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nika2105 [10]
3 years ago
10

KINEMATICS: MOTION ALONG STRAIGHT LINE

Physics
1 answer:
Vinil7 [7]3 years ago
6 0

Explanation:

hope this helps, cheers!

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3 years ago
A fish is hooked on a line that is wound around the spool of a fishing reel. The reel has a mass of 652 g and a diameter of 10 c
iren2701 [21]

The magnitude of the force exerted by the fish is 3,052.5 N.

The given parameters;

  • mass of the reel, m = 652 g = 0.652 kg
  • diameter of the reel, d = 10 cm
  • radius of the reel, r= 5 cm = 0.05 m
  • time of motion, t = 5 s
  • angular speed of the reel, ω = 306 rad/s

According to Newton's third law, the force exerted by the reel is equal in magnitude to the force fish exerted on the reel.

The centripetal force of the reel = force exerted by the fish

F_c = m\omega^2 r\\\\F_c = 0.652 \times (306)^2 \times 0.05\\\\F_c = 3,052.5 \ N

Thus, the magnitude of the force exerted by the fish is 3,052.5 N.

Learn more here: brainly.com/question/14905888

3 0
3 years ago
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
GuDViN [60]

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

7 0
3 years ago
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