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Andrej [43]
3 years ago
9

Señale la veracidad (V) o falsedad (F) de las siguientes proposiciones: I. Una partícula está en equilibrio cuando se encuentra

en reposo. II. Una partícula se encuentra en equilibrio, el observador no registra fuerza alguna sobre la partícula. III. Un cuerpo que se mueve con rapidez constante se encuentra en equilibrio. FFV VFF FFF VVF
Physics
1 answer:
Alex Ar [27]3 years ago
3 0

Answer:

(i) V, (ii) F, (iii) F

Explanation:

Se comprueba la veracidad de cada proposición:

(i) Una partícula está en equilibrio cuando se encuentra en reposo.

Respuesta: Según la Primera Ley de Newton, toda partícula está en equilibrio si está en reposo o desplazándose a velocidad constante. Por ende, la proposición es verdadera.

(ii) Una partícula se encuentra en equilibrio, el observador no registra fuerza alguna sobre la partícula.

Respuesta: Según la Primera Ley de Newton, toda partícula está en equilibrio si está en reposo o desplazándose a velocidad constante.

Además, la Segunda Ley de Newton establece que la fuerza neta sobre la partícula como resultado de fuerzas externas es igual a cero. Lo cual quiere decir que una partícula puede estar en equilibrio y estar sometidas a fuerzas externas. En conclusión, la proposición es falsa.

(iii) Un cuerpo que se mueve con rapidez constante se encuentra en equilibrio.

Respuesta: Según la Primera Ley de Newton, toda partícula está en equilibrio si está en reposo o desplazándose a velocidad constante.

La rapidez es la magnitud de la velocidad, para que la velocidad sea constante debe ser constante tanto en magnitud como en dirección. La afirmación no garantiza que la dirección de la partícula sea constante. Por ende, la proposición es falsa.

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Time taken by the blast waves to reach the ground = 2\ minutes\ 30\ seconds=2\times 60+30=150\ s

Spedd of the wave would be

Speed=\dfrac{Distance}{Time}\\\Rightarrow v_b=\dfrac{23500}{150}\\\Rightarrow v-b=156.67\ m/s

The velocity of the blast wave is 156.67 m/s

v = Velocity of sound = 343 m/s

\dfrac{v_b}{v}=\dfrac{156.67}{343}\\\Rightarrow v_b=v\dfrac{156.67}{343}\\\Rightarrow v_b=0.45676v

The blast wave is 0.45676 times the speed of sound

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An element has the following natural abundances and isotopic masses: 90.92% abundance with 19.99 amu, 0.26% abundance with 20.99
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<u>Answer:</u> The average atomic mass of the given element is 20.169 amu.

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Average atomic mass of an element is defined as the sum of masses of the isotopes each multiplied by their natural fractional abundance.

Formula used to calculate average atomic mass follows:

\text{Average atomic mass }=\sum_{i=1}^n\text{(Atomic mass of an isotopes)}_i\times \text{(Fractional abundance})_i     .....(1)

We are given:

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Percentage abundance of isotope 1 = 90.92 %

Fractional abundance of isotope 1 = 0.9092

  • For isotope 2:

Mass of isotope 2 = 20.99 amu

Percentage abundance of isotope 2 = 0.26%

Fractional abundance of isotope 2 = 0.0026

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Putting values in equation 1, we get:

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