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lbvjy [14]
3 years ago
5

A flute player hears four beats per second when she compares her note to an 880 Hz tuning fork (note A). She can match the frequ

ency of the tuning fork by pulling out the "tuning joint" to lengthen her flute slightly. What was her initial frequency?
Physics
1 answer:
ludmilkaskok [199]3 years ago
7 0

Answer:

884Hz

Explanation:

Beats is the absolute difference between two frequencies therefore

Beats = f1-f2

4=f1-880

F1=880+4

F1=884Hz

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Un objeto de 200 gramos está amarrado del extremo de una cuerda y gira describiendo un círculo horizontal de 1.20 m de radio a r
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Answer:

La tensión es 85.3 N.

Explanation:

Cuando el objeto gira en dirección horizontal, la sumatoria de fuerzas se puede calcular usando la segunda ley de Newton:  

\Sigma F_{x} =ma_{c}

T = ma_{c}  

Dado que el movimiento es horizontal, el peso (que está en el eje y) no contribuye en la sumatoria de fuerzas en el eje x. Por lo que la única fuerza actuando sobre el objeto en la dirección del movimiento es la tensión.  

En donde:                                          

m: es la masa del objeto = 200 g = 0.200 kg

a_{c}: es la aceleración centrípeta

La aceleración centrípeta viene dada por:  

a_{c} = \omega^{2} r

En donde:    

ω: es la velocidad angular del objeto = 3 rev/s

r: es el radio = 1.20 m

Entonces, la tensión es:

T = m\omega^{2} r = 0.200 kg(3\frac{rev}{s}*\frac{2\pi rad}{1 rev})^{2}*1.20 m = 85.3 N

   

Por lo tanto, la tensión es 85.3 N.  

Espero que te sea de utilidad!                                                                          

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