The orbital period, p, of the planet's distance from the sun, a, in astronomical units is related by the formula p=a^3/2. if saturn's orbital periodt n 29.5 years what is its distance from the sun?

Distance of Saturn from sun is 9.5473  astronomical units

Step-by-step explanation:

We have given,

A formula relating the planet's orbital period and planet's distance from the sun. i.e p =

since we have, orbital period of Saturn = 29.5 years

And we need to find the distance of Saturn from the sun.

so we can use the given formula: p =

Here, Orbital period , p = 29.5 , Let a be the distance between Saturn and sun.

29.5 =

Squaring both sides, we get :

29.5² =

Taking cube root both sides, we get:

a = 9.5473  astronomical units

Distance of Saturn from sun is 9.5473  astronomical units

option a, c,d are correct.

step-by-step explanation:

from the given figure, it is given that z is equidistant from the sides of the triangle rst, then from triangle tzb and triangle szb, we have

tz=sz(given)

bz=zb(common)

therefore, by rhs rule,δtzb ≅δszb

by cpctc, sz≅tz

also, from δctz and δasz,

tz=sz(given)

∠tcz=∠saz(90°)

by rhs rule, δctz ≅ δasz, therefore by cpctc, ∠ctz≅∠asz

also,from δasz and δzsb,

zs=sz(common)

∠zbs=∠saz=90°

by rhs rule, δasz ≅δzsb, therefore, by cpctc, ∠asz≅∠zsb

hence, option a, c,d are correct.

8

step-by-step explanation: