Recent comments in /f/explainlikeimfive

KaareKanin t1_ja6v2a5 wrote

The only thing is, I'm fully aware of this. With his answer Fried validated the initial question, and I still maintain that for normal people using cars normally, they don't require much power at all. They may use it in short bursts because it's there, but I've had rentals with only double digit horsepower figures, and I was never a hindrance to normal flow of traffic.

A tractor probably comes close to using what their engines can output frequently, and if they were to do ploughing a lot faster, power requirements would go up. It's almost all about speed.

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rwkgaming t1_ja6udh7 wrote

Would like to add to the not analysing everything all the time.

I dont know the english term for it but due to a condition i do. And i have had to learn how to filter out useless information because it just made me absurdly tired all the time and made me want to dont for 40% of the days i was awake.

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KaareKanin t1_ja6u1a0 wrote

I don't really need to. Horsepower is the measure of work that can be done. An engine with more horsepower will be able to do more work, hence most car engines could easily do the work a tractor does. And with regards to torque, that's just a question of gearing.

Speed up stuff = do more work.

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GalFisk t1_ja6temu wrote

If you live in a place that gets cold enough in the winter, you can see this phenomenon on roads when it starts to get cold. The roadway on bridges will be frost-covered in the morning, because they have had cold air all around them, while roadways the ground is still clear.

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manofredgables t1_ja6rnnq wrote

Probably the best perpetual motion machines we've built are things we've launched into space. Satellites that aren't in low earth orbit are gonna keep going for a really long time. Not to mention probes heading out of the solar system. The odds of them crashing into something is astronomically(heh) low.

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manofredgables t1_ja6rdg4 wrote

>Don’t listen to these guys regurgitate what their science teacher in grade 9 told them about energy.

You know... I'm pretty anti-establishment in general and mildly anarchist. I'm also an engineer. Know what I work with? Electrical motors, hybrids and generators! In semi trucks! Do you have any idea what a legend I'd be if I just did this "one simple trick" to make our semi trucks suddenly have unlimited range and energy? Yeah I'd be legendary and rich as fuck.

You think I wouldn't at least try it of there was any chance physics worked like that? There's a good reason anyone who isn't schizophrenic or mildly dumb sticks to saying it's impossible... That's because it is.

You're basically proposing a mechanism where a rock rolls down a ramp, which pushes a lever that tilts the ramp in the other direction. You just threw magnets into the equation, hoping they'd do something magical.

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Riegel_Haribo t1_ja6rdaz wrote

NASA: How long does it take light to get out from the inside of the Sun?

According to the famous 'drunkard's walk' problem, the distance a drunk, making random left and right turns, gets from the lamp post is his typical step size times the square root of the number of steps he takes. For the sun, we know how far we want to go to get out....696,000 kilometers, we just need to know how far a photon travels between emission and absorption, and how long this step takes. This requires a bit of physics!

The interior of the sun is a seathing plasma with a central density of over 100 grams/cc. The atoms, mostly hydrogen, are fully stripped of electrons so that the particle density is 10^26 protons per cubic centimeter. That means that the typical distance between protons or electrons is about (10^26)^1/3 = 2 x 10^-9 centimeters. The actual 'mean free path' for radiation is closer to 1 centimeter after electromagnetic effects are included. Light travels this distance in about 3 x 10^-11 seconds. Very approximately, this means that to travel the radius of the Sun, a photon will have to take (696,000 kilometers/1 centimeter)^2 = 5 x 10^21 steps. This will take, 5x10^21 x 3 x10^-11 = 1.5 x 10^11 seconds or since there are 3.1 x 10^7 seconds in a year, you get about 4,000 years. Some textbooks refer to 'hundreds of thousands of years' or even 'several million years' depending on what is assumed for the mean free patch. Also, the interior of the sun is not at constant density so that the steps taken in the outer half of the sun are much larger than in the deep interior where the densities are highest. Note that if you estimate a value for the mean free path that is a factor of three smaller than 1 centimeter, the time increases a factor of 10!

Typical uncertainties based on 'order of magnitude' estimation can lead to travel times 100 times longer or more. Most astronomers are not too interested in this number and forgo trying to pin it down exactly because it does not impact any phenomena we measure with the exception of the properties of the core region right now. These estimates show that the emission of light at the surface can lag the production of light at the core by up to 1 million years.

The point of all this is that it takes a LONG time for light to leave the sun's interior!!

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PerturbedHamster t1_ja6r40s wrote

To a photon, there's no such thing as time. From its point of view, it would be instant. One way of thinking about this is that in special relativity, distances shrink by a factor of sqrt(1-v^2/c^2). That is zero for a photon where v equals c, so from the photon's point of view, the distance between the Earth and the Sun is zero.

More technically, "age" is a funny concept in relativity. Time is seen differently by different observers. You need to specify both where and when something happened (not just "Alice met Bob at the corner of Main and Elm, but Alice met Bob at the corner of Main and Elm at 4:30 on Friday). If you take two different events at different times and different places, different observers won't agree on how far apart they are or when they happened, but they will always agree on the difference between distance squared and time multiplied by the speed of light squared (dx^2-c^2 dt^2). Since a photon moves at the speed of light the distance it moves dx in a time dt is just speed times dt, or dx=cdt, so dx^2-c^2 dt^2=0. As far as the universe is concerned, the distance between a photon leaving the sun and that photon hitting the earth is exactly zero. In our frame, that means dx is 93 million miles and dt is 8 minutes, but to the photon dx=0 and dt=0. There's no "right" answer for the age of the photon, since every frame is valid, but if you ask the photon, you'll get zero.

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