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podryga [215]
3 years ago
11

1.

Physics
2 answers:
Firdavs [7]3 years ago
4 0

1). The forces inside the atom are always, totally, completely, electrostatic forces. Those are so awesomely stronger than the gravitational forces that the gravitational ones are totally ignored, and it doesn't change a thing.

Parts 2 and 3 of this question are here to show us how the forces compare.

Part-2). The electrostatic force between a proton and an electron.

The constant in the formula is 9x10^9, and the elementary charge is 1.602 x 10^-19 Coulomb ... same charge on both particles, but opposite signs.

I worked through it 3 times and got 0.000105 N every time. So the best choice is 'C', even though we disagree by a factor of ten times. You'll see in part-3 that it really doesn't make any difference.

Part-3). Gravitational force between a proton and an electron.

The constant in Newton's gravity formula is 6.67x10^-11 . You'll have to look up the masses of the proton and the electron.

I got 2.163 x 10^-55 N ... exactly choice-C. yay !

Now, after we've slaved over a hot calculator all night, the thing that really amazes us is not only that the electrostatic force is stronger than the gravitational force, but HOW MUCH stronger ... 10^51 TIMES stronger. That's a thousand trillion trillion trillion trillion times stronger !

That's why it has no effect on the measurements if we just forget all about the gravitational forces inside the atom.

bearhunter [10]3 years ago
4 0

Answer:

The correct answer is Fel > Fg

Explanation:

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lesya [120]

Answer:

Sound Intensity at microphone's position is 9.417\times 10^{- 4} W/m^{2}

The amount of energy impinging on the microphone is 9.417\times 10^{- 8} W/m^{2}

Solution:

As per the question:

Emitted Sound Power, P_{E} = 32.0 W

Area of the microphone, A_{m} = 1.00 cm^{2} = 1.00\times 10^{- 4} m^{2}

Distance of microphone from the speaker, d = 52.0 m

Now, the intensity of sound, I_{s} at a distance away from the souce of sound follows law of inverse square and is given as:

I_{s} = \frac{P_{E}}{Area} = \frac{P_{E}}{4\pi d^{2}}

I_{s} = \frac{32.0}{4\pi (52.0)^{2}} = 9.417\times 10^{- 4} W/m^{2}

Now, the amount of sound energy impinging on the microphone is calculated as:

If I_{s} be the Incident Energy/m^{2}/s

Then

The amount of energy incident per 1.00 cm^{2} = 1.00\times 10^{- 4} m^{2} is:

I_{s}(1.00\times 10^{- 4}) = 9.417\times 10^{- 4}\times 1.00\times 10^{- 4} = 9.417\times 10^{- 8} J

7 0
3 years ago
A hollow spherical planet is inhabited by people who live inside it, where the gravitational is zero. When a very massive space
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Answer:

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Answer:

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