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Vikki [24]
3 years ago
7

un avión aterriza en la superficie de un portaaviones a 50 m/s y se detiene por completo en 120 metros, ¿cuál es la aceleración

necesaria para detenerlo?
Physics
1 answer:
sergejj [24]3 years ago
5 0

Answer:

La aceleración necesaria para detener el avión es - 10.42 m/s².

Explanation:

Un movimiento uniformemente acelerado (M.U.A) es aquél cuya aceleración es constante y la velocidad de un objeto cambia a medida que el movimiento evoluciona.

Siendo la aceleración "a" el cambio de velocidad al tiempo transcurrido en un punto A a B, la velocidad inicial la velocidad que tiene un cuerpo al iniciar su movimiento en un período de tiempo y la velocidad final la velocidad que tiene un cuerpo al finalizar su movimiento en un período de tiempo, entonces en  M.U.A se cumple:

Vf² - Vo² = 2*a*d

donde:

  • Vf: Velocidad final
  • Vo: Velocidad inicial
  • a: Aceleración
  • d: Distancia recorrida

En este  caso:

  • Vf: 0 m/s, porque el avión se detiene
  • Vo: 50 m/s
  • a: ?
  • d: 120 m

Reemplazando:

(0 m/s)² - (50 m/s)² = 2*a*120 m

Resolviendo:

a=\frac{(0 m/s)^{2} -(50 m/s)^{2} }{2*120 m}

a= - 10.42 m/s²

<u><em>La aceleración necesaria para detener el avión es - 10.42 m/s².</em></u>

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2) The angular acceleration is C) 15.7 rad/s^2

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4) The period of vibration is B) 0.6 s

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1)

The period of a simple pendulum is given by

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B) III. only (the length of the pendulum)

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The angular acceleration of the rotating disc is given by the equation

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For the compact disc in this problem we have:

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Substituting, we find

\alpha = \frac{31.4-0}{2}=15.7 rad/s^2

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For a simple harmonic oscillator, the acceleration and the displacement of the system are related by the equation

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a is the acceleration

x is the displacement

\omega is the angular frequency of the system

For the oscillator in this problem, we have the following relationship

a=-100 x

which implies that

\omega^2 = 100

And so

\omega = \sqrt{100}=10 rad/s

Also, the angular frequency is related to the frequency f by

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Therefore, the frequency of this simple harmonic oscillator is

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When the mass is hanging on the sping, the weight of the mass is equal to the restoring force on the spring, so we can write

mg=kx

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x = 8.0 cm = 0.08 m is the stretching of the spring

We can re-arrange the equation as

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According to the equation of continuity, the volume flow rate must remain constant, so we can write

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A_1 = \pi r_1^2 is the cross-sectional area of the hose, with r_1 = 5 mm being the radius of the hose

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A_2 = \pi r_2^2 is the cross-sectional area of the nozzle, with r_2 being the radius of the nozzle

v_2 = 16 m/s is the speed in the nozzle

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So, the diameter of the nozzle will be

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F_1 is the force that must be applied to the small piston

A_1 = \pi r_1^2 is the area of the first piston, with r_1= 2 cm being its radius

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A_2 = \pi r_2^2 is the area of the bigger piston, with r_2= 15 cm being its radius

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