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ruslelena [56]
4 years ago
5

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
Physics
1 answer:
arlik [135]4 years ago
6 0

Answer:

8.97 days

Explanation:

Assuming we are asked what will be the period of revolution, this planet's year, in Earth days?

The period of revolution T is given as

T=2\pi\sqrt{\frac{r}{g} }

here, r= distance between planet and its sun which is same as distance between Earth-Sun = 149.7\times10^9 m

g= acceleration due to gravity = 9.81 m/s^2

therefore,

T=2\pi\sqrt{\frac{149.7\times10^9}{9.81} }

=775580 seconds

= 8.97 days

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To solve this problem it is necessary to apply the concepts related to Young's Module and its respective mathematical and modular definitions. In other words, Young's Module can be expressed as

\Upsilon = \frac{F/A}{\Delta L/L_0}

Where,

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A = Area

\Delta L= Compression

L_0= Original Length

According to the values given we have to

\Upsilon_{steel} = 200*10^9Pa

\Delta L = 5.6*10^{-7}m

L_0 = 3.2m

r= 0.59m \rightarrow A = \pi r^2 = \pi *0.59^2 = 1.0935m^2

Replacing this values at our previous equation we have,

\Upsilon = \frac{F/A}{\Delta L/L_0}

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3 years ago
What do you mean by velocity ratio of a wheel and axle​
IgorC [24]

Answer:

Explanation:

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What net external force is exerted on a 1100-kg artillery shell fired from a battleship if the shell is accelerated at 2.40×104
pickupchik [31]

Answer:

Force exerted, F = 2.64 × 10⁷ Newton

Explanation:

It is given that,

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We need to find the magnitude of force exerted on the ship by the artillery shell. It can be determined using Newton's second law of motion :

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F = 2.64 × 10⁷ Newton

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6 0
4 years ago
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3 years ago
An open pipe is 1.42 m long
lora16 [44]

Answer:

the fundamental frequency produced by the open pipe is 120.78 Hz

Explanation:

Given;

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The length of the open pipe for the fundamental frequency is equivalent to half of wavelength;

L = \frac{\lambda}{2} \\\\\lambda = 2L

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Therefore, the fundamental frequency produced by the open pipe is 120.78 Hz

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