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Gennadij [26K]
3 years ago
15

What happens to the force between charged and uncharged objects as the distance between them decreases?

Physics
1 answer:
MaRussiya [10]3 years ago
4 0

Answer:Thus, if each of the charges were reduced by one-half, the repulsion would be reduced to one-quarter of its former value. Also In electrostatics, the electrical force between two charged objects is inversely related to the distance of separation between the two objects. ... And decreasing the separation distance between objects increases the force of attraction or repulsion between the objects.

Hope this helps have a awesome night/day❤️✨

Explanation:

Nothing, until those two objects physically touch each other; contact aligns polarity among the now single shared mass.

(Your question never states if both objects are unique or similar polar charges, so I just assumed they were both neutral objects existing within an electric field.)

So a better question would then be, what is gravity’s relationship with an electric field?

You could solve this with the following: confine the electric field’s volume to a set variable (never increasing nor decreasing in size or shape); density is variable and easily definable. This creates the limit to build upon. This density has to be fluid and has electron mass (full of electrons at rest mass, so with substance but no movement). Within, create a closed system (the hard part in this equation; outside interference like ambient light will eschew results) where each variable of kinetic energy then is accounted for or measurable (including heat and light, and the physical movement of the two objects)

Determine the mass for both objects, calculate gravity for both and each body’s inertia on the other as a sum over distance. record results. Polarity is shared across the masses until there is no longer inertia (one mass or contact).

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In a discussion person A is talking 1.2 dB louder than person B, and person C is talking 3.2 dB louder than person A. What is th
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Answer: 3.84dB

Explanation:

Since person A is talking 1.2dB louder than B, we will have

A = 1.2B... (1)

Similarly, person C is talking 3.2 dB louder than person A, we have

C = 3.2A... (2)

From equation 1, B = A/1.2... (3)

To get the ratio of the sound intensity of person C to the sound intensity of person B, we will divide equation 2 by 3 to give

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A dolphin can swim at a constant speed of 12.5 m/s. How
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Answer:

\boxed {\tt 3.6 \ seconds}

Explanation:

Time can be found by dividing the distance by the speed.

t=\frac{d}{s}

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d= 45 \ m \\s= 12.5 \ m/s

t=\frac{45 \ m}{12.5 \ m/s}

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t=\frac{45 }{12.5 \ s}

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