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kow [346]
3 years ago
9

A 56 kg astronaut stands on a bathroom scale inside a rotating circular space station. The radius of the space station is 250 m.

The bathroom scale reads 42 kg. At what speed does the space station floor rotate?
Physics
1 answer:
Zielflug [23.3K]3 years ago
6 0

Answer:

The speed of space station floor is 49.49 m/s.

Explanation:

Given that,

Mass of astronaut = 56 kg

Radius = 250 m

We need to calculate the speed of space station floor

Using centripetal force and newton's second law

F=mg

\dfrac{mv^2}{r}=mg

\dfrac{v^2}{r}=g

v=\sqrt{rg}

Where, v = speed of space station floor

r = radius

g = acceleration due to gravity

Put the value into the formula

v=\sqrt{250\times9.8}

v=49.49\ m/s

Hence, The speed of space station floor is 49.49 m/s.

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Two particles with masses m and 3m are moving toward each other along the x axis with the same initial speeds Vi. Particle m is
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Answer:

v1 = \sqrt{(2)}Vi

v2 = \sqrt{(2/3)}Vi

ANGLE is 35.3 degree celcius

Explanation:

Given data:

mass m and 3m

initial speed Vi

particle with mass m is moving toward left while particle with mass 3m is moving toward right

By using conservation of momentum :

mVi + 3m(-Vi) = mv1 +3mv2

-2mVi = m(v1 + 3v2)

-2Vi = v1 + 3v2

conservation of energy :

m(Vi^2) + 3m(-Vi^2) = mv1^2 + 3mv2^2

4mVi^2 = m(v1^2+3v2^2)

4Vi^2 = v1^2+3v2^2

After collision, particle with mass m moves at right angles, thus by considering conservation of momentum in x & y direction,

x direction : -2mVi = 3m.v2i

-2Vi = 3v2i

y direction : 0 = m(v1)j+3m(v2)j

-v1j = 3v2j

subsitute these value in energy conservation

v1 = \sqrt{(2)}Vi

v2 = \sqrt{(2/3)}Vi

angle = tan^{-1}(\frac{\sqrt{(2)}}{2}) =35.3 degree from x-axis

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The steps of the use of the laser technique is explained below:

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<h3>What is a Laser Technique?</h3>

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Hence, we can see that the laser technique is considered safer than conventional surgical methods.

Laser techniques include:

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Read more about laser technique here:

brainly.com/question/14091177

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