An asteroid is in circular orbit around the Sun. Its distance from the Sun is 7.3 times the average distance of Earth from the Sun. The period of this asteroid is C : 20 Earth years .

Explanation:

Kepler's law

T² ∝ R³

T = orbital period of the planet

R = semi major axis of the orbit.

Now

( T₁ / T₂ )² = ( R₁ / R₂ )³

T₁ = Earth time period = 1 year

T₂ = Asteroid time period

R₁ = The distance between Earth and Sun

R₂ = The distance between Asteroid and Sun

From the question ,

R₂ = 7.3 R₁

applying on the above equation ,

( T₁ / T₂ )² = ( R₁ / R₂ )³

( 1 / T₂ )² = ( R₁ / 7.3 R₁ )³

( 1 / T₂ )² = ( 1 / 7.3 )³

T₂ = 19.7 year

Hence ,

approximately T₂ = 20 years.

The period of this asteroid is C : 20 Earth years .

8 0

3 years ago

Find the center of the low, which is marked with the letter L. What type of front has formed here? What happens to the maritime

kenny6666 [7]

A stationary front is depicted in the picture at L. A stationary from is a narrow zone separating two air masses in which neither air mass is displacing the other. When this occurs there is a stall in the motion of the front. The blue triangles point away from the cold air and the red semicircles point away from the warm air.

8 0

3 years ago

The builder of a house on a cliff overlooking the ocean should be concerned about what

Anna11 [10]

Answer:

Erosion

Explanation:

Building a house on a cliff is a very trick thing. The cliffs usually are not composed of hard rocks, but instead are usually composed of sedimentary rocks which are much softer and less stable on the long run. The cliffs are also exposed to heavy erosion, in the lower parts because of the waves, in the higher parts because of the winds. The erosion makes the cliffs slowly fall apart, and occasionally even crumble down, so a builder that would build a house on a cliff should be wary of all these things and construct the house where it is the safest for the long run.

6 0

3 years ago