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andreev551 [17]
3 years ago
11

Since all objects are ‘weightless’ for an astronaut in orbit, is it possible for astronauts to tell whether an object is heavy o

r light? Think about it, think of newton’s laws, the equations and then explain your reasoning.
Physics
2 answers:
ivanzaharov [21]3 years ago
4 0
W=gm
where g - gravitation 
m - mass
w - weight
as gravitation equals to zero, multiplying by 0 gives W=0
It is not possible to tell whether and object is heavy or light 

ValentinkaMS [17]3 years ago
4 0
Well, while they're in orbit, nothing is 'heavy' and nothing is 'light'.
They're all weightless.

I think you're really asking:  Can the astronaut tell which objects will be
heavy and which ones will be light if they go back down to Earth ?

The answer is a resounding 'Sure' !

The weight of an object on Earth is (its mass) x (acceleration of gravity).
So the objects with a lot of mass will be heavy, and the objects with less
mass will be lighter.

How can the astronaut determine the mass of the object hanging there
in the air in front of him ? 
All he has to do is give it a push.  It'll accelerate away from him (and he
will accelerate away from the object). 
Newton's 2nd law tells us that  F=mA, so the acceleration will be

                             (the force he exerts) / (the mass).

An object with small mass will zip away from him, and it'll be light
down on the surface. 

An object with large mass will accelerate slightly, start moving slowly,
and it'll be heavy down on the surface.
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What is the potential energy of a 5kg rock that is 7m high on a hill
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Explanation:

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Atoms contain empty space true or false​
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If a mass of .343 kg moves down a 5m ramp in 5.8 seconds, what is the velocity and KE developed by the moving mass
Soloha48 [4]

Velocity 0.86m/s

0.13J

Explanation:

Given parameters:

Mass = 0.343kg

distance = 5m

time taken = 5.8s

Unknown:

Velocity of mass = ?

Kinetic energy = ?

Solution:

Velocity is the rate of change of displacement with time. It is a vector quantity that shows magnitude and direction.

 Mathematically;

   Velocity = \frac{displacement}{time}

  Velocity  = \frac{5}{5.8 } = 0.86m/s

Kinetic energy is the energy due to the motion of a body. It is expressed mathematically as:

  Kinetic energy  = \frac{1}{2}  m v^{2}

m is the mass

v is the velocity

  Kinetic energy  = \frac{1}{2 }   x    0.343  x    0.86^{2} = 0.13J

learn more:

Kinetic energy brainly.com/question/6536722

#learnwithBrainly

7 0
3 years ago
Earth has a magnetic dipole moment of 8.0 × 1022 J/T. (a) What current would have to be produced in a single turn of wire extend
Lelechka [254]

Answer:

Answer is

A. I = 6.3×10^8 A

B. Yes

C. No

Refer below.

Explanation:

Refer to the picture for brief explanation.

7 0
3 years ago
A child throws a ball with an initial speed of 8.00 m/s at an angle of 40.0° above the horizontal. The ball leaves her hand 1.00
vivado [14]

Answer:

7.5 m

Explanation:

v = initial speed of the ball = 8 m/s

\theta = angle of launch = 40° deg

Consider the motion along the vertical direction :

v_{oy} = initial velocity along vertical direction = v Sin\theta = 8 Sin40 = 5.14 m/s

a_{y} = acceleration along vertical direction = - 9.8 m/s²

t = time of travel

y = vertical displacement = - 1 m

Using the kinematics equation

y = v_{oy}t + (0.5)a_{y}t^{2}

- 1 = (5.14)t + (0.5)(- 9.8)t^{2

t = 1.22 sec

Consider the motion along the horizontal direction :

v_{ox} = initial velocity along horizontal direction = v Sin\theta = 8 Cos40 = 6.13 m/s

a_{x} = acceleration along vertical direction = 0 m/s²

t = time of travel = 1.22 sec

x = horizontal displacement = ?

Using the kinematics equation

x = v_{ox}t + (0.5)a_{x}t^{2}

x = (6.13)(1.22) + (0.5)(0)(1.22)^{2

x = 7.5 m

4 0
3 years ago
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