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anzhelika [568]
3 years ago
7

Which physicist first identified the electron as an independent part of the atom?

Physics
1 answer:
Simora [160]3 years ago
7 0
<h2>Answer: J. J. Thomson </h2>

Thomson conducted a series of experiments on cathode ray tubes, which led him to the discovery of electrons.

This is how on April 30th, 1897 Thomson announced the discovery of the electron (although at that time he called it a corpuscle)

This discovery was considered one of the milestones in science of the late nineteenth century, since it allowed to have a new conception of the structure of matter and its interaction with energy.

It should be noted that from this discovery, Thomson proposed a new model of the atom that although erroneous, remained valid for years and served as the basis for new postulates and research on the subject.

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Answer:

Explanation:

There is no set way to do this. All you have to do is define left and right. Left will be minus and right will be the opposite --- plus.

That is completely arbitrary. It could be the other way around. It does not matter.

Left is minus so: - 600 N   is the force going left.

Right plus so: + 500 N

Now just add.

Net Force = +500 - 600

Net Force = - 100 N

So the Net Force is - 100 N going to the left.

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Explanation:

average speed = total distance travelled / total time travelled

time to travel the first 6km: 6 / 50 = 3/25 (h)

time to travel the next 6km: 6 / 90 = 1/15 (h)

[I think there's problem in the question 'cause 900km/h sounds impossible for normal person to travel in normal condition]

The total time: 3/25 + 1/15 = 14/75 (h)

Average speed over the 12 km drive will be:

\frac{12}{ \frac{14}{75} }  =  \frac{450}{7}  = 64.3 \: km{h}^{ - 1}

8 0
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Suppose you wanted to be able to see astronauts on the moon. What is the smallest diameter of the objective lens required to res
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A:1.94

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Un contratista colocará azulejo importado en la pared de una cocina, que mide 6 metros de ancho y 4 metros de alto.
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The pupil of your eye
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What is the question?

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If you tell me the question I can help.

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