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sladkih [1.3K]
1 year ago
9

Suppose we were to attempt to use a similar machine to measure the charge-to-mass ratio of protons, instead. Suppose, for simpli

city, that we can get a "source" for a beam of protons as we do here for electrons (with a similar setup). What complications (if any) do you expect?
Physics
1 answer:
Igoryamba1 year ago
3 0

Answer:

In summary, the biggest difference in the experiment is that the proton mass is much more than the electron mass, so the voltages used are high and very dangerous.

Explanation:

The machine to measure the ratio of charge / mass of the electron, has two parts: a part where it accelerates the electrons in an electric field and a second section to charter the beam and measure its radius of curvature calculated from here the q / m ratio

In the case of having protons, the charge has the same value as that of the electrons, but with a positive charge, so the polarities of the fields should be changed.

The mass of the protons is much greater than the mass of the electrons, which introduced a significant difference in the excrement, since similar electric fields the speed of the protons is much less than the speed of the electrons, so the magnetic field through which the voters pass to have equivalent deflations in many cases this small values ​​of the magnetic field are not desirable due to the interference of the Earth's magnetic field.

Another solution could be to increase the electric field to have the protons with speeds similar to the electors, this possibility is not easy either, because the field of trunking of more than 5000 V would be needed, which are very dangerous.

In summary, the biggest difference in the experiment is that the proton mass is much more than the electron mass, so the voltages used are high and very dangerous.

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t is the time

The angular frequency of the motion in the problem is given by
\omega = 2 \pi f= 2 \pi (8.38 Hz) = 52.7 rad/s

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0.0059

Explanation:

According to the question the seismic activity density is given by

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Here,

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So,

\text{Seismic activity density}=\frac{424}{71300}\\\Rightarrow \text{Seismic activity density}=0.0059

The seismic activity density is 0.0059

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Answer:

The current is changing at the rate of 0.20 A/s

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Given;

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current in the inductor, I = 3.0 A

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The energy stored in inductor is given as;

E = ¹/₂LI²

E = ¹/₂(5)(3)²

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This energy is increased by 3.0 J/s

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Determine the new current at this given energy;

25.5 = ¹/₂LI²

25.5 = ¹/₂(5)(I²)

25.5 = 2.5I²

I² = 25.5 / 2.5

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I = √10.2

I = 3.194 A/s

The rate at which the current is changing is the difference between the final current and the initial current in the inductor.

= 3.194 A/s - 3.0 A/s

= 0.194 A/s

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By substituting the numbers into the formula, we find

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