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3 years ago

PLEASE HELP!!! BRAINLIEST ANSWER OFFERED!!!

Physics

1 answer:

Setler [38]3 years ago

3 0

A would be the correct answer.

If you look at radio waves of an electromagnetic spectrum, one can see the frequency is lower and the length is longer. They have less energy then the higher end of the spectrum.

Therefore, the answer is A!

Hope this helps!! Good luck!!

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you would multiply 30 by 15. because its the weight times the distance.

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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

aleksklad [387]

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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Answer:

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Explanation:

Velocity:40 km/h 2

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hypotenuse * Sin(n°) = Vy
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Vx:
Cos(n°) = Vx/hypotenuse
Hypotenuse * cos(n°) = Vx
100N*cos(30°) =Vx
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Answer:

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