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Oksanka [162]
3 years ago
12

a waves amplitude is 0.5 meters. If the amplitude is increased to 1 metro, how does its energy change

Physics
1 answer:
marysya [2.9K]3 years ago
5 0

Answer:

The energy becomes 4 times greater.

Explanation:

We know that the energy of a wave is proportional to the square of its amplitude

E ∝ Amplitude^2

Since the original amplitude = 0.5 m

and the new amplitude becomes = 1 m

We are doubling the amplitude. This means that the new energy will be affected by a factor of 4

E_new  ∝ (2*Amplitude)^2  =

E_new ∝ 4*(Amplitude)^2  

E_new  = 4*E

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La diferencia entre las tensiones máxima y mínima es de 19.614 newtons.

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T_{max} - m\cdot g = m\cdot \frac{v^{2}}{L}

En cambio, la mínima tensión aparece cuando la bola se encuentra en el cénit (o la cima) del trayecto circular, descrita por la misma ley de Newton:

T_{min} + m\cdot g = m\cdot \frac{v^{2}}{L}

Donde:

T_{min}, T_{max} - Tensiones mínima y máxima, medidas en newtons.

m - Masa de la bola, medida en kilogramos.

g - Constante gravitacional, medida en metros por segundo al cuadrado.

L - Distancia con respecto al eje de rotación, medida en metros.

v - Rapidez tangencial, medido en metros por segundo.

Se elimina la aceleración centrípeta de ambas expresiones por igualación:

T_{min} + m\cdot g = T_{max} - m\cdot g

Ahora, la diferencia entre las tensiones máxima y mínima es:

T_{max} - T_{min} = 2\cdot m \cdot g

Si m = 1\,kg y g = 9.807\,\frac{m}{s^{2}}, entonces:

T_{max} - T_{min} = 2\cdot (1\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)

T_{max}-T_{min} = 19.614\,N

La diferencia entre las tensiones máxima y mínima es de 19.614 newtons.

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