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Wikipedia:Administrator elections/October 2024/MMS/Voting phase: Difference between revisions

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The voting phase of the October 2024 administrator elections is officially open. As a reminder, the schedule of the election is:

The voting phase of the October 2024 administrator elections . As a reminder, the schedule of the election is:

*October 25–31 – SecurePoll voting phase

*October 25–31 – SecurePoll voting phase

*November 1–? – Scrutineering phase

*November 1–? – Scrutineering phase


Latest revision as of 05:21, 18 October 2024

The voting phase of the October 2024 administrator elections has started and continues until 24:00 31st October 2024 UTC. As a reminder, the schedule of the election is:

  • October 25–31 – SecurePoll voting phase
  • November 1–? – Scrutineering phase

In the voting phase, the candidate subpages will close to public questions and discussion, and everyone who qualifies for a vote will have a week to use the SecurePoll software to vote, which uses a secret ballot. You can see who voted, but not who they voted for. Please note that the vote tallies cannot be made public until after voting has ended and as such, it will not be possible for you to see an individual candidate’s tally during the election. The suffrage requirements are different from those at RFA. You can participate in the voting phase at Wikipedia:Administrator elections/October 2024/Voting phase.

Once voting concludes, we will begin the scrutineering phase, which will last for an indeterminate amount of time, perhaps a week or two. Once everything is certified, the results will be posted on the main election page. In order to be granted adminship, a candidate must have received at least 70.0% support, calculated as Support / (Support + Oppose). As this is a vote and not a consensus, there are no bureaucrat discussions (“crat chats”).

Any questions or issues can be asked on the election talk page. Thank you for your participation. Happy electing.

Wikipedia:Administrator elections/October 2024/MMS/Voting phase: Difference between revisions

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Atomic clocks on the Moon could create ‘lunar positioning system’ – Physics World

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32.4 Food Irradiation – College Physics

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32.4 Food Irradiation – College Physics

Summary

  • Define food irradiation low dose, and free radicals.

Ionizing radiation is widely used to sterilize medical supplies, such as bandages, and consumer products, such as tampons. Worldwide, it is also used to irradiate food, an application that promises to grow in the future. Food irradiation is the treatment of food with ionizing radiation. It is used to reduce pest infestation and to delay spoilage and prevent illness caused by microorganisms. Food irradiation is controversial. Proponents see it as superior to pasteurization, preservatives, and insecticides, supplanting dangerous chemicals with a more effective process. Opponents see its safety as unproven, perhaps leaving worse toxic residues as well as presenting an environmental hazard at treatment sites. In developing countries, food irradiation might increase crop production by 25.0% or more, and reduce food spoilage by a similar amount. It is used chiefly to treat spices and some fruits, and in some countries, red meat, poultry, and vegetables. Over 40 countries have approved food irradiation at some level.

Food irradiation exposes food to large doses of [latex]{\gamma}[/latex] rays, x-rays, or electrons. These photons and electrons induce no nuclear reactions and thus create no residual radioactivity. (Some forms of ionizing radiation, such as neutron irradiation, cause residual radioactivity. These are not used for food irradiation.) The [latex]{\gamma}[/latex] source is usually [latex]{^{60} \text{Co}}[/latex] or [latex]{^{137} \text{Cs}}[/latex], the latter isotope being a major by-product of nuclear power. Cobalt-60 [latex]{\gamma}[/latex] rays average 1.25 MeV, while those of [latex]{^{137} \text{Cs}}[/latex] are 0.67 MeV and are less penetrating. X-rays used for food irradiation are created with voltages of up to 5 million volts and, thus, have photon energies up to 5 MeV. Electrons used for food irradiation are accelerated to energies up to 10 MeV. The higher the energy per particle, the more penetrating the radiation is and the more ionization it can create. Figure 1 shows a typical [latex]{\gamma}[/latex] -irradiation plant.

32.4 Food Irradiation – College Physics
Figure 1. A food irradiation plant has a conveyor system to pass items through an intense radiation field behind thick shielding walls. The γ source is lowered into a deep pool of water for safe storage when not in use. Exposure times of up to an hour expose food to doses up to 104 Gy.

