please answer at least one
1.the angular diameter of jupiter is 36". if the distance of jupiter from earth is 8.25x10^8 km. Find the diameter of Jupiter.
2. the distance of a planet from earth is 3.66x10^5 km. if the angular diameter of the planet is 1.5. calculate the diameter of the planet
3. calculate the time taken by light to pass through a nucleus of diameter 1.56x10^ -16m. speed of light is 3x10^8 m/sec.
1.the angular diameter of jupiter is 36". if the distance of jupiter from earth is 8.25x10^8 km. Find the diameter of Jupiter.
2. the distance of a planet from earth is 3.66x10^5 km. if the angular diameter of the planet is 1.5. calculate the diameter of the planet
3. calculate the time taken by light to pass through a nucleus of diameter 1.56x10^ -16m. speed of light is 3x10^8 m/sec.
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1) 36" = (36/3600)º = 0.01º = (0.01/360)rev x 2π rad .. θ = 1.745^-4 rad
From radian measure .. arc length(S) = radius(R) x θ(radian)
Putting S ≡ jupiter diameter and R ≡ 8.25^8 km
S = 8.25^8km x 1.745^-4rad .. .. ►S = 1.44^5 km .. (1.44^8m)
2) Using above method .. S = Rθ
If the angular diameter is meant to be .. 1.50º (?) ..
θ = (1.5/360)rev x 2π rad = 0.0262 rad
S = 3.66^5 km x 0.0262rad .. .. ►S(diameter) = 9.60^3 km
3) v = d/t .. t = d/v .. 1.56^-16m / 3.0^8 .. .. ►t = 5.20^-25 s
From radian measure .. arc length(S) = radius(R) x θ(radian)
Putting S ≡ jupiter diameter and R ≡ 8.25^8 km
S = 8.25^8km x 1.745^-4rad .. .. ►S = 1.44^5 km .. (1.44^8m)
2) Using above method .. S = Rθ
If the angular diameter is meant to be .. 1.50º (?) ..
θ = (1.5/360)rev x 2π rad = 0.0262 rad
S = 3.66^5 km x 0.0262rad .. .. ►S(diameter) = 9.60^3 km
3) v = d/t .. t = d/v .. 1.56^-16m / 3.0^8 .. .. ►t = 5.20^-25 s