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barxatty [35]
3 years ago
5

Cual es la masa de los planetas, y la distancia que hay de la tierra a cada uno de ellos.

Physics
1 answer:
qwelly [4]3 years ago
8 0

Answer:

Mercurio

Masa

3,285 × 10^23 kg

Mercurio está a una distancia promedio de 48 millones de millas (77 millones de kilómetros) de la Tierra.

Venus

Masa

4.867 × 10^24kg

En su punto más cercano a la Tierra, Venus está a unos 61 millones de kilómetros (38 millones de millas) de distancia.

tierra

Masa

5.972 × 10^24kg

Marte

Masa

6,39 × 10^23 kg

La distancia mínima de la Tierra a Marte es de unos 54,6 millones de kilómetros (33,9 millones de millas).

Júpiter

Masa

1.898 × 10^27kg

Debido a que ambos planetas viajan en una trayectoria elíptica alrededor del sol, la distancia de Júpiter a la Tierra cambia constantemente. Cuando los dos planetas están en su punto más cercano, la distancia a Júpiter es de solo 365 millones de millas (588 millones de kilómetros).

Saturno

Masa

5.683 × 10^26kg

Su distancia más cercana a la Tierra es de aproximadamente 1.200 millones de kilómetros (746 millones de millas)

Urano

Masa

8.681 × 10^25kg

la distancia entre la Tierra y Urano cambia diariamente. Lo más cerca que están los dos es 1.600 millones de millas (2.600 millones de kilómetros).

Neptuno

Masa

1.024 × 10^26kg

Cuando Neptuno y la Tierra se alinean en el mismo lado del sol, en su punto más cercano, están a solo 2,700 millones de millas (4,300 millones de kilómetros)

masa de Plutón =

1.30900 × 10^22 kilogramos

Plutón se encuentra a 4670 millones de millas (7500 millones de kilómetros) de la Tierra. En su punto más cercano, los dos están a solo 2660 millones de millas (4280 millones de km) de distancia.

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adell [148]
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P = 100 * 5 Kg-m/s towards south
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Hope this helps!
6 0
3 years ago
A beam of white light passes through a uniform thickness of air. If the intensity of the scattered light in the middle of the gr
k0ka [10]

This question is incomplete, the complete question is;

A beam of white light passes through a uniform thickness of air. If the intensity of the scattered light in the middle of the green part of the visible spectrum (λ=520nm) is I, find the intensity (in terms of I) of scattered light

a) In the middle of the red part of the spectrum (λ= 665 nm)

b) In the middle of the violet part of the spectrum (λ = 420 nm)

Answer:

a) the intensity of scattered light ( in terms of I ) In the middle of the red part of the spectrum is 0.3739I

b) the intensity of scattered light ( in terms of I ) In the middle of the violet part of the spectrum is 2.3497I  

Explanation:

Given the data in the question,

the visible spectrum (λ=520nm) = I

we know that; intensity of scattered light is proportional to 1 / λ⁴  

I ∝ ( 1 / λ⁴ )

so

a)

I_R / I = ( λ / λ_R )⁴

As the middle of the green part of the visible spectrum λ is 520nm and middle of the red part of the spectrum λ_R is 665 nm

we substitute

I_R / I = ( 520 / 665 )⁴

I_R / I = ( 0.781954887 )⁴

I_R / I = 0.3739

I_R  =  0.3739I  { in terms of I  }

Therefore, the intensity of scattered light ( in terms of I ) In the middle of the red part of the spectrum is 0.3739I

b)

I_V / I = ( λ / λ_V )⁴

As the middle of the green part of the visible spectrum λ is 520nm and middle of the red part of the spectrum λ_R is 420 nm

we substitute

I_V / I = ( 520 / 420 )⁴

I_V / I = ( 1.238095 )⁴

I_V / I = 2.3497

I_V = 2.3497I  { in terms of I  }

Therefore, the intensity of scattered light ( in terms of I ) In the middle of the violet part of the spectrum is 2.3497I  

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3 years ago
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Dovator [93]
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Answer:

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An asteroid, whose mass is 3.6 × 10-4 times the mass of Earth, revolves in a circular orbit around the Sun at a distance that is
frosja888 [35]

Answer:

a) 2.02 years

b) 8.1 x 10⁻⁸.

Explanation:

Time period of a rotating body T² is proportional to radius of orbit R³ So

T₁² / T₂² = R₁³ /R₂³ ( T₁ and R₁ is time period and radius of orbit of the earth .)

1² / T₂² =( 1/1.6)³

T₂ = 2.02 years.

Kinetic energy of an orbiting body = 1/2 m v₀² ( v₀ is orbital speed)

= 1/2 m x 2 g R = m x G m/R² X R= m² x G /R

Kinetic energy of asteroid K₁ / kinetic energy of earth K₂ =

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