Owing to the fact that food irradiation seeks to destroy organisms such as insects and bacteria, much larger doses than those fatal to humans must be applied. Generally, the simpler the organism, the more radiation it can tolerate. (Cancer cells are a partial exception, because they are rapidly reproducing and, thus, more sensitive.) Current licensing allows up to 1000 Gy to be applied to fresh fruits and vegetables, called a low dose in food irradiation. Such a dose is enough to prevent or reduce the growth of many microorganisms, but about 10,000 Gy is needed to kill salmonella, and even more is needed to kill fungi. Doses greater than 10,000 Gy are considered to be high doses in food irradiation and product sterilization.

The effectiveness of food irradiation varies with the type of food. Spices and many fruits and vegetables have dramatically longer shelf lives. These also show no degradation in taste and no loss of food value or vitamins. If not for the mandatory labeling, such foods subjected to low-level irradiation (up to 1000 Gy) could not be distinguished from untreated foods in quality. However, some foods actually spoil faster after irradiation, particularly those with high water content like lettuce and peaches. Others, such as milk, are given a noticeably unpleasant taste. High-level irradiation produces significant and chemically measurable changes in foods. It produces about a 15% loss of nutrients and a 25% loss of vitamins, as well as some change in taste. Such losses are similar to those that occur in ordinary freezing and cooking.

How does food irradiation work? Ionization produces a random assortment of broken molecules and ions, some with unstable oxygen- or hydrogen-containing molecules known as free radicals. These undergo rapid chemical reactions, producing perhaps four or five thousand different compounds called radiolytic products, some of which make cell function impossible by breaking cell membranes, fracturing DNA, and so on. How safe is the food afterward? Critics argue that the radiolytic products present a lasting hazard, perhaps being carcinogenic. However, the safety of irradiated food is not known precisely. We do know that low-level food irradiation produces no compounds in amounts that can be measured chemically. This is not surprising, since trace amounts of several thousand compounds may be created. We also know that there have been no observable negative short-term effects on consumers. Long-term effects may show up if large number of people consume large quantities of irradiated food, but no effects have appeared due to the small amounts of irradiated food that are consumed regularly. The case for safety is supported by testing of animal diets that were irradiated; no transmitted genetic effects have been observed. Food irradiation (at least up to a million rad) has been endorsed by the World Health Organization and the UN Food and Agricultural Organization. Finally, the hazard to consumers, if it exists, must be weighed against the benefits in food production and preservation. It must also be weighed against the very real hazards of existing insecticides and food preservatives.

Conceptual Questions

1: Does food irradiation leave the food radioactive? To what extent is the food altered chemically for low and high doses in food irradiation?

2: Compare a low dose of radiation to a human with a low dose of radiation used in food treatment.

3: Suppose one food irradiation plant uses a [latex]{^{137} \text{Cs}}[/latex] source while another uses an equal activity of [latex]{^{60} \text{Co}}[/latex]. Assuming equal fractions of the [latex]{\gamma}[/latex] rays from the sources are absorbed, why is more time needed to get the same dose using the [latex]{^{137} \text{Cs}}[/latex] source?

Glossary

food irradiation
treatment of food with ionizing radiation
free radicals
ions with unstable oxygen- or hydrogen-containing molecules
radiolytic products
compounds produced due to chemical reactions of free radicals

NIST publishes first set of ‘finalized’ post-quantum encryption standards – Physics World

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User:Kinsley Estrada 1 Fan – Wikipedia

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User:Kinsley Estrada 1 Fan – Wikipedia

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Superconductivity appears in nickelate crystals under pressure – Physics World

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electrostatics – Can external charges affect charges within a cacity inside a conductor?

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electrostatics – Can external charges affect charges within a cacity inside a conductor?

