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2 years ago

Differences between weightlessness in space and weightlessness in earth

Physics

2 answers:

zhuklara [117]2 years ago

6 0

Answer:

it depends on a person's own weight

ExtremeBDS [4]2 years ago

3 0

Answer:

The gravity is roughly the same. The inverse square law applies to gravity and that means that being 60 miles above Earth is about the same as being on Earth. The thing is, they are going really fast in an orbit and thus falling around the Earth. The weightlessness comes from a sort of cancellation of gravity pulling them down and their going perpendicular to that force. The forces are cancelling and making them relatively “weightless”. It’s not really weightlessness but just an equilibrium reached between the two accelerations of gravity and their motion against the pull of the Earth.

Explanation:

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A _______ is a very massive object in space that does not allow any object or radiation to escape its gravitational hold.

marysya [2.9K]

Black hole, it sucks in pretty much everything in its path

5 0

3 years ago

Read 2 more answers

A load of mass 5kg is raised through a height of 2m. calculate the work done against (g=10mls)

Illusion [34]

The work done against gravity is 100 J

Explanation:

The work done against gravity in order to lift an object is equal to the change in gravitational potential energy of the object:

$W=mg\Delta h$

where

m is the mass of the object

g is the acceleration of gravity

$\Delta h$ is the change in height of the object

For the object in this problem, we have:

m = 5 kg

$g=10 m/s^2$

$\Delta h = 2 m$

Substituting into the equation,

$W=(5)(10)(2)=100 J$

Learn more about work:

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3 0

3 years ago

Vector ????⃗ has a magnitude of 16.6 and is at an angle of 50.5∘ counterclockwise from the +x‑axis. Vector ????⃗ has a magnitude

natka813 [3]

Answer:

For vector u, x component = 10.558 and y component =12.808

unit vector = 0.636 i+ 0.7716 j

For vector v, x component = 23.6316 and y component = -6.464

unit vector = 0.9645 i-0.2638 j

Explanation:

Let the vector u has magnitude 16.6

u makes an angle of 50.5° from x axis

So $u_x=ucos\Theta =16.6\times cos50.5=10.558$

Vertical component $u_y=usin\Theta =16.6\times sin50.5=12.808$

So vector u will be u = 10.558 i+12.808 j

Unit vector $u=\frac{10.558i+12.808j}{\sqrt{10.558^2+12.808^2}}=0.636i+0.7716j$

Now in second case let vector v has a magnitude of 24.5

Making an angle with -15.3° from x axis

So horizontal component $v_x=vcos\Theta =24.5\times cos(-15.3)=23.6316$

Vertical component $v_y=vsin\Theta =24.5\times sin(-15.3)=-6.464$

So vector v will be 23.6316 i - 6.464 j

Unit vector of v $=\frac{23.6316i-6.464}{\sqrt{23.6316^2+6.464^2}}=0.9645i-0.2638j$

8 0

3 years ago

A rock weighing 98 newtons is pushed off the edge of a bridge 50 meters above the ground. What was the potential energy of the r

lidiya [134]

Ep = 4900 because Ep = wh

4 0

3 years ago

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A projectile is launched from ground level at an angle of 30 degrees above the horizontal. Neglect air resistance and consider t

Oduvanchick [21]

Answer:

just before landing the ground

Explanation:

Let the velocity of projection is u and the angle of projection is 30°.

Let T is the time of flight and R is the horizontal distance traveled. As there is no force acting in horizontal direction, so the horizontal velocity remains constant. Let the particle hits the ground with velocity v.

initial horizontal component of velocity, ux = u Cos 30

initial vertical component of velocity, uy = u Sin 30

Time of flight is given by

$T = \frac{2u Sin\theta }{g}$