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Molodets [167]
3 years ago
7

The radius of the earth's very nearly circular orbit around the sun is 1.5 1011 m. find the magnitude of the earth's velocity, a

ngular velocity, and centripetal acceleration as it travels around the sun. assume a year of 365 days. (a) the earth's velocity
Physics
1 answer:
kotykmax [81]3 years ago
5 0
The radius of Earth's circular trajectory is
R=1.5*10^{11} m
The time for the Earth to travel around the Sun is
t=365(days)*24(hours)*3600(seconds) =3.1536*10^7 (s)
Thus the velocity is velocity
v=2\pi R/t =2\pi *1.5*10^{11}/(3.1536*10^{7})=2.99*10^4 (m/s)
From this we can deduce the centripetal acceleration
a=v^2/R =(2.99*10^4)^2/(1.5*10^{11}) =5.96*10^{-3}  (m/s^2)
Angular velocity is
\omega =v/R =(2.99*10^4)/(1.5*10^{11})=2*10^{-7} (rad/s)

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Explanation:

1) Los parámetros dados son;

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r = √ (656100 / π) ≈ 456,994 cm

El diámetro = 2 × r ≈ 2 × 456.994 ≈ 913.987 cm

El diámetro ≈ 913,987 cm

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La fuerza del cilindro = 5576850 kgf

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