The h Bar

2:44 AM

@JohnRennie , since the roller coaster is being accelerated greater than a speed of g, I expect the boxes in it will leave the car, when released and rise higher. This is similar to how if boxes are placed in a forward moving car on the road, and when they are released they will move backwards. A similar situation: If you have travelled in a forward moving bus, and you are standing and holding the handle; if you stop holding the handle you will be pushed backwards.

123

$\Large{m \Large{\int_{t_a}^{t_b}}\frac{dv}{dt}vdt}$

Hi guys pls help me out how to write bit size integral symbol here

Q: What does it mean to assign group operations to distinct sets for space groups?

@JohnRennie, I made a mistake here, g is not a speed, its an acceleration due to gravity. So if boxes are accelerated downwards greater than the acceleration of g, when they are released they should go upwards, since the impending motion is upwards.

Do my ideas make sense? I'm willing to learn

I hope to learn something and leave this site with a better understanding of Physics

123

@ShashankVM Whenever any object accelerating it feel backward force due to property of inertia. Because objects don't want to change its velocity.

In order to keep boxes in roller coaster car when at some height it downward/invert it is necessary to create enough centrifugal force which is greater that gravity and forces acting on it downward. So, it feel its upward force.

Hi @SirCumference

$\Large{m \int_{t_a}^{t_b}\frac{\mathrm{d} }{\mathrm{d} t}\left ( \frac{1}{2} v^{2} \right ) \mathrm{d}t}$

How to solve above relationship?

First integrate or differentiate???

Nike Dattani

3:22 AM

@ZeroTheHero There's another space groups question, is it something with which you have familiarity?

8

I am trying to understand space groups in crystallography. In International tables for crystallography, for a nonsymmorphic space group, they list some symmetry operations. 8 of them are listed under the (0,0,0)+ set and 8 in the (1/2, 1/2, 1/2)+ set. What does this mean? Are there 16 operations ...

4:48 AM

Good Morning!

Happy Diwali y'all

Just In: New record created at #Ayodhya Deepotsav. 5,84,572 diyas lit today.

Ayodhya Beats own record

Prateek Mourya

5:13 AM

Happy diwali everyone

6

5:41 AM

@123 The integral will reduce to Kinetic Energy

$KE|_{t_b} - KE |_{t_a}$

5:57 AM

@ShashankVM suppose the roller coaster car and the boxes are all falling freely. So the observer in the ground frame sees all three of them accelerating downwards at g.

If you were sitting in the car then you, the car and the boxes would all be weightless and stationary with respect to each other.

If we now accelerate the car downwards, e.g. attach a rocket motor to it, then you and the boxes would still be weightless and falling freely but now you'd see the car accelerating away from you.

For the person watching from the ground you and the boxes would still be falling freely as before, so you and the boxes would only move apart if you and the boxes fell freely at different rates.

In a vacuum you and the boxes would stay together because you'd all be accelerating downwards at the same acceleration of g.

In air we need to include the drag due to air resistance, and that would make you and the boxes fall at different speeds so all three objects would separate from each other.

So the question comes down to what effect the air resistance has on the rate the boxes fall.

Kenzo Tenma

6:41 AM

Happy Diwali!

JingleBells

8:31 AM

We should make a genie lamp and put Alexa in it and to activate it you have to rub the lamp

Mauro Giliberti

9:11 AM

@JingleBells That would be awesome

9:38 AM

@JohnRennie In the question I asked, the boxes have the same size and shape. So they will experience the same air resistance. But when they are released, they will experience a force of friction, since the car faces downwards. The body of the car is perpendicular to the ground. Since these boxes have different masses, they will experience different frictional forces. So I feel the frictional force has more effect than air resistance

@ShashankVM Why will the boxes experience any friction? The car and the boxes are all falling freely so they are all weightless. There are no forces acing between them.

The car is accelerating faster than g

So it is not falling freely

then the car will accelerate away from the boxes so they will lose contact with it.

Initially the boxes are fastened to the car. They are release in the middle of the descent. When they are released the car accelerates away from the boxes. But when this happens the contact friction between the car and box comes into play

You'll have to draw a diagram to show what is happening.

Q: Gravitational mass vs inertial mass: A problem in classical mechanics

9:47 AM

The car is accelerating greater than g because it is on a roller coaster and it is propelled by electromagnetic force.

@JohnRennie So the question comes down to which box experiences greater frictional force, am I right?

@JohnRennie, Are you able to undertand my question now?

I also added the diagram to my question

-1

There are 2 boxes wooden boxes of the same size and shape. One is empty and another is filled with a heavy material, like Gold. These boxes are placed on top of an open car of a roller coaster ride and fastened to the car. The car goes over the top of the roller coaster and starts its descent ver...

homework-and-exercises newtonian-mechanics mass

Yashas

10:24 AM

Why do people say that if we know $(x(t), p(t))$, we know everything about a classical system. What about the potential and external forces?

ACuriousMind

@Yashas why would you be interested in the individual forces if you already know the entire motion of the system?

also, you can get the total force on the system at each instant from $F=m\ddot{x}$

Yashas

I am not sure if I am overthinking but if you have an electrical wire loop, the even if you know $(x(t), p(t))$, you won't know the emf induced.

But I think this might not be the right question to ask because the whole system is a collection of particles. Maybe having individual $(x(t), p(t))$ does give all the information about the system.

ACuriousMind

@Yashas when people say that $(x(t), p(t))$ suffices, they mean a classical mechanical system, not a circuit

once electromagnetism becomes involved the electric and magnetic fields or rather their potentials become additional variables you need to know to know the entire state of the system

@ACuriousMind Correct!

Yashas

@ShashankVM "Undergraduate student at St Joseph Engineering College, Mangaluru" (from your GitHub) => I think we're less than 30km from each other.

12:59 PM

Just In: SUPER EXCLUSIVE SCENES FROM AYODHYA CELEBRATION, BROUGHT TO YOU BY @RewCie

Secret

arxiv.org/pdf/cs/0611100.pdf

ultrafinitism turns out to be quite interesting in a predicative mathematics perspective

you ask for the smallest number which takes as many operations as itself to define

that is, given any exponential operator EXP on the natural numbers, you want to find the natural n such that EXP(n)=n

Ultrafinitism is just representation of numbers by repeated exponents of $e$ na?

Secret

the trouble for ultrafinitism is you cannot use induction, so you do not have the luxury of the fundamental theorem of arithmetic to show it is impossible

yes, they deny finitist natural set

Secret

$ e^{e^{e^{79}}} $ is one of the proposed limits

1:08 PM

okay, gotcha...

ZeroTheHero

1:58 PM

@NikeDattani The question a bit too vague. If I had a definite example I could look it up and sort it out but in this form it’s difficult to say much.

ZeroTheHero

2:12 PM

@user6232128. Here’s the article what I was originally looking for: ams.org/publicoutreach/feature-column/fcarc-latinii1

4:08 PM

I am doing an experiment packing disks to determine the packing fraction. I am trying to calculate ...

... pi of course.