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3 years ago

A proposed space station includes living quarters in a circular ring 45.0 m in diameter.

Physics

1 answer:

siniylev [52]3 years ago

6 0

Answer:

Angular speed, $\omega=0.65\ rad/s$

Explanation:

Given that,

Diameter of the circular ring, D = 45 m

Radius, r = 22.5 m

Let $\omega$ is the angular speed should the ring rotate so the occupants feel that they have the same weight as they do on Earth. It can be given by providing the centripetal force to the normal force as :

$mg=m\omega^2r$

$\omega=\sqrt{\dfrac{g}{r}}$

$\omega=\sqrt{\dfrac{9.8\ m/s^2}{22.5\ m}}$

$\omega=0.65\ rad/s$

So, the angular speed of the ring is 0.65 rad/s. Hence, this is the required solution.

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Suppose the original segment of wire is stretched to 10 times its original length. How much charge must be added to the wire to

Debora [2.8K]

Here we want to study how the linear charge density changes as we change the measures of our body.

We will find that we need to add 9*Q of charge to keep the linear charge density unchanged.

I will take two assumptions:

The charge is homogeneous, so the density is constant all along the wire.

As we work with a linear charge density we work in one dimension, so the wire "has no radius"

Originally, the wire has a charge Q and a length L.

The linear charge density will be given by:

λ = Q/L

Now the length of the wire is stretched to 10 times the original length, so we have:

L' = 10*L

We want to find the value of Q' such that λ' (the linear density of the stretched wire) is still equal to λ.

Then we will have:

λ' = Q'/L' = Q'/(10*L) = λ = Q/L

Q'/(10*L) = Q/L

Q'/10 = Q

Q' = 10*Q

So the new charge must be 10 times the original charge, this means that we need to add 9*Q of charge to keep the linear charge density unchanged.

If you want to learn more, you can read:

brainly.com/question/14514975

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2 years ago

Zeros are never significant digits true or false

liraira [26]

This affirmative is false

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3 years ago

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inessss [21]

Here are the answers to the question. Make sure to give a valid reason, please.

A. the sum of the protons and neutrons in one atom of the element.

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3 years ago

A 1.90-m-long barbell has a 25.0 kg weight on its left end and a 37.0 kg weight on its right end. if you ignore the weight of th

gregori [183]

Answer: The center of gravity is 1.1338 m away from the left side of the barbell

Explanation:

Length of the barbell = 1.90 m

The distance center of gravity from left = x

Mass on the left side = 25 kg

The distance center of gravity from right = 1.90 - x

Mass on the right side = 37 kg

At the balance point: $m_1x_1=m_2x_2$

$25 kg\times x=37 kg\times (1.90-x)$

$x=1.1338 m$

The center of gravity is 1.1338 m away from the left side of the barbell

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Kisachek [45]

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3 years ago