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2 years ago

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A 5. 0-c charge is 10 m from a small test charge. what is the magnitude of the force experienced by a 1. 0 nc charge placed at t

he location of the test charge?

1 answer:

sveta [45]2 years ago

4 0

The magnitude of the force experienced by a 1. 0 c charge placed at the location of the test charge is 4.5*10^8N.

To find the answer, we have to know more about the force experienced between two charges.

<h3>How to find the force experienced between two charges?</h3>

It is given that,

$Q_1=5C\\Q_2=1C\\r=10m.$

We have to find the force experienced between two charges,

$F=\frac{kQ_1Q_2}{r^2}$

where, k=8.99*10^9.

Thus, by substitution, the force will be,

$F=\frac{8.99*10^9*5*1}{10^2} =4.5*10^8N$

Thus, we can conclude that, the magnitude of the force experienced by a 1. 0 c charge placed at the location of the test charge is 4.5*10^8N.

Learn more about the force here:

brainly.com/question/4086258

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Answer:

a

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Explanation:

From the question we are told that

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Generally the condition for two slit interference is

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Where m is the order which is given from the question as m = 2

=> $\theta = sin ^{-1} [\frac{m \lambda}{d} ]$

substituting values

$\theta = 0.0022 rad$

Now on the second question

The distance of separation of the slit is

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The intensity at the the angular position in part "a" is mathematically evaluated as

$I = I_o [\frac{sin \beta}{\beta} ]^2$

Where $\beta$ is mathematically evaluated as

$\beta = \frac{\pi * d * sin(\theta )}{\lambda }$

substituting values

$\beta = \frac{3.142 * 3*10^{-4} * sin(0.0022 )}{550 *10^{-9} }$

$\beta = 0.06581$

So the intensity is

$I = I_o [\frac{sin (0.06581)}{0.06581} ]^2$

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3 years ago

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Answer:

The speed of space station floor is 49.49 m/s.

Explanation:

Given that,

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$F=mg$

$\dfrac{mv^2}{r}=mg$

$\dfrac{v^2}{r}=g$

$v=\sqrt{rg}$

Where, v = speed of space station floor

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Put the value into the formula

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Answer:

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