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AleksAgata [21]
3 years ago
14

An electron at Earth's surface experiences a gravitational force meg. How far away can a proton be and still produce the same fo

rce on the electron?

Physics
2 answers:
kondaur [170]3 years ago
7 0
I will discuss what is a gravitational force since no figures are attached or given. An objects weight is dependent upon its location in the universe because they exhibit gravitational waves. For example, the earth is a massive planet. Because of its massiveness, it exhibits a strong gravitational force within it. In turn, the objects near the earth will be attracted to it and thereby feels a much stronger gravity on earth. That is why bodies of water, despite its liquid features, stick to the earth. The heavier the body is, the stronger its gravitational pull. Another example is the Milky Way Galaxy, there is a gravitational pull because it is to other galaxies. Also, other galaxies are heavier than the earth and therefore, it is attracted to the Milky Way galaxy because of its gravitational pull. 

Nady [450]3 years ago
3 0

The proton could be 5 m far away from electron.

<h3>Further explanation</h3>

Newton's gravitational law states that the force of attraction between two objects can be formulated as follows:

\large {\boxed {F = G \frac{m_1 ~ m_2}{R^2}} }

<em>F = Gravitational Force ( Newton )</em>

<em>G = Gravitational Constant ( 6.67 × 10⁻¹¹ Nm² / kg² )</em>

<em>m = Object's Mass ( kg )</em>

<em>R = Distance Between Objects ( m )</em>

Let us now tackle the problem!

<u />

<u>Given:</u>

me = 9.11 × 10⁻³¹ kg

qp = qe = 1.6 × 10⁻¹⁹ kg

<u>Unknown:</u>

R = ?

<u>Solution:</u>

F_e = F_p

m_e \times g = k \times \frac{q_e \times q_p}{R^2}

9.11 \times 10^{-31} \times 9.81 = 9 \times 10^9 \times \frac{(1.6 \times 10^{-19})^2}{R^2}

R \approx 5 ~ m

<h3>Learn more</h3>
  • Impacts of Gravity : brainly.com/question/5330244
  • Effect of Earth’s Gravity on Objects : brainly.com/question/8844454
  • The Acceleration Due To Gravity : brainly.com/question/4189441

<h3>Answer details</h3>

Grade: High School

Subject: Physics

Chapter: Gravitational Field

Keywords: Gravity , Unit , Magnitude , Attraction , Distance , Mass , Newton , Law , Gravitational , Constant

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When a stone is in free fall, it experiences a gravitational acceleration. This acceleration from gravity, though, only affects the vertical velocity. Since gravity is vertical as well, it's essentially impossible for the horizontal velocity to be changed at all.

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Since we rounded the time, it makes sense that our final answer's a little bit off to the options. The closest one is option B, which is only 0.6m off, a tiny difference that may have come from the test maker's use of '10 m/s^{2}' as the gravitational acceleration (while we stayed as accurate as possible with 9.80) as well as our rounding of the final time.

Option B, the stone will have travelled 47.85 meters.

If you have any questions on how I got to the answer or if you're still confused on any topic I attempted to explain, just ask in the comments and I'll try to answer it to the best of my ability! Good luck!

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