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insens350 [35]
3 years ago
11

Una profesora de física que se encuentra en la tierra aplica un examen a sus estudiantes que están en una nave espacial que se d

esplaza a velocidad v = 0,6c con respecto a la tierra. En el instante en que la nave pasa frente a la profesora, ésta indica el inicio del examen. La profesora quiere que sus alumnos cuenten con dos horas (tiempo de la nave espacial) para completar el examen.
Physics
1 answer:
GuDViN [60]3 years ago
7 0

Answer:

t´ =  2,5 h

Explanation:

Dado que a velocidades "comparables" a la velocidad de la luz, los principios de la física deben ser tratados a la luz de los conceptos relativistas, es necesario aplicar el factor de Lorentz al tiempo que los estudiantes tomaran como base para la elaboración de su examen-

El factor de Lorentz  γ =  1 / √1 - (v²/c²)

En ese factor c es la velocidad de la luz y v es la velocidad a la que se mueve el móvil ( en este caso la nave espacial que va a 0,6 *c )

La otra consideración que hay que tomar en cuenta es que cuando se tienen dos sistemas inerciales con velocidades relativas comparables a la velocidad de la luz, las coordenadas espaciales ( y el tiempo que es la cuarta coordenada ) se relacionan según:

(Movimiento exclusivo en la dirección del eje x )

x´  =  ( x - vₓt )*γ      y´= y    z´ = z      t´  = ( t - vₓ/c² ) *γ

Donde las coordenadas s distinguen por ´ y sin ´ para identificar a los dos sistemas, luego entonces:

t´ ( tiempo en la nave espacial )

t´  = [ t  - (0,6*c/c² ] *γ    ⇒    t´ =  t - 0,36/c / √1 - (0.6*c)²/c²

Dado que   t = 2 h   y que la velocidad de la luz c es  300000 km/s expresada en horas es  300000*3600 = 108*10⁷,el numerador se puede aproximar a 2 horas

t´ =  2 /√ 64*10⁻²

t´ = 2 / 0,8  h

t´ =  2,5 h

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The question is incomplete, below is the complete question "A particle moves through an xyz coordinate system while a force acts on it. When the particle has the position vector r with arrow = (2.00 m)i hat − (3.00 m)j + (2.00 m)k, the force is F with arrow = Fxi hat + (7.00 N)j − (5.00 N)k and the corresponding torque about the origin is vector tau = (4 N · m)i hat + (10 N · m)j + (11N · m)k.

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T=(18-14)i-(-12-2F_{x})j+(12+3F_{x})k\\T=4i-(-12-2F_{x})j+(12+3F_{x})k\\\\

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(4Nm)i+(10Nm)j+(11Nm)k=4i-(-12-2F_{x})j+(12+3F_{x})k\\

if we directly compare the j coordinate we have

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