Math Is Fun Forum

Thanks, Bob

So do you just get to know by heart the ones that you use frequently, without trying to memorise them; and consult the list when there's one you've either forgotten or didn't know by heart in the first place?

Thanks, ktesla39.

Thanks, Bob, for going above (literally; to the loft) and beyond.

Yes, there does seem to be a lack of consensus regards the concept of inertia.

I've heard more than one person say that it's not a well-defined technical term in physics. And others saying it tend to come up early on for students and

then they hear of it less and less.

As for a unit (for inertia regards linear motion) most seem to be saying there isn't a unit, with a few saying that it's kg, as inertia is equivalent to mass.

This is what I as after.

(I'll abbreviate Gm1m2 to simply 'G'. And I’ll abbreviate the denominators to simply 'o' and 'e'.
I think this will work just as well)

(G/o)/(G/e)

= (G/o)*(e/G)

simplify by crossing out the G's;

(1/o)*(e/1)

= e/o

And, replacing the original terms, gives us;

numerator;    (re)^2
denominator; (ro)^2

Is that correct?

Cheers, Bob.

And lol

**

You said emphasis is bold; I got sloping lines (same as italics) when I used the 'em' brackets etc

Also, is heading text just the same as bold?

Thanks, ktesla39, thanks, Bob, really helpful, as ever.

“To divide by a fraction, invert it and multiply.”

Ah, of course. Thanks, Bob. KFC (the mnemonic).

Keep. Flip. Change.

Keep the first fraction as it is.
Flip the second fraction.
Change the sign.

“Do you know why this works?”

I think I’m getting there.

Example.

(10/1)/(2/1) = (10/1)(1/2)
(10/1)(1/2) = 10/2
10/2=5

In words;

Ten divided by two...
...is the same as...
...a half of ten.

(To half something is to divide it by two)

Is that the basic idea?

Ah, Ns, of course.
Thanks, Bob.

Thanks, Bob.

“Bring the ball to rest by embedding it in the frame of reference...”

Thanks, Bob. What does that mean, to embed it in the frame of reference? I did Google it but I’m not sure that I’ve got it.

“...and it's now the Earth that moves and at the same (but sign reversed) magnitude.”

Do you mean; in my scenario the ball moved 2m downwards towards Earth; but in your scenario Earth moved 2m upwards towards the ball?

“Should we worry about this?”

I’m not sure. When I’m about to carry out the actual experiment I will give you notice, maybe you can strap everything down, warn the neighbours, reassure nervous dogs, etc.

I did a crude calculation, ignoring the distance of 2m (I think I just put a distance to illustrate that it was ‘near the surface of Earth’).

Earth on ball;

F=ma
F=5kg(10m/s/s)
F=50N downwards

From there I inferred that if Earth is exerting 50N on the ball, pulling it with a force of 50N, the ball must be exerting a force of 50N on Earth, pulling it with a force of 50N.

Then;

Ball on Earth;

F=ma
50N=6x10^24kg(a)
a=50N/(6x10^24kg)
a=8.3recurring x 10^-23m/s/s

So if I drop a bowling ball I cause Earth (with you, Commander Bob, and the jittery poodle at no.42, on it) to accelerate at approx 1 ten billionth of trillionth m/s/s ?