Which stage of the grieving process does closure occur?*

Physics

1 answer:

Alecsey [184]4 years ago

4 0

Remorse is the answer

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Why on earth the sky is in blue colour but it is black on the moon?

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In order to attain orbit around earth, the ATLAS-V rocket must accelerate up to a

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Consider a positive charge Q and a point B twice as far away from Q as point A. What is the ratio of the electric field strength

Vikentia [17]

Answer:

$\frac{E_{A}}{E_{B}}=4$

Explanation:

The electric field is defined as the electric force per unit of charge, this is:

$E=\frac{F}{q}$ .

The electric force can be obtained through Coulomb's law, which states that the electric force between to electrically charged particles is inversely proportional to the square of the distance between them and directly proportional to the product of their charges. The electric force can be expressed as

$F=\frac{kQq}{r^{2}}$ .

By substitution we get that

$E=\frac{kQq}{qr^{2}}\\\\E=\frac{kQ}{r^{2}}$

Now, letting $E_{A}$ be the electric field at point A, letting $E_{B}$ be the electric field at point B, and letting R be the distance from the charge to A:

$E_{A}=\frac{kQ}{R^{2}}\\\\E_{B}=\frac{kQ}{(2R)^{2}}$ .

The ration of the electric fields is

$\frac{E_{A}}{E_{B}}=\frac{\frac{kQ}{R^{2}}}{\frac{kQ}{(2R)^{2}}}\\\\\frac{E_{A}}{E_{B}}=\frac{\frac{1}{R^{2}}}{\frac{1}{(2R)^{2}}}\\\\\frac{E_{A}}{E_{B}}=\frac{\frac{1}{R^{2}}}{\frac{1}{(4)R^{2}}}\\\\\\\frac{E_{A}}{E_{B}}=\frac{1}{\frac{1}{(4)}}\\\\\frac{E_{A}}{E_{B}}=4$