Math Is Fun Forum

Thank, Bob.

What's the difference between ADC and ACD?

I'm wondering why they aren't two ways of labelling the same thing, like we could call/label a square ABCD, going from, say, top left, clockwise through B,C and D. Or we could call/label the same square DCBA, going from bottom left, anti-clockwise through C,B and A. Is that correct?

What am I missing regards ADC v ACD

They look to me like the same shape, the same triangle, the former being labelled anti-clockwise, the latter clock-wise. But obviously I'm missing something

Or maybe I'm just confused about the very basics of this kind of thing?

Thanks, Bob.

I’m struggling a bit with the concept of kinetic energy.
Having always thought of energy as THE CAPACITY TO DO WORK (or cause change) to discover that kinetic energy is the energy an object has BY VIRTUE OF ITS MOTION puzzles me.
If I’m sitting at home, with potential chemical energy from the food I had earlier, I have the capacity to do work, e.g., go for a walk.
When I go for the walk I think of it in terms of spending that potential energy, by doing some work.
But I’m now told that energy can’t be destroyed, that it has to go somewhere, in this case, converted into kinetic energy. It seems more intuitive to me to think of it being converted into work. I had energy, and I spent it doing some work. And the more work I do, the less energy I have.

I’m not questioning it, as such, just trying to get my head around it.

@ktestla39

Cheers

Thanks, Bob, although I'm having second thoughts;

Two examples later and my method didn't seem to work, and theirs did;

6x^2 + 11x -10

P.S. Is there a BBCode for superscript that works here?

Thanks

Thanks.

So the nth root of x = x^1/n ?

Thanks, Bob. When you say ‘effectively weightless’ are you referring to his apparent weight as opposed to his true weight (terms I’ve just come across)?
So if the cable breaks his apparent weight, the weight he ‘feels’(?), the weight he ‘seems’ to be(?); but his true weight his what it always is regardless of other factors, like acceleration, i.e, the gfs of 10N/kg times his mass of 90kg?

I’m told this isn’t actually freefall because of the force of friction involved; is that correct?

So the air, being matter, stuff with mass, accelerates too, at -10m/s/s?
You said he’d have no way of telling that he wasn’t able to float; so he isn’t able to float?

Do you mean 20m/s/s on top of the 10m/s/s due to gravity? Or 20m/s/s in total (10 due to gravity, 10 due to it’s motor/whatever)?

Ah, so he is floating, floating up. Why? Why isn’t he being pulled down at 10m/s/s due to the gfs?

10 due to the gfs, 10 due to the motor/whatever?

The force on him from the ceiling is 90(20-10)?

I think I’m partly getting that, but I’ll need to try harder to grasp it better.

I’m not quite sure I’ve followed the above to get here but 10-2(90)=720 which I think was the correct answer to my original question.

Rise up in the reference frame of your body? From the reference frame of a viewing platform would your organs remain stationary for a moment while you plummet, like a gory episode of The Road Runner?

Thanks, Bob.

I asked chatgpt about it.

I said,

"If inertia is 'an object's tendency to resist acceleration', and increasing velocity alters that tendency, then inertia must depend, at least partially, on velocity."

It replied with,

"And you're not wrong — if we use that broad, intuitive definition of inertia.

And then it got complicated, with how Newton actually defined things like mass, and acceleration, in terms of F=ma, etc.

I think I'll leave it for now, let it brew, and come back to it later. Thanks for your help with this one.

Fascinating. And mind-bending. For me at least.

So you, Bob, are driving a train, east at 6m/s.

And I’m driving a trolley east at 6m/s, parallel to you.

From the reference frame of your train I’m at rest.

And in your reference frame no force is required to stop me, because I’m already at rest? Even though we both know a force would be required to stop me in the reference frame of the ground beneath me?

Thanks, Bob

So now we view the trolley in the second example in the reference frame of the train moving at 6m/s in the same direction as the trolley.
To stop the trolley in 2s;

F=ma
F=100kg(-3m/s/s)
F=-300N

But if we view the trolley in the second example in the reference frame of the train why not view the trolly in the first example from that reference frame?
It was moving at 6m/s with reference to the ground. Now it’s moving at 0m/s with reference to the train.
But we still want to stop it in 2s.
So,

F=ma
F=100kg(a)

What do we put for ‘a’?

A=(v-u)/t
A=(0-0)/2
A=0/2
A=0

So,

F=100kg(0)
F=0N
But we know that a force of >0N would be required to stop the trolley, don’t we?

“I just used those as examples.  There's no limit to the number of possible frames.”

Thanks, Bob. I thought that might be the case, but wanted to be sure.

“If you set up a frame of reference that travels at 6 m/s in the second case, then the trolley is only travelling at 6 m/s relative to that frame”

Do you mean something like;

Train heading west at 6m/s relative to ground.
Trolley on platform heading west at 12/ms relative to ground.
Trolley in reference frame of train = 6m/s west
(12m/s – 6m/s)?