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3 years ago

1. ¿Cuál es el aporte a la electrostática que se le atribuye a Thales de Mileto?

1 answer:

8 0

Answer: Descubrio una forma de observar la corriente estática, y quizas fue uno de los primeros en observarla y escribir sobre ella.

Explanation:

Alrededor del 600 a.C. Tales se da cuenta de que frotando pieles con otros materiales, lograba que estos atrajeran motas de polvo, y luego escribió sobre esto que hoy en día conocemos como electricidad estática. De hecho los romanos descubrieron que frotando pieles con ámbar podían ver como pequeñas chispas saltando de estos.

Esto sucede por que al frotar los dos materiales, uno de ellos tiene tendencia a perder cargas mientras el otro (usualmente conductor o aislante) tendrá tendencia a absorberlos. Luego esas cargas absorbidas son una corriente estática que queda "almacenada" en uno de los materiales.

Entonces podemos decir que el aporte a la electrostática que se le atribuye a Tales de Mileto es "descubrir" la electricidad estática.

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Two spherical asteroids have the same radius R. Asteroid 1 has mass M and asteroid 2 has mass 1.97·M. The two asteroids are rele

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Answer:

$0.536\sqrt{\frac{GM}{R}}$

Explanation:

We are given that

Mass of one asteroid 1, $m_1=M$

Mass of asteroid 2, $m_2=1.97 M$

Initial distance between their centers,d=13.63 R

Radius of each asteroid=R

d'=R+R=2R

Initial velocity of both asteroids

$u=0$

We have to find the speed of second asteroid just before they collide.

According to law of conservation of momentum

$(m_1+m_2)u=m_1v_1+m_2v_2$

$(M+1.97 M)\times 0=Mv_1+1.97Mv_2$

$Mv_1=-1.97 Mv_2$

$v_1=-1.97v_2$

According to law of conservation of energy

$Gm_1m_2(\frac{1}{d'}-\frac{1}{d})=\frac{1}{2}m_1v^2_1+\frac{1}{2}m_2v^2_2$

$GM(1.97M)(\frac{1}{2R}-\frac{1}{13.63R})=\frac{1}{2}M(-1.97v_2)^2+\frac{1}{2}(1.97M)v^2_2$

$1.97M^2G(\frac{13.63-2}{27.26R})=\frac{1}{2}Mv^2_2(3.8809+1.97)$

$1.97MG(\frac{11.63}{27.26 R})=\frac{1}{2}(5.8509)v^2_2$

$v^2_2=\frac{1.97GM\times11.63\times 2}{27.26R\times 5.8509}$

$v_2=\sqrt{\frac{1.97GM\times11.63\times 2}{27.26R\times 5.8509}}$

$v_2=0.536\sqrt{\frac{GM}{R}}$

Hence, the speed of second asteroid = $0.536\sqrt{\frac{GM}{R}}$

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