/sci/ - Science & Math » Thread #11728988

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Anonymous Fri 29 May 05:07:00 2020 No.11728988 View Reply Original Report

Quoted By: >>11729034 >>11729055 >>11730289

Why does distance matter when calculating the energy required for a constant acceleration in a vacuum? I don't get it.

Anonymous

Anonymous Fri 29 May 2020 05:12:35 No.11728992 Report

Quoted By: >>11729017

what are your calculations?

Anonymous

Anonymous Fri 29 May 2020 05:22:58 No.11729017 Report

Quoted By: >>11729539

>>11728992
Trying to figure out how many MWh I need to reach sub relativistic speeds in 2 years at a constant acceleration of 1 m/s, given an 8000 ton object
anyway apparently I need something less than 13 MWh capacity? That can't be right

Anonymous

Anonymous Fri 29 May 2020 05:29:58 No.11729034 Report

Quoted By: >>11729037

>>11728988
Because otherwise you get the coronavirus

Anonymous

Anonymous Fri 29 May 2020 05:30:57 No.11729037 Report

Quoted By: >>11729081

>>11729034
i need to know how badly i fucked up because there's no way 13MWh sustained for a year is enough to annihilate a civilization.

Anonymous

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Anonymous Fri 29 May 2020 05:40:00 No.11729055 Report

Quoted By:

>>11728988
from newtons second law F=MA, you can multiply both sides by the distance the force was applied to get FD=MAD, if you use average acceleration and write the distance in terms of initial and final velocities you get force times distance = the change in energy
FD=change in energy

So if you apply a force over a distance you change the energy of an object. In your face force time distance = kinetic energy

Anonymous

Anonymous Fri 29 May 2020 05:56:03 No.11729081 Report

Quoted By: >>11729135

>>11729037
roughly 63 million seconds in two years.

so you're accelerating to 63million m/s

the work required is 8 mil kg x 63 mil m/s/s multiplied by the distance covered

which is roughly the arithmetic series 1 + 2 + ... 63 million ~= (63 million)^2 / 2

which works out to 10^30 Joules = 10^30 W

so 9.5 x 10^23 W/h

or 9.5 x 10^17 MW/h

which is pretty far from what you get. i'm high tho so check my calculations

Anonymous

Anonymous Fri 29 May 2020 06:27:43 No.11729135 Report

Quoted By: >>11730280

>>11729081
I got as far as 14400000000000000000J and gave up. I'm not sure how to take that and figure out how much power would need to be output at a constant rate to produce that in 2 years.

Anonymous

Anonymous Fri 29 May 2020 06:38:36 No.11729152 Report

Quoted By: >>11729536

So why would it take more energy for an object to go from a velocity of 1600 to 1601 m/s than from 0 to 1 m/s?

Anonymous

Anonymous Fri 29 May 2020 07:01:43 No.11729185 Report

Quoted By:

Why the fuck is it (F * D)/ T and not F * T? I'm so fucking wasted holy shit.

Anonymous

Anonymous Fri 29 May 2020 09:54:38 No.11729533 Report

Quoted By:

>constant acceleration
>so the system requires a constant input of energy
>the closer you get to relativistic speeds the more energy is required to increase the acceleration

what exactly is the question anon? what exactly is 'sub relativistic speeds'
Do you just simply mean: "how much energy do you need to constantly accelerate with 1ms an 8000 ton object for 2 years long?"

Anonymous

Anonymous Fri 29 May 2020 09:55:55 No.11729536 Report

Quoted By: >>11729697

>>11729152
because einstein said E=MC^2 so the more energy is in the system the more mass it has, the more energy is required to produce a force to accelerate this mass. (this feedback-loop is what makes it impossible to reach lightspeed using our current understanding of physics... i think..)

Anonymous

Anonymous Fri 29 May 2020 09:58:16 No.11729539 Report

Quoted By:

>>11729017
0 Mwh because 0 m/s is a sub-relativistic speed.

Anonymous

Anonymous Fri 29 May 2020 11:19:05 No.11729697 Report

Quoted By: >>11730136

>>11729536
>sci not understanding physics
That's for near relativistic velocities. One way to think of it is a force does the same acceleration no matter what, but once something is already moving, it can do more work in the same time because greater distance, so the energy difference is greater. It doesn't really get more intuitive then that, if you can't wrap your head around how that works then idk what to say, just maybe fo into biology or something where you'll be safe from the big scary formulæ (for the most part).

Anonymous

Anonymous Fri 29 May 2020 14:56:57 No.11730136 Report

Quoted By:

>>11729697
why the FUCK are you insulting my smarts :( i'm a second year engineering student, you know we atleast try. I'm just recalling how my physics teacher explained it. Which in my brainlet mind made sense because f = ma, a stays constant, m increases, thus force required for the same a is bigger. Now you say this is wrong, which I believe, and I think I understand your explanation.
just
>It doesn't really get more intuitive then that
it very probably does, i wouldnt recommend you becoming a teacher. Also biology, I don't know why you would compare your highschool biology experience to an actual university study of biology. I'm pretty sure it's not that easy and that even in bio you're not safe from big scary formulae.
dont be so goddamn condesending.

Anonymous

Anonymous Fri 29 May 2020 16:00:03 No.11730280 Report

Quoted By:

>>11729135
Bruh, power is j/s

Anonymous

Anonymous Fri 29 May 2020 16:05:19 No.11730289 Report

Quoted By:

>>11728988
$\displaystyle \left \{ \begin{array}{l} P~power \\ U~voltage \\ I~current \\ W~work,energy \\ t~time \\ F~force \\ s~distance \\ m~mass \\ a~acceleration \\ v~speed \\ \\ P=UI\\ P= \dfrac{W}{t}= \dfrac{F \cdot s}{t}= \dfrac{ma \cdot s}{t} = \dfrac{m \dfrac{v}{t} \cdot s}{t}= \dfrac{m \dfrac{s/t}{t} \cdot s}{t} = \dfrac{m \cdot s^2}{t^3} \end{array} \right. \\\\ \left \{ \begin{array}{l} W~watt \\ V~volt \\ A~ampere \\ J~joule \\ s~second \\ N~newton \\ m~meter \\ kg~kilogram \\ \\ W=V \cdot A \\ W= \dfrac{J}{s}= \dfrac{N\cdot m}{s}= \dfrac{kg \cdot \frac{m}{s^2} \cdot m}{s} = \dfrac{kg \cdot m^2}{s^3} \end{array} \right. $

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