Checking a table of star sizes, we'll see that our Sun is a medium sized star, so the answer is D, average.
Answer:
1.324 × 10⁷ m
Explanation:
The centripetal acceleration, a at that height above the earth equal the acceleration due to gravity, g' at that height, h.
Let R be the radius of the orbit where R = RE + h, RE = radius of earth = 6.4 × 10⁶ m.
We know a = Rω² and g' = GME/R² where ω = angular speed = 2π/T where T = period of rotation = 1 day = 8.64 × 10⁴s (since the shuttle's period is synchronized with that of the Earth's rotation), G = gravitational constant = 6.67 × 10⁻¹¹ Nm²/kg², ME = mass of earth = 6 × 10²⁴ kg. Since a = g', we have
Rω² = GME/R²
R(2π/T)² = GME/R²
R³ = GME(T/2π)²
R = ∛(GME)(T/2π)²
RE + h = ∛(GMET²/4π²)
h = ∛(GMET²/4π²) - RE
substituting the values of the variables, we have
h = ∛(6.67 × 10⁻¹¹ Nm²/kg² × 6 × 10²⁴ kg × (8.64 × 10⁴s)²/4π²) - 6.4 × 10⁶ m
h = ∛(2,987,477 × 10²⁰/4π² Nm²s²/kg) - 6.4 × 10⁶ m
h = ∛75.67 × 10²⁰ m³ - 6.4 × 10⁶ m
h = ∛(7567 × 10¹⁸ m³) - 6.4 × 10⁶ m
h = 19.64 × 10⁶ m - 6.4 × 10⁶ m
h = 13.24 × 10⁶ m
h = 1.324 × 10⁷ m
The speed of a car travelling over a hill that has a radius of curvature should not exceed a certain speed other it will topple. This speed is related to the radius of curvature and the gravitational acceleration as shown below:
V^2 = Rg, where V = maximum speed, R = Radius of curvature, g = gravitational acceleration.
Substituting;
V = Sqrt (Rg) = Sqrt (120*9.81) = 34.31 m/s
We have: K.E. = mv² / 2
here, m = 4 Kg
v = 9 m/s
Substitute their values into the expression:
K.E. = (4)(9)² / 2
K.E. = (4)(81) / 2
K.E. = 324 / 2
K.E. = 162 Joules
In short, Your Answer would be 162 J
Hope this helps!
It would mean that only one side of earth would be light and the other dark all the time also we would only see the sun on one side and on the other we see the moon
