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3 years ago

Why are stars and plantets spherical

Physics

1 answer:

belka [17]3 years ago

6 0

Answer:

When a forming planet is big enough, it starts to clear its path around the star it orbits. ... A planet's gravity pulls equally from all sides. Gravity pulls from the center to the edges like the spokes of a bicycle wheel. This makes the overall shape of a planet a sphere, which is a three-dimensional circle.

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What do deltas and natural levees have in common?

Tpy6a [65]

They are both formed by deposited river sediment.

7 0

3 years ago

Un gallo se pavonea en línea recta en el techo. Su movimiento se muestra en la siguiente gráfica de la posición horizontal, x, c

Svetllana [295]

Answer:

Entonces, velocidad instantánea = | v | = 0,25 m / s

Explanation:

pregunta completa

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Solución

La velocidad instantánea está dada por,

v = dx / dt = pendiente de la curva x-t

aquí, del gráfico:

en x = 1 m ---> t = 8 s

en x = 3 m ---> t = 0 s

entonces, dx = 1-3 = -2 m

dt = 8-0 = 8 s

Entonces,

v = -2/8

v = -0,25 m / s

Dado que, pseed es una cantidad escalar.

Entonces, velocidad instantánea = | v | = 0,25 m / s

8 0

3 years ago

Satellite A orbits a planet at a distance d from the planet’s center with a centripetal acceleration a0. A second identical sate

leva [86]

To solve this problem it is necessary to use the concepts related to the Gravitational Force and Newton's Second Law, as far as we know:

$F_g = \frac{GMm}{r^2}$

Where,

G = Gravitational constant

M = Mass of earth (in this case)

m = mass of satellite

r = radius

In the other hand we have the second's newton law:

$F = ma$

Where,

m = mass

a = acceleration

Equation both equations we have,

$ma = \frac{GMm}{r^2}$

For the problem we have that,

<em>Satellite A:</em>

$ma_A = \frac{GMm}{r^2}$

<em>Satellite B:</em>

$ma_B = \frac{GMm}{(2r)^2}$

The ratio between the two satellites would be,

$\frac{ma_A}{ma_B}= \frac{\frac{GMm}{r^2}}{\frac{GMm}{(2r)^2}}$

Solving for a_B,

$a_B = \frac{a_A}{4}$

Therefore the centripetal acceleration of $A_B$ is a quarter of $a_A$

7 0

3 years ago

A horizontal circular platform (m = 119.1 kg, r = 3.23m) rotates about a frictionless vertical axle. A student (m = 54.3kg) walk

Murrr4er [49]

Answer:

$\omega_2=5.1rad/s$

Explanation:

Since there is no friction angular momentum is conserved. The formula for angular momentum thet will be useful in this case is $L=I\omega$ . If we call 1 the situation when the student is at the rim and 2 the situation when the student is at $r_2=1.39m$ from the center, then we have:

$L_1=L_2$

Or:

$I_1\omega_1=I_2\omega_2$

And we want to calculate:

$\omega_2=\frac{I_1\omega_1}{I_2}$

The total moment of inertia will be the sum of the moment of intertia of the disk of mass $m_D=119.1 kg$ and radius $r_D=3.23m$ , which is $I_D=\frac{m_Dr_D^2}{2}$ , and the moment of intertia of the student of mass $m_S=54.3kg$ at position $r$ (which will be $r_1=r=3.23m$ or $r_2=1.39m$ ) will be $I_{S}=m_Sr_S^2$ , so we will have: