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adoni [48]
4 years ago
10

If a magnet could swing freely, what would happen to its north-seeking pole?

Physics
1 answer:
maksim [4K]4 years ago
8 0

You clearly identified the pole you're talking about as the
"north-seeking" pole.  Assuming your integrity and sincerity,
we would then naturally expect that pole to seek north, and
point to Earth's north magnetic pole. 

I'm confident in this answer also because I have several of
these devices hanging from the ceiling of my office, and I can
attest to the fact that on most clear days, they do in fact point
toward Earth's north magnetic pole.

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Answer:

82780.42123 m/s

14.45 days

Explanation:

m = Mass of the planet

M = Mass of the star = 0.85\times 1.989\times 10^{30}\ kg=1.69065\times 10^{30}\ kg

r = Radius of orbit of planet = 0.11\times 149.6\times 10^{9}\ m=16.456\times 10^{9}\ m

v = Orbital speed

The kinetic and potential energy balance is given by

\frac{GMm}{r^2}=\frac{mv^2}{r}\\\Rightarrow v=\sqrt{\dfrac{Gm}{r}}\\\Rightarrow v=\sqrt{\dfrac{6.67\times 10^{-11}\times 1.69065\times 10^{30}}{16.456\times 10^{9}}}\\\Rightarrow v=82780.42123\ m/s

The orbital speed of the star is 82780.42123 m/s

The orbital period is given by

t=\frac{2\pi r}{v}\\\Rightarrow t=\dfrac{2\pi \times 16.456\times 10^{9}}{82780.42123}\\\Rightarrow t=1249040.48419\ seconds=\dfrac{1249040.48419}{24\times 60\times 60}=14.45\ days  

The orbital period is 14.45 days

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Answer:

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