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lubasha [3.4K]
1 year ago
6

An astronaut leaves a spaceship that is moving through free space. Will the spaceship move off and leave the astronaut behind?

Physics
1 answer:
Paraphin [41]1 year ago
9 0

When an astronaut leaves a spaceship that is moving through free space, the spaceship does not move off and leave the astronaut behind.

This is because when the astronaut is inside the spaceship, which is moving in free space, the astronaut also has the same velocity as the spaceship. Hence, when he leaves the spaceship, there is no relative velocity between the spaceship and the astronaut. This is the reason why the spaceship will not move off and leave the astronaut behind.

A spaceship or a spacecraft is an especially designed vehicle that is used to travel to outer space. It can be used for space exploration, transportation of humans and cargo, and meteorological purposes.

An astronaut is a person who has undergone specific training to travel to space. The duties of an astronaut aboard a spaceship can vary. Usually, a pilot and a commander work together to lead the mission. Other positions could be science pilot, mission expert, payload commander, and flight engineer.

Refer to more about astronaut here

brainly.com/question/11244838

#SPJ4

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A bowling ball is dropped off the top of the Eiffel Tower. If the Eiffel Tower is 300 meters
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7.82 s

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Read 2 more answers
The force of magnitude F acts along the edge of the triangular plate. Determine the moment of F about point O. Find the general
Sever21 [200]

Answer:

The moment is 81.102 k N-m in clockwise.

Explanation:

Given that,

Force = 260 N

Side = 580 mm

Distance h = 370 mm

According to figure,

Position of each point

O=(0,0)

A=(0,-b)

B=(h,0)

We need to calculate the position vector of AB

\bar{AB}=(h-0)i+(0-(-b))j

\bar{AB}=hi+bj

We need to calculate the unit vector along AB

u_{AB}=\dfrac{\bar{AB}}{|\bar{AB}|}

u_{AB}=\dfrac{h\hat{i}+b\hat{j}}{\sqrt{h^2+b^2}}

We need to calculate the force acting along the edge

\hat{F}=F(u_{AB})

\hat{F}=F(\dfrac{h\hat{i}+b\hat{j}}{\sqrt{h^2+b^2}})

We need to calculate the net moment

\hat{M}=\hat{F}\times OA

Put the value into the formula

\hat{M}=F(\dfrac{h\hat{i}+b\hat{j}}{\sqrt{h^2+b^2}})\times(-b\hat{j})

\hat{M}=\dfrac{F}{\sqrt{h^2+b^2}}((h\hat{i}+b\hat{j})\times(-b\hat{j}))

\hat{M}=\dfrac{F}{\sqrt{h^2+b^2}}(-bh\hat{k})

\hat{M}=-\dfrac{bhF}{\sqrt{h^2+b^2}}

Put the value into the formula

\hat{M}=-\dfrac{580\times10^{-3}\times370\times10^{-3}\times260}{\sqrt{(370\times10^{-3})^2+(580\times10^{-3})^2}}

\hat{M}=-81.102\ \hat{k}\ N-m

Negative sign shows the moment is in clockwise.

Hence, The moment is 81.102 k N-m in clockwise.

5 0
3 years ago
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