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Anarel [89]
3 years ago
8

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

Por favor, encontrar el archivo adjunto

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:

\omega_2=\frac{(I_D+I_{S1})\omega_1}{(I_D+I_{S2})}

or:

\omega_2=\frac{(\frac{m_Dr_D^2}{2}+m_Sr_{S1}^2)\omega_1}{(\frac{m_Dr_D^2}{2}+m_Sr_{S2}^2)}

which for our values is:

\omega_2=\frac{(\frac{(119.1kg)(3.23m)^2}{2}+(54.3kg)(3.23m)^2)(3.1rad/s)}{(\frac{(119.1kg)(3.23m)^2}{2}+(54.3kg)(1.39m)^2)}=5.1rad/s

6 0
3 years ago
Any two importance of measurement in points (i am giving you 30 points)​
erma4kov [3.2K]

Answer:

1.It helps in trade and business.

2. It helps to perform scientific calculations.

5 0
3 years ago
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