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3 years ago

Why did Thomson's results from experimenting with cathode rays a big change in scientific thought about atoms?

1 answer:

3 0

Thomson's results from experiments with cathode ray tubes produced a big change in scientific thought about atoms because <u>his results gave the first evidence that atoms were made of smaller particles.</u>

Particularly, he proved the existence of negatively charged sub-atomic particles that are called as electrons which together with a positively charged nucleus form an atomic unit.

Thomson's plum pudding model of the atom had negatively-charged electrons embedded within a positively-charged "soup." Thomson also made the following conclusions from his experiment:

The cathode ray is composed of negatively-charged particles.
The particles must be as part of an atom because the mass of each particle is only ~ 1/2000 compared to one hydrogen atom mass.

Within the atoms of all elements, these subatomic particles are found.

Initially it was controversial, but later Thomson's discoveries were gradually accepted by scientists. After then, his cathode ray particles were given a name called as electrons.

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The vertical weight carried by the builder at the rear end is F = 308.1 N

<h3>Calculations and Parameters</h3>

Given that:

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The torque equilibrium condition can be used to solve

We can note that the torque due to the force of the rear person about the position of the front person = Torque due to the weight of the block about the position of the front person

This would lead to:

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2 years ago

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4 years ago

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3 years ago

A vertical cylindrical tank 10 ft in diameter, has an inflow line of 0.3 ft inside diameter and an outflow line of 0.4 ft inside

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Answer:

$\frac{dh}{dt} = 1.3 \times 10^{-3} \frac{ft}{s}$ , level is rising.

Explanation:

Since liquid water is a incompresible fluid, density can be eliminated of the equation of Mass Conservation, which is simplified as follows:

$\dot V_{in} - \dot V_{out} = \frac{dV_{tank}}{dt}$

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By replacing all known variables:

$\frac{dh}{dt} = \frac{(0.3 ft)^{2}\cdot (5 \frac{ft}{s} ) - (0.4 ft)^{2} \cdot (2 \frac{ft}{s} )}{(10 ft)^{2}}\\\frac{dh}{dt} = 1.3 \times 10^{-3} \frac{ft}{s}$

The positive sign of the rate of change of the tank level indicates a rising behaviour.

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4 years ago