Can you really square a second?

[CuriosOne]

 

Assuming something squared assumes a cycle of completion obviously for our second per second squared..Is that how it works?? 

Example of squaring the second would be the accelleration of earth's gravity at its surface 9.8 m/s-1...

The moon "falling around" earth in a cycle from point a to point b, month to month..."example"

Or even earth's seasons mark reference points "example" Summer to Winter..

Now I don't want to mention atomic orbitals becuase we are not sure how the electron orbits the nucleus, but caclulations of such does use pi ratio in QM so I guess that's another topic..

Anyway, when i thnk of a second squared "regardless" of the units involved" I see 1 complete cycle "almost independant of time itself" ie magnetic attraction versus the speed of light, IE Gravitational Energy" good example or maybe even KE whom knows at this point.

Or the refractional index, or how light travels slower in a medium etc etc...

These are just my thoughts..

But really how can you square a second??

Or is it the "unit" that's squared as "time" and energy conservation???

 

Edited October 20, 2020 by CuriosOne

[OldChemE]

So, acceleration is one example where we see seconds squared,  but we are not really squaring a second-- its mathematical language. When we say, for example, that the acceleration is 10 meters per second squared we are really saying that the velocity is increasing by 10 meters per second each second.  Mathematically this looks likes we are squaring a second but in reality we are not.

[CuriosOne]

9 minutes ago, OldChemE said:

So, acceleration is one example where we see seconds squared,  but we are not really squaring a second-- its mathematical language. When we say, for example, that the acceleration is 10 meters per second squared we are really saying that the velocity is increasing by 10 meters per second each second.  Mathematically this looks likes we are squaring a second but in reality we are not.

What I don't get is how, accelleration, velocity and speed are related with one of them being positive, the other being a vector and the other staring at "zero."

Math language in base 10 uses..

0 9 8 7 6 5 4 3 2 1

With 0 =1 second ????

And how about the square root usage of 1/2?

Z^2 = [A^2+B^2]^1/2<------------- does that mean second per second squared??

I'm tempted to think everything is constant^2...But not sure..

Edited October 20, 2020 by CuriosOne

[studiot]

17 minutes ago, CuriosOne said:

Assuming something squared assumes a cycle of completion obviously for our second per second squared..Is that how it works?

No it doesn't work like that.

 

Either "per second squared "

or

"seconds per second"

note the plural in the first seconds in the second phrase.

Both are correct and mean the same thing.

but not second per second squared.

[michel123456]

After acceleration, you have even worse:

https://en.wikipedia.org/wiki/Jerk_(physics)

 

[Charles 3781]

What does "squaring" actually mean?  Isn't it just a shorthand linguistic expression for "multiplying a number by the same number"?

It's just multiplication of one number by another.  I can't see how it differs from, say, multiplying any two numbers.

[MigL]

Make an attempt to live beyond your fingertips and eyes; not everything has to be seen/touched and be physical.
Usually it means 'per second, per second', but it could have other meanings.
Does velocity squared have a physical representation ?
The root of -1, i, appears in many equations; what exactly does it represent physically ?

Would you be happier if we gave acceleration a new unit. say a ( where a=m/s^2 ), such that there is no squaring of the time unit ?

Edited October 20, 2020 by MigL

[CuriosOne]

48 minutes ago, MigL said:

Make an attempt to live beyond your fingertips and eyes; not everything has to be seen/touched and be physical.
Usually it means 'per second, per second', but it could have other meanings.
Does velocity squared have a physical representation ?
The root of -1, i, appears in many equations; what exactly does it represent physically ?

Would you be happier if we gave acceleration a new unit. say a ( where a=m/s^2 ), such that there is no squaring of the time unit ?

So what's the ""time unit""???

And can a see a simple real world example?

55 minutes ago, Charles 3781 said:

What does "squaring" actually mean?  Isn't it just a shorthand linguistic expression for "multiplying a number by the same number"?

It's just multiplication of one number by another.  I can't see how it differs from, say, multiplying any two numbers.

My point exactly, adding squares then taking the root confuses me greatly..

 

[MigL]

1 hour ago, CuriosOne said:

So what's the ""time unit""???

And can a see a simple real world example?

What's wrong with the unit you used ?
The second.
Is this heading in the same irrational and nonsensical direction as your thread on the 3rd dimension ?
None of us have time for that.

[CuriosOne]

22 minutes ago, MigL said:

What's wrong with the unit you used ?
The second.
Is this heading in the same irrational and nonsensical direction as your thread on the 3rd dimension ?
None of us have time for that.

