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Gre4nikov [31]
3 years ago
7

Two massive, positively charged particles are initially held a fixed distance apart. When they are moved farther apart, the magn

itude of their mutual gravitational force changes by a factor of n. Which of the following indicates the factor by which the magnitude of their mutual electrostatic force changes?
a. 1/n2
b. 1/n
c. n
d. n2
Physics
1 answer:
Aneli [31]3 years ago
6 0

c. n

The magnitude of the gravitational force is given by Newton's law of universal gravitation:

F_g=-G\frac{m_1m_2}{d^2}

Here G is the gravitational constant, m_1, m_2 are the masses of the particles and d is the distance between them.

The magnitude of the electrostatic force is given by Coulomb's law:

F_e=k\frac{q_1q_2}{d^2}

Here k is the Coulomb constant, q_1, q_2 are the chargues of the particles and d is the distance between them.

As can be seen, both forces are inversely proportional to the square of the distance. Thus, the factor is the same in both cases.

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The concept required to solve this problem is the optical relationship that exists between the apparent depth and actual or actual depth. This is mathematically expressed under the equations.

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The electric field strength E₀ is measured at a perpendicular distance R from an infinitely large, thin sheet that contains a un
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Answer:

Explanation:

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7 0
3 years ago
a spring gun initially compressed 2cm fires a 0.01kg dart straight up into the air. if the dart reaches a height it 5.5m determi
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Answer:

2697.75N/m

Explanation:

Step one

This problem bothers on energy stored in a spring.

Step two

Given data

Compression x= 2cm

To meter = 2/100= 0.02m

Mass m= 0.01kg

Height h= 5.5m

K=?

Let us assume g= 9.81m/s²

Step three

According to the principle of conservation of energy

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Making k subject of formula we have

kx²= 2mgh

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Hence the spring constant k is 2697.75N/m

7 0
3 years ago
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