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Fudgin [204]
3 years ago
6

An astronaut stands on the surface of an asteroid. The astronaut then jumps such that the astronaut is no longer in contact with

the surface. The astronaut falls back down to the surface after a short time interval. Which of the following forces CANNOT be neglected when analyzing the motion of the astronaut?

Physics
1 answer:
Anna71 [15]3 years ago
6 0

(D) The gravitational force between the astronaut and the asteroid.

Reason :

All the other forces given in the options, except (D), doesn't account for the motion of the astronaut. They are the forces that act between nucleons or atoms and neither of them accounts for an objects motion.

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Molodets [167]

The work-energy theorem says that the total work done on the block is equal to the difference of its kinetic energies at points B and A. Then the total work done on the block is

W_{\rm total} = K_B - K_A = 4.0\,\mathrm J - 5.0\,\mathrm J = -1.0\,\mathrm J

Friction acts on the block to oppose its motion, so it does negative work on the block, -4.5 J.

The only other force acting on the block as it moves is the force <em>P</em>. Let W_P be the work done by the force <em>P</em>. Then the total work done on the block is

W_{\rm total} = W_P + W_{\rm friction} \iff -1.0\,\mathrm J = W_P - 4.5 \,\mathrm J \implies W_P = \boxed{3.5\,\mathrm J}

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3 years ago
It is proposed that future space stations create an artificial gravity by rotating. Suppose a space station is constructed as a
Archy [21]

To solve this problem, apply the concepts related to the centripetal acceleration as the equivalent of gravity, and the kinematic equations of linear motion that will relate the speed, the distance traveled and the period of the body to meet the needs given in the problem. Centripetal acceleration is defined as,

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

Here,

v = Tangential Velocity

r = Radius

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v = \sqrt{a_c r}

But at this case the centripetal acceleration must be equal to the gravitational at the Earth, then

v = \sqrt{gr}

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The perimeter of the cylinder would be given by,

\phi = 2\pi r

\phi = 2\pi (500m)

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Therefore now related by kinematic equations of linear motion the speed with the distance traveled and the time we will have to

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t = 44.9s

Therefore the period will be 44.9s

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