A 9.0 × 10 3 kg satellite orbits the earth at the distance of 2.56 × 10 7 m from Earth's surface. What is its period?

1.6537 * 10^9

Explanation:

Given that :

Distance (r) = 2.56 × 10^7 m

Using the relation from Kepler's third law of motion :

T2 = [ 4* pi^2/ G (M1 + M2) ]r^3

M1 = mass of earth = 5.97 *10^24

M2 = Mass of satellite = 9.0 × 10^3 kg

G = Gravitational constant = 6.67 *10^-11

T2 = [ (4 x 3.14^2) / 6.67 x 10^-11 ( 9 x 10^3 + 5.97 x 10^24)]/ (2.56 x 10^7)^3

= 1.6537 * 10^9

ice to water 1 water to vapour 2 latent heat

we have been given a quadratic function and we need to restrict the domain such that it becomes a one to one function.

we know that vertex of this quadratic function occurs at (5,2).

further, we know that range of this function is .

if we restrict the domain of this function to either or , it will become one to one function.

let us know find its inverse.

upon interchanging x and y, we get:

let us now solve this function for y.

hence, the inverse function would be if we restrict the domain of original function to and the inverse function would be if we restrict the domain to .