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3 years ago

What is a substance

Physics

1 answer:

Ulleksa [173]3 years ago

4 0

Well, if i am correct. a substance is:

a particular kind of matter with uniform properties

or...

the real physical matter of which a person or thing consists and which has a tangible, solid presence.

choose which ever you like better:)

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The force that contributed to the formation of planets, determines the motion of bodies in the solar system, and pulls objects t

kykrilka [37]

Answer:

The force of gravity

Explanation:

Gravity was studied, by early scientists such as Copernicus and others, Galileo was the first to ensure that planets moved according to a physical equation that depended on a force that caused celestial bodies to move and interact with each other. But years later Newton based on studies conducted deciphering what Galileo assumed, he was able to find the equation of the force of gravity in any body in the universe. This equation depends on the masses of the two interacting bodies, the distance between them and a constant, which I call universal gravitation constant.

$F_{g}=G*\frac{m_{1}*m_{2}}{r^2}$

Fg = gravity force [N]

G = universal gravitation constant = 6.67*10^(-11) [N*m^2/kg^2]

m1 = mass of the 1st body [kg]

m2 = mass of the 2nd body [kg]

r = distance between the bodies [meters]

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3 years ago

Why are there so few working wave power stations?

Bond [772]

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3 years ago

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zalisa [80]

Answer:

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Explanation:

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How does a boulder change

Minchanka [31]

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3 years ago

Se lanza verticalmente una esfera con una rapidez de 30m/se. Determinar la rapidez de la esfera a una altura de 40m (g=10m/s2)

sergeinik [125]

$v^2-{v_0}^2=2a(x-x_0)$

dónde $v$ es la velocidad de la esfera, $v_0$ es suya velocidad inicial, $a=-g$ la aceleración debida a la gravedad, $x$ la posición, y $x_0$ la posición inicial. Tomamos $x_0=0\,\mathrm m$ a referirse a la posición de la esfera en el momento que la esfera fue lanzada.

Entonces

$v^2-\left(30\,\dfrac{\mathrm m}{\mathrm s}\right)^2=2\left(-10\,\dfrac{\mathrm m}{\mathrm s^2}\right)(40\,\mathrm m)$

$\implies v^2=100\,\dfrac{\mathrm m^2}{\mathrm s^2}\implies v=\pm10\,\dfrac{\mathrm m}{\mathrm s}$

Esto nos dice que la esfera alcanza una altura de 40 m en dos momentos - una vez hacia arriba y una vez hacia abajo. Sin embargo, independientemente del signo de la velocidad, sabemos que suya magnitud es 10 m/s, y así tenemos una rapidez de 10 m/s también en ambos momentos.

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3 years ago