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Deffense [45]
2 years ago
6

Astronauts are weightless when in orbit in space. are they also weightless during the launch of the station? How about during th

eir return to earth? explain
Physics
1 answer:
mel-nik [20]2 years ago
7 0

<u>Astronauts are not weightless during either launch or return to Earth.</u>

<u></u>

<h3>Brief explanation</h3>

Astronauts become weightless when they stop accelerating. Basically that means when the engines cut out and they begin to coast in orbit. They will remain “weightless” for as long as they are in orbit. To get out of orbit, they have to decelerate (i.e. Accelerate in the opposite direction) and so they begin to feel a force that feels very much like gravity as they are falling back to Earth.

One of the cool things is that you can't tell the difference between gravity and acceleration. They're the same thing as far as your body is concerned. Einstein used a variety of somewhat related thought experiments while he has working out the details of the special theory of relativity.

Hence, with this explanation , we can conclude that astronauts are not weightless during either launch or return to Earth.

Learn more about astronauts being weightless

brainly.com/question/2231528

#SPJ4

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A banked circular highway curve is designed for traffic moving at 58 km/h. The radius of the curve is 201 m. Traffic is moving a
skelet666 [1.2K]

Answer:0.077

Explanation:

Given

banked designed for traffic moving at 58 km/h\approx 16.11 m/s

Radius of the curve 201 m

Actual traffic velocity =37 km/h approx  10.27 m/s

For banking of road

tan\theta =\frac{v^2}{rg}

tan\theta =\frac{16.11^2}{201\times 9.81}

\theta =7.49^{\circ}

Centripetal acceleration is given by

a=\frac{v^2}{r}

Taking component of centripetal acceleration

along and perpendicular to surface

a_{parallel}=\frac{v^2cos\theta }{r}

a_{perpendicular}=\frac{v^2sin\theta }{r}

From FBD

mgsin\theta -f_s=ma_{parallel}

f_s=mgsin\theta -ma_{parallel}----1

where f_s is frictional force

N-mgcos\theta =ma_{perpedicular}

N=mgcos\theta +ma_{perpedicular}----2

and we know coefficient of friction is given by

\mu =\frac{f_s}{N}

\mu =\frac{mgsin\theta -ma_{parallel}}{mgcos\theta +ma_{perpedicular}}

\mu =\frac{gsin\theta -\frac{v^2cos\theta }{r}}{gcos\theta +\frac{v^2sin\theta }{r}}

\mu =\frac{1.2804-0.5202}{9.726+0.068}

\mu =0.077

7 0
3 years ago
THE WINNER WILL BE AWARDED THE BRAINLIEST PLEASE HELPP
Ivenika [448]

Answer:Correct answer: 15.85 kg·m/s

Explanation:

A 30 kg gun is standing on a frictionless sur-face. The gun fires a 50 g bullet with a muzzlevelocity of 317 m/s.The positive direction is that of the bullet.Calculate the momentum of the bullet im-mediately after the gun was fired.

7 0
3 years ago
A crane lifts up two boxes.<br><br> Which free body diagram shows the forces acting on block A?
Temka [501]

Answer:

The 3rd graph

Explanation:

A free body diagram is a diagram which shows all the forces acting on an object.

The problem asks us to find the free body diagram of block A, so we must find all the forces acting on block A.

We have 3 forces acting on block A in total:

- The force of gravity (its weight), which pushes the block downward (in the diagram, it is the force represented with F_{gA}

- The tension in the rope 1, which pulls block A upwards: this force is represented with F_{T1}

- The tension in the rope 2, due to the weight of block 2, which pulls block A downwards: this force is represented with F_{T2}

Based on the direction of these 3 forces, the correct diagram is the 3rd one.

7 0
3 years ago
Read 2 more answers
A wire as a length of 1.50m, diameter 0.60mm and resistance of 2ohms. calculate the resistance R of a wire of the same materials
Otrada [13]
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ρ=R×A/L

A=pie(r²)=pie(0.6²)=pie(0.36)=1.13

ρ=2ohm×1.13mm²/1500mm

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R=ρ×L/A

R=0.0015ohm.mm×500mm/0.09mm²

R=8.3'ohm

so our resistance for the second wire is :

<em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em>R</em><em>=</em><em>8</em><em>.</em><em>3</em><em>'</em><em>ohm</em>

3 0
3 years ago
The device that demonstrates the presence of static electricity is called a(n) _____.
galben [10]
<span>The device that demonstrates the presence of static electricity
is called a electroscope .</span>
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3 years ago
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