I am aware that the (net) electric field within an empty cavity inside a conductor is 0 regardless of the charge on the conductor and the presence of any external fields outside the conductor. I have seen this described as contents of the cavity being “sheilded” from external influence. However, I feel like the word “shielding” would be meaningful only if the cavity actually contained some charge that could interact with the external field in the first place. To that end, I would like to know whether there is a difference in the electric field due to all sources excluding the charge within the cavity, $\vec{E}_\text{other}$, and thus a difference in the force on the charge, $\vec{F}=q\vec{E}_\text{other}$, depending on the presence of charge outside the conductor.

I have come up with the following argument but I am not convinced of its correctness:

electrostatics – Can external charges affect charges within a cacity inside a conductor?

Suppose that an external charge distribution $Q$ is present outside a (possibly charged) and there is an empty cavity inside it. Then the electric field inside inside the cavity is $0$. Now add the charge $q$ by infinitismal amounts inside the cavity. At each step, there the charge added and the charge subsequently induced on the innerwall remain unaffected by any other fields (as they add to $0 $ within the cavity and conductor) and thus distribute themselves as they would in the case with no external fields, which means that the charge in the cavity and the innerwalll produce no field and thus force on external sources of charge once the transient phase is over. Moreover, these sources remain unaffected even in the transient phase due to the infinitesimal nature of the charge added in each step.

This determines a valid solution of $\vec{E}$ inside the cavity which is bounded by a conductor with a fixed amount of charge which is simply the vector sum of the field due to the charge and the innerwalls. By the uniqueness theorem, this is the only solution. Therefore, the net electric field inside the cavity and thus the force on the charge within the cavity (produced by $\vec{E}_\text{net}-\vec{E}_\text{charge}$ is identical to the case where there are no external sources of charge.

Talk:Tuner (radio)/Gallery RFC: Difference between revisions

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User:Kinsley Estrada 1 Fan – Wikipedia

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Should the article contain a [[WP:GALLERY|gallery]] of images (as shown below) illustrating the history of radio [[tuner (radio)|tuners]]?

Should the article contain a [[WP:GALLERY|gallery]] of images (as shown below) illustrating the history of radio [[tuner (radio)|tuners]]?

====Reason for Image Gallery====

====Reason against Image Gallery====


Latest revision as of 05:13, 18 October 2024

{{rfc}}

Should the article contain a gallery of images (as shown below) illustrating the history of radio tuners?

[edit]

[edit]

Please state Yes or No with a brief statement. Please do not reply to the statements of other editors in this section. That is what the Discussion section is for.

[edit]

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optics – Green transparent acrylic—effective in blocking red light noise of 840nm-LED IR floodlights?

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optics – Green transparent acrylic—effective in blocking red light noise of 840nm-LED IR floodlights?

Since green absorbs red,…

Not entirely accurate. When a material or a filter appears green, it does not necessarily mean that it absorbs the red, it’s definitely possible that it reflects the red light. Look at the images below, this is a bandpass filter (transmitting range: 450-640nm) I’ve used, which is a type of interference filter. Obviously, the color of this filter appears different color when viewed from different directions/angles of light.
optics – Green transparent acrylic—effective in blocking red light noise of 840nm-LED IR floodlights?

What said that ‘green absorbs red’ may be referring to an absorptive filter, which block a certain band of wavelengths by absorping them. The most common used absorptive filters are color glass and absorptive neutral density filter.


acrylic transmits IR

Not entirely accurate. Transparent acrylic can be designed to transmit visible light but block infared light. See this video or this product of MidOpt as examples. (Note that the second example is exactly a green transparent acrylic as you’re interested.)


That correct?

It’s possible. But it depends on the specific transmitting spectrum of your ‘green transparent acrylic’. Because [a normal, uncoated acrylic] + [green and transparent] these two conditions are not sufficient to determine its actual spectral transmittance. As @SolomonSlow commented, it’s a complex topic.


I’m not sure about the specific composition of the ‘green transparent acrylic’ you mentioned, but I suggest you purchase professional filters to meet your needs.

Fermilab is ‘doomed’ without management overhaul claims whistleblower report – Physics World

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