Ma = 1 kg
Mb = 4 kg
V1 = 1 m/s
V2 = ? (What would be the value of V2? How many meters per second would V2 be equal to?)
V3 = ? (What would be the value of V3? How many meters per second would V3 be equal to?)
Looking forward to your two answers for the 11th time.
Looking forward to your two answers for the 11th time. --willthanks for keeping count. It will make it easier when I ban you from this forum. I will just look at that count and say "Will repeated himself 11 times, violating the rule against repetitious posting."
Have you noticed that no one cares about your thread anymore? Maybe it’s time to move on and find new victims elsewhere? If it makes you happy, I could declare that you won this thread and give you a prize.
All that effort to bury this politely, out the window. ?
To @lausten.
=======================================
@will34ab maybe no one has the answers or the answers you want. I can’t see how there would be any legal violation if someone did know the answer and told people. Seeking and getting info is not illegal.
@will34ab What is your answer? And please show your work, including units of measure.
Watching the Units saved my butt countless times while at University.
To @mriana.
<hr />
Hi dear colleague,
Thank you for your reply.
I am answering you immediately.
Ma = 1 kg
Mb = 4 kg
V1 = 1 m/s
V2 = 0.6 m/s = theoretical value
V2 = 0.5999992 m/s = mean experimental value
<hr />
V3 = 0.1 m/s = theoretical value
V3 = 0.0999997 m/s = mean experimental value
<hr />
Friction is reduced enough so that the experimental error due to friction is much smaller than 1 %.
Please look again at the link Perpetual motion and reactionless drive - YouTube and if something is unclear again, then please ask your next questions. I am ready to answer.
Looking forward to your reply.
To @mrmhead.
<hr />
Hi dear colleague,
Thank you for your reply.
Ok, this is already another song!
This is already a dialogue!
<hr />
I am fulfilling your request immediately.
Ma = 1 kg
Mb = 4 kg
V1 = 1 m/s
———————-
V2 = 0.6 m/s = theoretical value
V2 = 0.5999992 m/s = mean experimental value
<hr />
V3 = 0.1 m/s = theoretical value
V3 = 0.0999997 m/s = mean experimental value
<hr />
Friction is reduced enough so that the experimental error due to friction is much smaller than 1 %.
Please look again at the link Perpetual motion and reactionless drive - YouTube and if something is unclear again, then please ask your next questions. I am ready to answer.
Looking forward to your reply.
“top-experts”Thanks for the promotion. We're just a few random people from the internet. This is not a panel of experts.
Why? Because whatever values to choose for V2 and for V3 the result will be always one and same.OMG, you actually got to the point. Well, how you can keep taunting us now that you told us what you wanted to say? I hope you understand this sarcasm. I'm pointing out the childishness of the game you are playing.
And here’s the question that every reasonable forum participant who has ever lived wants to know; If you can prove the law of conversation of energy is incorrect, what are you doing on a little forum on the edge of the internet, throwing a tantrum about it. I don’t think that’s how scientists act. Funny thing, if you use the main menu here, the good people at CFI have a way of taking ideas like this and responding to them. I’ve never done it, so don’t know exactly how it works, but I know they have looked at people’s “experiments” and given scientific responses.
https://centerforinquiry.org/programs/ It’s geared more toward the paranormal, but hey, there’s 250K prize, give it a try.
So I’m still having trouble reconciling the zig-zag.
What is the velocity pre-zig-zag (Va), during zig-zag (Vb) and post zig-zag (Vc)?
And During zig-zag we would need to know the absolute velocity and the velocity along the Y-axis. Call them Vb and Vb’.
I’ll give you “take gravity and friction out of the equation and consideration” … that’s what we always did in high-school.
“take gravity and friction out of the equation and consideration...”Guess I must have been blind to that detail. Now it all makes all the sense.
The word “Real” in the title must has thrown me off the scent.
Hi dear colleague,Thank you for your reply.
I am answering you immediately.
I didn’t know I asked a question. I was referring to those who are more interested in this thread than I am.
then please ask your next questions.
What was my first question that I never asked?
I will say this: This thread has far too many numbers for someone who has dyscalculia (AKA numbers dyslexia). All your numbers are all over the place, so as a moderator, I’m just making sure the thread stays civil.
The word “Real” in the title must has thrown me off the scent.Reality is relatively subjective ... but that's another topic ?
To @mrmhead.
<hr />
Hi dear colleague,
Thank you for your post of June 11, 2021 at 2:57 pm. It is reasonable and smart. Please give me some time to consider it carefully. I will write to you in the nearest future.
Regards,
… And, I don’t get the purpose of the two tubes for each apparatus. Can’t you do the same experiment with just one straight tube, and one zig-zag tube?
To @mrmhead.
<hr />
Hi dear colleague,
Please look again at PART 3 of the link Perpetual motion and reactionless drive - YouTube . Please focus on the “upper” zigzag case.
Ma = 1 kg.
Mb = 4 kg.
Va’ = pre-zig-zag velocity of the blue component = 1 m/s = const.
Vb’ = pre-zig-zag velocity of the black component = 0 m/s; the black component is motionless.
