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natta225 [31]
3 years ago
12

Suppose that a spring with a larger spring constant was used in this same apparatus. If a given mass were rotated at the same ra

dius at which it had been rotated with the original spring, would the new period of rotation using the new spring be greater or less than the period of rotation using the original spring
Physics
1 answer:
malfutka [58]3 years ago
6 0

Answer:

The new period of rotation using the new spring would be less than the period of rotation using the original spring

Explanation:

Generally the  period of rotation of the mass is mathematically represented as

       T = 2 \pi \sqrt{\frac{I}{k} }

Here I is the moment of inertia of the mass about the rotation axis and  k is the spring constant

Now looking at the equation we can tell that T is inversely proportional to the square root of the spring constant which means that for a larger spring constant the time period would be lesser

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Calcula el valor de la velocidad de las ondas sonoras en el agua sabiendo que su
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  1. La velocidad de las ondas sonoras es aproximadamente 1469,694 metros por segundo.
  2. La longitud de onda de las ondas sonoras es 1,470 metros.

1) Inicialmente, debemos determinar la velocidad de las ondas sonoras a través del agua (v), en metros por segundo:

v = \sqrt{\frac{K}{\rho} } (1)

Donde:

  • K - Módulo de compresibilidad, en newtons por metro cuadrado.
  • \rho - Densidad del agua, en kilogramos por metro cúbico.

Si sabemos que \rho = 1\times 10^{3}\,\frac{kg}{m^{3}} y K = 2,16\times 10^{9}\,\frac{N}{m^{2}}, entonces la velocidad de las ondas sonoras es:

v = \sqrt{\frac{2,16\times 10^{9}\,\frac{N}{m^{2}}}{1\times 10^{3}\,\frac{kg}{m^{3}} } }

v\approx 1469,694\,\frac{m}{s}

La velocidad de las ondas sonoras es aproximadamente 1469,694 metros por segundo.

2) Luego, determinamos la longitud de onda (\lambda), en metros, mediante la siguiente fórmula:

\lambda = \frac{v}{f} (2)

Donde f es la frecuencia de las ondas sonoras, en hertz.

Si sabemos que v\approx 1469,694\,\frac{m}{s} y f = 1000\,hz, entonces la longitud de onda de las ondas sonoras es:

\lambda = \frac{1469,694\,\frac{m}{s} }{1000\,hz}

\lambda = 1,470\,m

La longitud de onda de las ondas sonoras es 1,470 metros.

Para aprender más sobre las ondas sonoras, invitamos a ver esta pregunta verificada: brainly.com/question/1070238

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