Kepler's third law hypothesizes that for all the small bodies in orbit around the
same central body, the ratio of (orbital period squared) / (orbital radius cubed)
is the same number.
<u>Moon #1:</u> (1.262 days)² / (2.346 x 10^4 km)³
<u>Moon #2:</u> (orbital period)² / (9.378 x 10^3 km)³
If Kepler knew what he was talking about ... and Newton showed that he did ...
then these two fractions are equal, and may be written as a proportion.
Cross multiply the proportion:
(orbital period)² x (2.346 x 10^4)³ = (1.262 days)² x (9.378 x 10^3)³
Divide each side by (2.346 x 10^4)³:
(Orbital period)² = (1.262 days)² x (9.378 x 10^3 km)³ / (2.346 x 10^4 km)³
= 0.1017 day²
Orbital period = <u>0.319 Earth day</u> = about 7.6 hours.
Answer:
λ = 1.4 × 10^(-7) m
Explanation:
We are given;
distance of eye piece from the source;D = 1.5 m
distance between the virtual sources;d = 7.5 × 10^(-4) m
To find the wavelength, we will use the formula for fringe width;
X = λD/d
Where X is fringe width, λ is wavelength, while d and D remain as before.
Now, fringe width = eye-piece distance moved transversely/number of fringes
Eye piece distance moved transversely = 1.88 cm = 1.88 × 10^(-2) m
Thus,
Fringe width = (1.88 × 10^(-2))/10 = 1.88 × 10^(-3) m
Thus;
1.88 × 10^(-3) = λ(1.5)/(7.5 × 10^(-4))
λ = [1.88 × 10^(-3) × (7.5 × 10^(-4))]/1.5
λ = 1.4 × 10^(-7) m
Density formula: d = M/V
So I think the answer would be 5.67/ 835 I am not sure of the answer I got confused (´∀｀) but I hope it will help
A is the answer I really don’t know the answer to that but if u can u can help me on my work
