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How fast (in m^(2) s^(-1)) is area swept...

How fast (in `m^(2) s^(-1))` is area swept out by (a) the radius from sun to earth ? (b) the radius from the sun to earth `= 1.496 xx 10^(11) m` , Distance of earth to moon `= 3.845 xx 10^(8) m` and period of revolution of moon `= 27 (1)/(3)` days.

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How fast (in m^2//s ) is area swept out by (a) the radius from sun to earth ?

How fast (in m^2//s ) is area swept out by (b) the radius from earth to moon ? Given distance of sun to earth =1.496xx10^(11)m , distance of earth to moon =3.845xx10^8m and period of revolution of moon =271/3 days.

Calculate the area covered per second (m^(2)s^(-1)) by the Moon for one complete revolution round the Earth (distance of Moon from Earth =3.845 times 10^(8) and period of revolution of Moon =27""1/3 days).

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The distance of the sun from earth is 1.496xx10^(11)m. If the angular diameter of the sun is 2000'', find the diameter of the sun.

The sun's angular diameter is measured to be 1920''. The distance of the sun from the earth is 1.496xx10^(11)m. What is the diameter of the sun?

The sun's angular diameter is measured to be 1920''. The distance of the sun from the earth is 1.496xx10^(11)m. What is the diameter of the sun?

The sun's angular diameter is measured to be 1920''. The distance of the sun from the earth is 1.496xx10^(11)m. What is the diameter of the sun?