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Stolb23 [73]
4 years ago
14

A science-fiction tale describes an artificial "planet" in the form of a band completely encircling a sun. The inhabitants live

on the inside surface (where it is always noon). Imagine that this sun is exactly like our own, that the distance to the band is the same as the Earth-Sun distance (to make the climate temperate), and that the ring rotates quickly enough to produce an apparent gravity of g as on Earth. What will be the period of revolution, this planet's year, in Earth days?
Physics
1 answer:
pickupchik [31]4 years ago
4 0

Answer:

Refer below for the explanation.

Explanation:

With the Sun sparkling on that band without pause and expecting that an environment like Earth's could be kept up (which it can't), that temperature would soar and before long become dreadful.

So the main issue here is to discover the period T that yields

=2*pie*squareroot(93E6*5280/32.2)

= 775908.2472 = 77.6E4 seconds or 1.283 weeks or around 9 days.

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Okay please help me all my assignments are due tomorrow! I need to know 5 examples of all 6 simple machines that can be found in
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Review. (c) Assume the dipole is a compass needle-a light bar magnet-with a magnetic moment of magnitude μ . It has moment of in
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The frequency of oscillation is \frac{1}{2\pi }  \sqrt{\frac{\mu B}{I} }.

<h3>What is a magnetic moment?</h3>

The magnetic moment is the magnetism of a magnet or other item that creates a magnetic field, as well as its orientation and strength. Electromagnets, permanent magnets, elementary particles like electrons, different compounds, and a variety of celestial objects are examples of things that have magnetic moments (such as many planets, some moons, stars, etc). The phrase "magnetic moment" typically refers to a system's magnetic dipole moment, which can be represented by an analogous magnetic dipole, which has a magnetic north and south pole that are barely separated from one another. For sufficiently small magnets or at sufficiently wide distances, the magnetic dipole component is adequate. For extended objects, additional terms may be required in addition to the dipole moment, such as the magnetic quadrupole moment.

T = \frac{ Id^{2}\theta }{dt^{2} }

-\mu B\theta=\frac{ Id^{2}\theta }{dt^{2} }

\frac{d^{2}\theta }{dt^{2} } = -(\frac{\mu BI}{\theta} )

By Comparing the above equation with the SHM equation

\frac{d^2 \theta} {dt^{2} } = -\omega^{2} \theta

\omega^{2} =\frac{ \mu B}{I}

Frequency = \frac{\mu}{2\pi }

=\frac{\sqrt{\frac{\mu B}{I} } }{2\pi}

=\frac{1}{2\pi }  \sqrt{\frac{\mu B}{I} }

To learn more about a magnetic moment, visit:

brainly.com/question/17000031

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