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2 years ago

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what is the name of the area around a charged object where the object can exert a force on other charged objects?

1 answer:

shusha [124]2 years ago

4 0

Answer:

An electric field is a region around a charged object where the object's electric force is exerted on other charged objects. Electric fields get weaker the farther away they are from the charge. An electric field is invisible. You can use the field line to represent it.

Explanation:

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Answer:

Explanation:

Given

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we know

$a_c=\frac{v^2}{r}$

$v=\sqrt{r\times a_c}$

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$v=21.84\times 10^5 m/s$

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$r=100\times 5.3\times 10^{-11} m\approx 5.3\times 10^{-9} m$

$a_c=10^4\times 9\times 10^{22} m/s^2$

$a_c=9\times 10^{26} m/s^2$

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$v=\sqrt{9\times 10^{26}\times 5.3\times 10^{-9}}$

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3 years ago

What Is one of many factors that can determine the rate of alcohol absorption

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The percentage of the drink that finds the target and lands in your mouth.

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3 years ago

On a snowy day, max (mass = 15 kg) pulls his little sister maya in a sled (combined mass = 20 kg) through the slippery snow. max

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Work done by a given force is given by

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here on sled two forces will do work

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3 years ago

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St. John’s

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2 years ago

An experiment is conducted on a long straight wire of diameter d. A constant current is sent through the wire and the magnetic f

soldi70 [24.7K]

Answer:

D.

Explanation:

To solve the exercise it is necessary to apply the concepts related to the Magnetic Field described by Faraday.

The magnetic field is given by the equation:

$B = \frac{\mu_0 I}{2\pi d}$

Where,

$\mu =$ Permeability constant

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For the given problem we have a change in the diameter, twice that of the initial experiment, therefore we define that:

$B_1 = \frac{\mu_0 I}{2\pi d}$

$B_2 = \frac{\mu_0 I}{2\pi 2d}$

The ratio of change between the two is given by:

$\frac{B_2}{B_1} = \frac{\frac{\mu_0 I}{2\pi d}}{\frac{\mu_0 I}{2\pi 2d}}$

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Therefore the correct answer is D.

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3 years ago