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AveGali [126]
4 years ago
6

An astronaut on a spacewalk accidentally drops a tool, and it floats away.Which of the following objects is exerting a gravitati

onal force on the floating tool? Choose all that apply
the astronaut
the Moon
the Earth
the Sun
Physics
2 answers:
castortr0y [4]4 years ago
4 0

Answer:

All of them are correct

Explanation:

They are all correct select all

a_sh-v [17]4 years ago
3 0

Gravitational force is given as

F = \frac{Gm_1m_2}{r^2}

so here gravitational force will be given by above formula where the force between two objects depends on the mass of two objects and distance between them.

Now here since the tool is at finite distance from Astronaut and Moon, Sun, Earth

So all the above will exert gravitational force on the tool but the magnitude of force will be different as all of the above are of different masses and situated at different distance.

So all options are correct here

The Astronaut

The Moon

The Sun

The Earth

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A piano wire with mass 2.95 g and length 79.0 cm is stretched with a tension of 29.0 N . A wave with frequency 105 Hz and amplit
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The concept needed to solve this problem is average power dissipated by a wave on a string. This expression ca be defined as

P = \frac{1}{2} \mu \omega^2 A^2 v

Here,

\mu = Linear mass density of the string

\omega =  Angular frequency of the wave on the string

A = Amplitude of the wave

v = Speed of the wave

At the same time each of this terms have its own definition, i.e,

v = \sqrt{\frac{T}{\mu}} \rightarrow Here T is the Period

For the linear mass density we have that

\mu = \frac{m}{l}

And the angular frequency can be written as

\omega = 2\pi f

Replacing this terms and the first equation we have that

P = \frac{1}{2} (\frac{m}{l})(2\pi f)^2 A^2(\sqrt{\frac{T}{\mu}})

P = \frac{1}{2} (\frac{m}{l})(2\pi f)^2 A^2 (\sqrt{\frac{T}{m/l}})

P = 2\pi^2 f^2A^2(\sqrt{T(m/l)})

PART A ) Replacing our values here we have that

P = 2\pi^2 (105)^2(1.8*10^{-3})^2(\sqrt{(29.0)(2.95*10^{-3}/0.79)})

P = 0.2320W

PART B) The new amplitude A' that is half ot the wavelength of the wave is

A' = \frac{1.8*10^{-3}}{2}

A' = 0.9*10^{-3}

Replacing at the equation of power we have that

P = 2\pi^2 (105)^2(0.9*10^{-3})^2(\sqrt{(29.0)(2.95*10^{-3}/0.79)})

P = 0.058W

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