You speak for the whole of science??

Plain and simple if x = 1, 2, 3 or "whatever" and it's time "dependant" ie varies, and you use changes in time  how do you multiply by "nothing" if x-> 0 or if x=0 and get a phyiscal quanity??

Besides if energy is conserved then whats the point of any of this??

I'm asking questions but "science and math are inconsistent."

I agree with its nonsense no doubt about it..

Edited October 20, 2020 by CuriosOne

[MigL]

3 hours ago, CuriosOne said:

You speak for the whole of science??

You certainly don't speak any science.

 

3 hours ago, CuriosOne said:

Plain and simple if x = 1, 2, 3 or "whatever" and it's time "dependant" ie varies, and you use changes in time  how do you multiply by "nothing" if x-> 0 or if x=0 and get a phyiscal quanity??

If  x is a function of time, it is usually written as x(t).
But do you understand the concept of a mathematical model describing a physical reality ?
And limits of applicability ( usually determined by boundary conditions where the function goes to zero or diverges to infinity ) ?

 

3 hours ago, CuriosOne said:

Besides if energy is conserved then whats the point of any of this??

I fail to understand what energy conservation has to do with any of this.
( am I just wasting energy and time replying to this nonsense ? )

 

3 hours ago, CuriosOne said:

I'm asking questions but "science and math are inconsistent."

Science and math are consistent.
As are you, with your misunderstanding.

[CuriosOne]

1 hour ago, MigL said:

You certainly don't speak any science.

 

If  x is a function of time, it is usually written as x(t).
But do you understand the concept of a mathematical model describing a physical reality ?
And limits of applicability ( usually determined by boundary conditions where the function goes to zero or diverges to infinity ) ?

 

I fail to understand what energy conservation has to do with any of this.
( am I just wasting energy and time replying to this nonsense ? )

 

Science and math are consistent.
As are you, with your misunderstanding.

Usually determined by boundary conditions.

What are these boundaries?? And who sets them?

where the function goes to zero and diverges to infinity...

0.1111111111111111111111-> Infinity???? X^infinity????

Does infinity have a number?

I'd place it myself but at this point I'd like to see a real physical calculus problem "real world example"  from someone whom knows what they are doing becuase obviously im not getting my point across the "standard" way..

Edited October 21, 2020 by CuriosOne

[studiot]

6 hours ago, MigL said:

And limits of applicability ( usually determined by boundary conditions where the function goes to zero or diverges to infinity ) ?

 

5 hours ago, CuriosOne said:

What are these boundaries?? And who sets them?

where the function goes to zero and diverges to infinity...

0.1111111111111111111111-> Infinity???? X^infinity????

Does infinity have a number?

 

Apart from the fact that you claimed to understand when I started to explain about derivatives and began with the answers to these questions,

I also suggested you should spend more of your time reading what was written properly and then digesting it.

I have highlighted the difference between MigL's correct statement and your all-too-quick snap response.

The difference between 'and'   &  'or'   is so huge they are nearly opposites.
Not only that but this difference appears in many many places in logic, maths, physis and science in general as well as being one of the few special words that are exactly the same in plain English as formal scientific language.

Please learn to walk before you try to run.

You will make much faster progress that way.

Another example would be that you did not pick up the difference between your opening statement and my correction for you.

13 hours ago, studiot said:

No it doesn't work like that.

 

Either "per second squared "

or

"seconds per second"

note the plural in the first seconds in the second phrase.

Both are correct and mean the same thing.

but not second per second squared.

 

Edited October 21, 2020 by studiot

[Sensei]

14 hours ago, CuriosOne said:

Can you really square a second?

Example of squaring the second would be the accelleration of earth's gravity at its surface 9.8 m/s-1...

(It should be 9.8 m/s^2 or 9.8 m * s^-2)

g (or a) is coefficient used in equation of distance traveled by body i.e.

d = 1/2 * a * t^2

"second square" unit of a, cancels out with "second square" unit of t^2. Simple cancellation of units.

m/s^-2 * s^2 = m

To calculate acceleration of the body, you transform this equation from

d = 1/2 * a * t^2

to

a=2*d/t^2

or you can derive time:

t=sqrt(2*d/a)

 

https://en.wikipedia.org/wiki/Equations_for_a_falling_body

 

How scientists get this equation? They were releasing objects from well known height, and measuring time of flight to the ground.

