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yawa3891 [41]
4 years ago
8

Tres móviles, A, B y C, con las mismas características tienen los siguientes estados: A se encuentra inicialmente en reposo y ca

e libremente, B se mueve hacia la derecha con una velocidad constante de 10m/s y C está en el suelo completamente en reposo
¿Algunos de los tres móviles tendría algún tipo de fuerza que actué sobre él? Si es afirmativa tu respuesta ¿Es o son las únicas fuerzas? Explica.
Physics
1 answer:
Morgarella [4.7K]4 years ago
6 0

Answer:

La respuesta es sí, hay una fuerza que actúa sobre el móvil A y es la única fuerza ya que A cae libremente bajo la influencia de la fuerza.

Explanation:

Según la primera ley de movimiento de Newton, un cuerpo continuará en su estado de reposo o en un movimiento uniforme en línea recta a menos que actúen sobre él fuerzas impresas.

Dado que el móvil A cae libremente, desde su estado de reposo inicial, según la primera ley de movimiento de Newton, experimenta una fuerza que actúa sobre él para hacer que caiga y continúe en caída libre.

El móvil B se mueve con una velocidad constante, por lo tanto, de acuerdo con la primera ley de movimiento de Newton, no hay fuerzas impresas que actúen sobre él.

El móvil C está completamente en reposo en el suelo, por lo tanto, tampoco hay fuerzas que actúen sobre él.

La respuesta es sí, hay una fuerza actuando sobre el móvil A y es la única fuerza cuando A cae libremente bajo la influencia de la fuerza.

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Answer:

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Explanation:

For this exercise let's calculate the energy of a single quantum of energy, use Planck's law

         E = h f

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Let's calculate

          E₀ = 6.6310⁻³⁴ 3 10⁸/1000 10⁻⁹

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This is the energy emitted by a photon let's use a proportions rule to find the number emitted in P = 100 w

                #_photon = P / E₀

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3 years ago
An apparatus like the one Cavendish used to find G has large lead balls that are 5.2 kg in mass and small ones that are 0.046 kg.
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Answer:

The magnitude of gravitational force between two masses is 4.91\times 10^{-9}\ N.

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F=G\dfrac{m_1m_2}{r^2}\\\\F=6.67259\times 10^{-11}\times \dfrac{5.2\times 0.046}{(0.057)^2}\\\\F=4.91\times 10^{-9}\ N

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Part A: t = v_0/a_0

Part B: t = v_0/a_0

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v = v_0 + at\\0 = v_0 - a_0t\\t = \frac{v_0}{a_0}

Part B:

We will use the same kinematics equation:

v = v_0 + at\\v_0 = 0 + a_0t\\t = \frac{v_0}{a_0}

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The total time takes is 2t.

So the train moves a distance of

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So the total distance that the car traveled is d = \frac{v_0^2}{a_0}

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Answer:

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