\(6 \times 10^6 km\)
\(3 \times 10^6 km\)
\(6 \times 10^7 km\)
\(3 \times 10^7 km\)
Using Kepler’s third law:
\( T^2 \propto R^3 \) or \( \frac{T_1^2}{T_2^2} = \frac{R_1^3}{R_2^3} \)
Given:
\( T_1 = 1 \, \text{year}, \quad R_1 = 1.5 \times 10^6 \, \text{km}, \quad T_2 = 2.83 \, \text{years} \)
Substitute:
\[ \frac{1}{(2.83)^2} = \frac{(1.5 \times 10^6)^3}{R_2^3} \]
\[ R_2^3 = \left[ (2.83)^2 \cdot (1.5 \times 10^6)^3 \right] \]
Taking the cube root:
\[ R_2 = 3 \times 10^6 \, \text{km} \]
Let \( y = f(x) \) be the solution of the differential equation
\[ \frac{dy}{dx} + 3y \tan^2 x + 3y = \sec^2 x \]
such that \( f(0) = \frac{e^3}{3} + 1 \), then \( f\left( \frac{\pi}{4} \right) \) is equal to:
Find the IUPAC name of the compound.
If \( \lim_{x \to 0} \left( \frac{\tan x}{x} \right)^{\frac{1}{x^2}} = p \), then \( 96 \ln p \) is: 32