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LiRa [457]
3 years ago
15

Un cuerpo se lanza verticalmente hacia arriba con una velocidad de 13 m/s. ¿Cuánto tiempo tarda en alcanzar la altura máxima? a)

-1,33 m/s b) 1,4 s c) 1,33 m/s d) 1,33 s
Physics
1 answer:
Elena L [17]3 years ago
6 0

Answer:

D. 1.33 segundos.

Explanation:

El cuerpo es experimenta un movimiento en caída libre al modificarse su velocidad por efecto de la gravitación terrestre. Este cuerpo alcanza instantáneamente el reposo cuando se encuentra a su altura máxima, el tiempo puede obtenerse sabiendo la aceleración y las velocidades incial y final a partir de la siguiente ecuación cinemática:

v = v_{o}+g\cdot t

Donde:

v - Velocidad final del cuerpo, medida en metros por segundo.

v_{o} - Velocidad inicial del cuerpo, medida en metros por segundo.

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

t - Tiempo, medido en segundos.

Ahora se despeja el tiempo:

t = \frac{v-v_{o}}{g}

Si v_{o} = 13\,\frac{m}{s}, v=0\,\frac{m}{s} y g = -9.807\,\frac{m}{s^{2}}, entonces:

t = \frac{0\,\frac{m}{s}-13\,\frac{m}{s}}{-9.807\,\frac{m}{s^{2}} }

t = 1.326\,s

Por ende, la respuesta correcta es D.

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The acceleration due to gravity, g , is constant at sea level on the Earth's surface. However, the acceleration decreases as an
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Answer:

  g    = g₀   [1- 2 h / Re + 3 (h / Re)²]

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        F = G m Me / Re²

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We call gravity acceleration a

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When the body is at a height h above the surface the distance is

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      F = G Me m / (Re + h)²

Let's take Re's common factor

      F = G Me / Re²  m / (1+ h / Re)²

As Re has a value of 6.37 10⁶ m and the height of the body in general is less than 10⁴ m, the h / Re term is very small, so we can perform a series expansion

         (1+ h / Re)⁻² = 1 -2 h / Re + 6/2 (h / Re) 2 + ...

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       F = G Me /Re²   m [1- 2 h / Re + 3 (h / Re)²]

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If we call the force of attraction at height

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3 0
3 years ago
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here we know that

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F = \frac{mg}{2}

F = \frac{145}{2} = 72.5 N

so our applied force is 72.5 N

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