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xz_007 [3.2K]
3 years ago
14

The average distance of Enceladus from Saturn is 238,000 km; the average distance of Titan from Saturn is 1,222,000 km. How much

longer does it take Titan to orbit Saturn compared to Enceladus?
Physics
1 answer:
aleksklad [387]3 years ago
8 0

Answer:

Titan takes 11.634 times longer to orbit Saturn as compared to Enceladus.

Explanation:

We have been given that the average distance of Enceladus from Saturn is 238,000 km; the average distance of Titan from Saturn is 1,222,000 km.

We will use Kepler's Law to solve our given problem.

\frac{(T_1)^2}{(T_2)^2}=\frac{(r_1)^3}{(r_2)^3}

Upon substituting our given values, we will get:

\frac{(T_1)^2}{(T_2)^2}=\frac{(238,000)^3}{(1,222,000)^3}

\frac{(T_1)^2}{(T_2)^2}=\frac{13481272000000000}{1824793048000000000}

\frac{(T_1)^2}{(T_2)^2}=\frac{13481272}{1824793048}

\frac{(T_1)^2}{(T_2)^2}=0.0073878361246365

Taking square root of both sides, we will get:

\frac{T_1}{T_2}=\sqrt{0.0073878361246365}

\frac{T_1}{T_2}=0.0859525225030452495

\frac{T_2}{T_1}=\frac{1}{0.0859525225030452495}

\frac{T_2}{T_1}=11.634329870476699\approx 11.634

T_2=11.634\cdot T_1

This implies that time period of Titan about Sturn is 11.634 times more compared to time period of Enceladus about Saturn.

So, basically Titan takes 11.634 times longer to orbit Saturn as compared to Enceladus.

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The net force acting on the crate is determined as 176 N to the left.

<h3>Net force acting on the crate</h3>

The net force acting on the crate is calculated as follows;

∑F = F1 + F2 + F3 + F4

F(net) = -440y + 176x + 440y - 352x

F(net) = -176 x

The resultant force is pointing in negative x direction.

Thus, the net force acting on the crate is determined as 176 N to the left.

Learn more about net force here: brainly.com/question/14361879

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Answer:

Parkour is a type of sport or “Hobby”

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You can perform parkour by jumping over certain objects, leaping and turning, etc.

Hope this helped!

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3 years ago
How many moles are in 95 g of Na?
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In a heat engine if 1000 j of heat enters the system the piston does 500 j of work, what is the final internal energy of the sys
nydimaria [60]

Answer : The final energy of the system if the initial energy was 2000 J is, 3500 J

Solution :

(1) The equation used is,

\Delta U=q+w\\\\U_{final}-U_{initial}=q+w

where,

U_{final} = final internal energy

U_{initial} = initial internal energy

q = heat energy

w = work done

(2) The known variables are, q, w and U_{initial}

initial internal energy = U_{initial} = 2000 J

heat energy = q = 1000 J

work done = w = 500 J

(3) Now plug the numbers into the equation, we get

U_{final}-(2000J)=(1000J)+(500J)

(4) By solving the terms, we get

U_{final}-(2000J)=(1000J)+(500J)

U_{final}-(2000J)=1500J

U_{final}=2000J+1500J

U_{final}=3500J

(5) Therefore, the final energy of the system if the initial energy was 2000 J is, 3500 J

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3 years ago
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