ago
7 0

Answer:

The resistance of the bulb is 1000 ohms.

Explanation:

What is the resistance of a bulb that is connected to a 20 v source through which a current of 20 mA passes

We have,

Voltage of the bulb is 20 V and the current passes is 20 mA.

It is required to find the resistance of the bulb. The relation between resistance, current and voltage is given by using Ohm's law as :

V =  IR

R is resistance

R=\dfrac{V}{I}\\\\R=\dfrac{20}{20\times 10^{-3}}\\\\R=1000\ \Omega

So, the resistance of the bulb is 1000 ohms.

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Answer:

84.4 %

Explanation:

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If the second harmonic of a certain string is 42 Hz, what is the fundamental frequency of the string?
sdas [7]
Data:
f_{2} = 42 Hz
n (Wave node)
V (Wave belly) 
L (Wave length)
<span>The number of bells is equal to the number of the harmonic emitted by the string.
</span>
f_{n} =  \frac{nV}{2L}

Wire 2 → 2º Harmonic → n = 2

f_{n} = \frac{nV}{2L}
f_{2} = \frac{2V}{2L} &#10;
2V =  f_{2} *2L
V =  \frac{ f_{2}*2L }{2}
V =  \frac{42*2L}{2}
V =  \frac{84L}{2}
V = 42L

Wire 1 → 1º Harmonic or Fundamental rope → n = 1


f_{n} = \frac{nV}{2L}
f_{1} = \frac{1V}{2L}
f_{1} =  \frac{V}{2L}

If, We have:
V = 42L
Soon:
f_{1} = \frac{V}{2L}
f_{1} = \frac{42L}{2L}
\boxed{f_{1} = 21 Hz}

Answer:

<span>The fundamental frequency of the string:
</span>21 Hz

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