Va" = during-zig-zag velocity of the blue component = variable and comparatively difficult (but not impossible) to calculate.
Vb" = during-zig-zag velocity of the black component = variable and comparatively difficult (but not impossible) to calculate.
Vy = during-zig-zag velocity of each couple blue rod-blue ball along the Y-axis = variable and comparatively difficult (but not impossible) to calculate.
Va’‘’ = post-zig-zag velocity of the blue component = 0.6 m/s = const.
Vb’‘’ = post-zig-zag velocity of the black component = 0.1 m/s = const.
According to the third Newton’s law and to the related law of conservation of linear momentum we can write down the equalities
((Ma) x (Va’)) + ((Mb) x (Vb’)) = ((Ma) x (Va’‘’)) + ((Mb) x (Vb’‘’)) <=>
<=> ((Ma) x (Va’)) + 0 = ((Ma) x (Va’‘’)) + ((Mb) x (Vb’‘’)) <=>
<=> (Ma) x (Va’) = ((Ma) x (Va’‘’)) + ((Mb) x (Vb’‘’)) <=>
<=> (1 kg) x (1 m/s) = ((1 kg) x (0.6 m/s)) + ((4 kg) x (0.1 m/s)) <=>
<=> 1 kg.m/s = 1 kg.m/s.
In one word, the values of Va", Vb" and Vy are actually of no interest to us. Actually only the values of Va’, Va’‘’ and Vb’‘’ are of interest to us as these three values determine the validity of the third Newton’s law and the related law of conservation of linear momentum.
You wrote: “…take gravity and friction out of equation and consideration…”. Perfectly agree with this.
The mass of each couple blue rod-blue ball is much smaller than the mass of the blue T-shaped component. For example if Ma = 1 kg, then the mass of each couple blue rod-blue ball must be equal to, let’s say, 0.0001 kg (and even smaller).
In our numerous real experiments we strongly reduce friction and the mean values of Va’‘’ and Vb’‘’ are equal to 0.5999992 m/s and to 0.0999997 m/s, respectively, that is, Va’‘’ = 0.5999992 m/s and Vb’‘’ = 0.0999997 m/s. The latter clearly shows that the experimental error (due to friction) is much smaller than 1 % and this experimental error is perfectly acceptable.
Looking forward to your comments.
P.S. Welcome to our team! (Not pressing, only suggesting. But we (our team) really need young and qualified people like you.)
13) You wrote: “…take gravity and friction out of equation and consideration…”. Perfectly agree with this.So basically this is just, shall we say this has been a philosophical journey into other dimensions, and not related to this physical reality we exist within. After all, friction and gravity may be the enemies of simple math formulas, but they are something you and I, or any living person, will never get away from.
Hmmm, seems that ignoring “gravity and friction” is simply a “trick” in every sense of the word? Or?
@mrmhead. Reality is relatively subjective … but that’s another topic ?Oh boy, am I ready for that thread. :-)
Although my compliments mrmhead on rescuing this one, well done, and sort of interesting.
Now if you could explain what all that up there is for, or about, that would be interesting too.
Oh boy, am I ready for that thread. ?(3.01) Diary - But, wait! There's more. Ten Learned Responses:
“Probing the interface theory of perception: Reply to commentaries, by Donald D. Hoffman, Manish Singh & Chetan Prakash"
Psychonomic Bulletin & Review. volume 22, pages1551–1576(2015)
Abstract
We propose that selection favors nonveridical perceptions that are tuned to fitness. Current textbooks assert, to the contrary, that perception is useful because, in the normal case, it is veridical. Intuition, both lay and expert, clearly sides with the textbooks. We thus expected that some commentators would reject our proposal and provide counterarguments that could stimulate a productive debate. … (HSP)https://confrontingsciencecontrarians.blogspot.com/p/case-for-reality.html
OK, I’ll bite.
No matter how you look at it, in the long run “natural selection” always favors pattern integrity. This is already apparent in pre-life chemistry where “weak” atomic links will eventually be replaced by “strong” atomic links, leading to greater integrity and stability of the pattern.
Alpha decay is one type of radioactive decay, in which an atomic nucleus emits an alpha particle, and thereby transforms (or “decays”) into an atom with a mass number decreased by 4 and atomic number decreased by 2
Radioactive decay (also known as nuclear decay, radioactivity, radioactive disintegration or nuclear disintegration) is the process by which an unstable atomic nucleus loses energy by radiation. A material containing unstable nuclei is considered radioactive. Three of the most common types of decay are alpha decay, beta decay, and gamma decay, all of which involve emitting one or more particles or photons. The weak force is the mechanism that is responsible for beta decay, while the other two are governed by the usual electromagnetic and strong forces.[1]
Radioactive decay is a stochastic (i.e. random) process at the level of single atoms. According to quantum theory, it is impossible to predict when a particular atom will decay, regardless of how long the atom has existed.[2][3][4] However, for a significant number of identical atoms, the overall decay rate can be expressed as a decay constant or as half-life. The half-lives of radioactive atoms have a huge range; from nearly instantaneous to far longer than the age of the universe.https://en.wikipedia.org/wiki/Radioactive_decay
This is a legitimate question: What is time and how does it become expressed?
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