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2 years ago

When in orbit, astronauts experience weightlessness. What is this caused by?

Physics

1 answer:

Ulleksa [173]2 years ago

7 0

Answer:

Earth-orbiting astronauts are weightless for the same reasons that riders of a free-falling amusement park ride or a free-falling elevator are weightless. They are weightless because there is no external contact force pushing or pulling upon their body. In each case, gravity is the only force acting upon their body.

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A 10-kg disk-shaped flywheel of radius 9.0 cm rotates with a rotational speed of 320 rad/s. Part A Determine the rotational mome

Leya [2.2K]

Answer:

(A). The rotational momentum of the flywheel is 12.96 kg m²/s.

(B). The rotational speed of sphere is 400 rad/s.

Explanation:

Given that,

Mass of disk = 10 kg

Radius = 9.0 cm

Rotational speed = 320 m/s

(A). We need to calculate the rotational momentum of the flywheel.

Using formula of momentum

$L=I\omega$

$L=\dfrac{1}{2}mr^2\omega$

Put the value into the formula

$L=\dfrac{1}{2}\times10\times(9.0\times10^{-2})^2\times320$

$L=12.96\ kg m^2/s$

(B). Rotation momentum of sphere is same rotational momentum of the flywheel

We need to calculate the magnitude of the rotational speed of sphere

Using formula of rotational momentum

$L_{sphere}=L_{flywheel}$

$I\omega_{sphere}=I\omega_{flywheel}$

$\omega_{sphere}=\dfrac{I\omega_{flywheel}}{I_{sphere}}$

$\omega_{sphere}=\dfrac{I\omega_{flywheel}}{\dfrac{2}{5}mr^2}$

Put the value into the formula

$\omega_{sphere}=\dfrac{12.96}{\dfrac{2}{5}\times10\times(9.0\times10^{-2})^2}$

$\omega_{sphere}=400\ rad/s$

Hence, (A). The rotational momentum of the flywheel is 12.96 kg m²/s.

(B). The rotational speed of sphere is 400 rad/s.

4 0

3 years ago

Please answer this question

belka [17]

ANSWER is c

HOPR THIS WILL BE HELPFUL

8 0

3 years ago

Margy is trying to improve her cardio endurance by performing an exercise in which she alternates walking and running 100.0 m ea

madreJ [45]

In order to answer this exercise you need to use the formulas

S = Vo*t + (1/2)*a*t^2

Vf = Vo + at

The data will be given as

Vf = final velocity = ?

Vo = initial velocity = 1.4 m/s

a = acceleration = 0.20 m/s^2

s = displacement = 100m

And now you do the following:

100 = 1.4t + (1/2)*0.2*t^2

t = 25.388s

and

Vf = 1.4 + 0.2(25.388)

Vf = 6.5 m/s

So the answer you are looking for is 6.5 m/s

7 0

3 years ago

Read 2 more answers

Calculate the height from which a body is released from rest if its velocity just before hitting the ground is 30ms

nexus9112 [7]