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algol [13]
3 years ago
9

If the tension in a string is doubled, then a natural frequency will (a) increase, (b) decrease by ________?

Physics
1 answer:
-Dominant- [34]3 years ago
6 0

Answer:

(a) increase by \sqrt{2} times

Explanation:

Natural frequency of a wave in a string is given by:

f = \frac{1}{2L}\sqrt{\frac{T}{\mu}}

where, L is the length of the string, T is the tension in the string and \mu is the linear density of the string.

Considering the length and linear density of the string are constant, if the tension in a string is doubled, the natural frequency of the string would:

f\propto \sqrt{T}

\frac{f_n}{f}=\sqrt{\frac{T_n}{T}}\\ \Rightarrow f_n =\sqrt{\frac{2T}{T}}f = \sqrt{2} f

Thus, the natural frequency of the string would increase by  \sqrt{2} times.

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Sindrei [870]

Answer:

13.2 x 10^5 J

Explanation:

Power = work/time

Given

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6 0
3 years ago
Read 2 more answers
Una ambulancia se desplaza por la calle a 34 m/s emitiendo un estruendoso sonido de sirena a 304 Hz, Si usted se ALEJA de la amb
Alik [6]

Answer:

f = 295.06 Hz

Explanation:

Este es un típico ejercicio de efecto doppler donde tenemos una fuente emisora de una frecuencia.

En este caso, vemos una ambulancia que se está desplazando a cierta velocidad emitiendo una frecuencia particular. Un observador que está en la calle decide alejarse, corriendo a una velocidad, que obviamente es menor que la de la ambulancia, pero el caso es que se está alejando de la fuente emisora del sonido, en este caso de la ambulancia.

Como se están alejando, podemos usar la siguiente expresión:

f = (v - v₀ / v) * f₀

Donde:

f: frecuencia percibida por la persona.

v: velocidad de la ambulancia

v₀: velocidad de la persona

f₀: frecuencia emitida por la ambulancia.

Teniendo todos esos valores, solo debemos reemplazar en la expresión:

f = (34 - 1 / 34) * 304

f = (33/34) * 304

<h2>f = 295.06 Hz</h2>

Espero te ayude

6 0
3 years ago
A kettle transfers 6,000j of energy electrically. 1,500j of this is wasted. what is the efficiency of this kettle?
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output energy=4500

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efficiency=0.75x100%

efficiency=75%

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