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Pepsi [2]
3 years ago
15

What is a circumpolar star? A)a star that is close to the north celestial pole a star that is close to the south celestial pole

C) a star that always remains above your horizon and appears to rotate around the celestial pole D) a star that makes a daily circle around the celestial sphere E) a star that is visible from the Arctic or Antarctic circles
Physics
1 answer:
Vedmedyk [2.9K]3 years ago
6 0

Answer:

A star that always remains above your horizon and appears to rotate around the celestial pole.

Explanation:

A) a star that is close to the north celestial pole: a circumpolar star could be close to the north celestial pole, but this answer is omitting the south celestial pole.

B) a star that is close to the south celestial pole: a circumpolar star could be close to the south celestial pole, but this answer is omitting the north celestial pole.

C) a star that always remains above your horizon and appears to rotate around the celestial pole: this is the definition of a circumpolar star.

D) a star that makes a daily circle around the celestial sphere: every star does this.

E) a star that is visible from the Arctic or Antarctic circles : there are many starts visible from there that are not circumpolar.

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1 year ago
The 480 g bar is rotating as shown what is the angular momentum of the bar about the axle?
Greeley [361]
On a similar problem wherein instead of 480 g, a 650 gram of bar is used:

Angular momentum L = Iω, where 
<span>I = the moment of inertia about the axis of rotation, which for a long thin uniform rod rotating about its center as depicted in the diagram would be 1/12mℓ², where m is the mass of the rod and ℓ is its length. The mass of this particular rod is not given but the length of 2 meters is. The moment of inertia is therefore </span>
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<span>The angular momentum ω = 2πf, where f is the frequency of rotation. If the angular momentum is to be in SI units, this frequency must be in revolutions per second. 120 rpm is 2 rev/s, so </span>
<span>ω = 2π * 2 rev/s = 4π s^(-1) </span>

<span>The angular momentum would therefore be </span>
<span>L = Iω </span>
<span>= 1/3m * 4π </span>
<span>= 4/3πm kg*m²/s, where m is the rod's mass in kg. </span>

<span>The direction of the angular momentum vector - pseudovector, actually - would be straight out of the diagram toward the viewer. </span>

<span>Edit: 650 g = 0.650 kg, so </span>
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