You release something from 1 meter. Measure time. Release something from 2 meters. Measure time. Release something from 4 meters. Measure time. etc.

(it is interesting experiment. In the modern times we can record it on camera, and see it in slow motion speed, and easily measure time using frames-per-second of video)

You will receive sequence of pair of data. Distance of flight and time. From them you can extract "g" (or "a") coefficient.

Then you can predict time of flight of something placed at any arbitrary height.

 

How long you will be flying from 3 km?

t=sqrt(2*3000/9.81)=24.731 seconds (assuming no air resistance, thus no terminal velocity)

Edited October 21, 2020 by Sensei

[joigus]

14 hours ago, CuriosOne said:

But really how can you square a second??

I've just squared a second and it gave me 1 second squared, so apparently you can.

[CuriosOne]

6 hours ago, joigus said:

I've just squared a second and it gave me 1 second squared, so apparently you can.

10^1/10^1=1

Oh I get it now thanXx 🙂

6 hours ago, Sensei said:

(It should be 9.8 m/s^2 or 9.8 m * s^-2)

g (or a) is coefficient used in equation of distance traveled by body i.e.

d = 1/2 * a * t^2

"second square" unit of a, cancels out with "second square" unit of t^2. Simple cancellation of units.

m/s^-2 * s^2 = m

To calculate acceleration of the body, you transform this equation from

d = 1/2 * a * t^2

to

a=2*d/t^2

or you can derive time:

t=sqrt(2*d/a)

 

https://en.wikipedia.org/wiki/Equations_for_a_falling_body

 

How scientists get this equation? They were releasing objects from well known height, and measuring time of flight to the ground.

You release something from 1 meter. Measure time. Release something from 2 meters. Measure time. Release something from 4 meters. Measure time. etc.

(it is interesting experiment. In the modern times we can record it on camera, and see it in slow motion speed, and easily measure time using frames-per-second of video)

You will receive sequence of pair of data. Distance of flight and time. From them you can extract "g" (or "a") coefficient.

Then you can predict time of flight of something placed at any arbitrary height.

 

How long you will be flying from 3 km?

t=sqrt(2*3000/9.81)=24.731 seconds (assuming no air resistance, thus no terminal velocity)

Understood, but question...

d = 1/2 * a * t^2

What is 1/2 ?

Is that an empty square root??

The whole thing looks similar The Pythagorean Thoerem, no one can say it does not.

z^2 = [a^2+b^2]^1/2 

In this set up does 1 = 100% ??

Afterall you need a base to start time from or atleast some form of default reference "point."

In your last example of 24.73 seconds, that is close the root of 5^2 , I see this all the time maybe x^2*5^2, means seconds per seconds squared and our 1 is "base 10"

Is this the correct way to think about it?

Edited October 21, 2020 by CuriosOne

[Sensei]

5 hours ago, CuriosOne said:

d = 1/2 * a * t^2

What is 1/2 ?

Is that an empty square root??

What?! It is 1 divided by 2. Half. 0.5 ....

5 hours ago, CuriosOne said:

The whole thing looks similar The Pythagorean Thoerem, no one can say it does not.

Absolute nonsense.

5 hours ago, CuriosOne said:

Is this the correct way to think about it?

Absolutely not..

Correct way, I told you already, would be playing with digital camera. Drop something from well known height e.g. 2 meters, record it on camera, load video file in some free video editing tool e.g. VirtualDub. Then analyze where used to be (d (distance) parameter) dropped object in time, at which frame, during flying to the ground. Then make spreadsheet of gathered data in e.g. OpenOffice SpreadSheet or Excel. Then figure out equation from data. And you end up with  a=2*d/t^2

[CuriosOne]

28 minutes ago, Sensei said:

What?! It is 1 divided by 2. Half. 0.5 ....

Absolute nonsense.

Absolutely not..

Correct way, I told you already, would be playing with digital camera. Drop something from well known height e.g. 2 meters, record it on camera, load video file in some free video editing tool e.g. VirtualDub. Then analyze where used to be (d (distance) parameter) dropped object in time, at which frame, during flying to the ground. Then make spreadsheet of gathered data in e.g. OpenOffice SpreadSheet or Excel. Then figure out equation from data. And you end up with  a=2*d/t^2

I will try then get back to you.

May take a while, thnXxxx

About the root question I was referring to

1/2 power of the root I should have mentioned that.

From.

Wikipedia..

The square root of 2, or the one-half power of 2.

 

 

 

 

Edited October 21, 2020 by CuriosOne