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Thanks, ktesla39.
Bob, you said,
“The Earth rotates on its axis, and around the Sun, which in turn, rotates around the centre of the Galaxy, which is moving relative to other galaxies. That's why we have the concept of frames of reference.”
Wouldn’t we have reference frames without all of that? Wouldn’t we have reference frames if the Universe consisted solely of a static Earth?
On a static Earth you could still, for example, be seated on a railway platform, at rest with reference to the person sharing the bench with you; while a person on a moving train, at rest with reference to the person sharing the seat with her, would be moving with reference to you; and they could see themselves as at rest with you moving away from them. And someone could be running along the aisle of the train to catch the buffet car before it closed, etc, etc. Wouldn’t that amount to several reference frames?
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.
Hello, ktesla39. Yes, it is fun, especially for me, as I've often never used any kind of code before.
Question for you, and Bob, and anyone else reading this. Have you memorised all the commands/tags/whatever it is you call the bracketed thingys, and now when you use it is as simple as typing any kind of text?
P.S. Why is this block of yellow text called 'inserted text'?
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?
Thanks, Bob
Thanks, Phil
(You are in bold and italics respectively because I'm practicing my BBcode)
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?
Writing my first BBcode sentence, hopefully in bold
Post preview suggests that it worked; now let's hope this sentence is in italics
Result. Bold, and italics seem straightforward enough; now for 'emphasis'
Interesting. What's the difference between italics and emphasis?
Now for strikethrough
This is fun. I'm using code. Next stop; Rule the Universe! Bob, you've created a MONSTER! Mwhahahahahahaha!
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?
Key;
o=radius when in orbit
E=radius when on surface of Earth
**
Simplify the following fraction to the point where the ‘Gm1m2’ part is simplified to ‘1’.
Numerator; (Gm1m2)/(ro)^2
Denominator; (Gm1m2/(rE)^2
Sorry for the unorthodox way of stating this but I haven’t yet mastered BBcode (I haven’t even started with it yet).
**
The only way I can think of simplifying Gm1m2 to 1 is to divide it by Gm1m2. But then I would also have to divide the denominators by Gm1m2 and that doesn’t look possible.
Is inertia solely dependent on mass?
This is what I’m hearing, and I pretty much accept it, but I’m wondering why it isn’t also dependent on velocity.
Example.
We want to stop a 100kg trolley with a constant v of 6m/s in 2s;
F=ma
F=100kg(-3m/s/s)
F=-300N
Compared with the same trolley with a constant v of 12m/s (again, stopping it in 2s);
F=ma
F=100kg(-6m/s/s)
F=-600N
**
So it takes more F to stop the trolley when it’s moving faster. Why doesn’t this translate to; the trolley’s tendency to resist acceleration depends on, among other things, the trolley’s velocity?
Thanks, Bob.
5.25x10^5 + 6.37x10^6
If it was 5.25x10^5 TIMES 6.37x10^6 I could rearrange it to;
5.25x6.37x10^5x10^6
And then add the powers;
5.25x6.37x10^5+^6
To give;
33.4425x10^11=3.34425x10^12
But it’s PLUS not TIMES
Can we manipulate the powers when it’s PLUS?
Or (ignoring using a calculator) do we have to do it the long way;
5.25x10^5=525,000
6.37x10^6=6,370,000
And then add the two.
?
Acceleration; Freefall v Car on a road etc
Is there anything fundamentally different about a man in freefall compared with objects accelerating horizontally, for example a car on a road (other than the rate of acc. due to g being around 10m/s/s near or on the surface of the Earth; and acc. for a car on a road, or box on a floor, etc, varies according to the context)?
If a car was accelerating at a constant 10m/s/s would the following values be the same for both the man in freefall and the car on the road (ignoring friction of any kind)?
0-1s; 0m/s-10m/s; average v = 5m/s; distance = 5m; total distance=5m
1-2s; 10m/s-20m/2; average v= 15m/s; distance = 15m; total distance=20m
2-3s; 20m/s-30m/s; average v= 25m/s; distance = 25m; total distance=45m
Ah, Ns, of course.
Thanks, Bob.
A baseball has a mass of 0.14 kg. It is moving with a velocity of 38 m/s LEFTWARDS towards the bat just before impact. The bat applies an average force of 17,000 N RIGHTWARDS for 0.0007 s until they lose contact.
Calculate the ball's velocity when it leaves the bat in the following scenario.
P of ball = 0.14(38)
=5.32kg.m/s LEFTWARDS
***
J=Ft
J=17,000(0.0007)
=11.9N RIGHTWARDS
J=delta P
∴ Delta P also = 11.9N RIGHTWARDS
***
Delta P=Pf-Pi
Delta P=Pf-5.32 RIGHTWARDS
Delta P=0.14v-5.32 RIGHTWARDS
I’ll now consider rightwards to be positive, leftwards to be negative
11.9N=0.14v-(-5.32kg.m/s)
11.9N=0.14v+5.32kg.m/s
0.14v=11.9N-5.32kg.m/s
I’m not sure if this feels right, subtracting kg.m/s from N, but I’ll try it for now.
0.14v=6.58
V=6.58/0.14
V=47m/s?
Thanks, Bob.
Thanks, Bob.
First second of freefall
According to physicsclassroom.com;
Using 10m/s/s for acceleration due to gravity near the surface of Earth;
Time interval=1s
Change in v=0-10m/s
Avg.v=5m/s
D=5m
I’m thinking the average v is arrived at using the mean of velocities;
(0+10)/2=5
Is that correct?
But to check, I’ve divided the first second into tenths
After 1/10sec, v=1m/s
After 2/10, v=2m/s
Etc, up to,
After 10/10, v=10m/s
But the mean of those is 5.5, I think;
1+2+3 etc, up to, +10 = 55
55 tenths = 5.5
Where have I gone wrong?
What you call it isn't important. The calculation is
Decide on a positive direction (either will do but then use it throughout).
As the objects are travelling towards each other one v will be + and the other - .
Bob
So;
90x2-45x2=(135) x v3
180-90=135 x v3
90 = 135 x v3
v3 = 90/135
v3 = 2/3m/s
And does the calculation mean;
Total p before = Total p after
“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 ?
Object A; 90kg, 2m/s east
Object B; 45kg, 2m/s west
Q. After a collision, if the two objects 'stick' together, and 'become one' (like when a car goes into the back of another car, etc) how do we refer to this new object? Do we call it AB, or something like that?
Q. What is the velocity of this combined object after the collision?
What is the acceleration due to gravity OF THE EARTH in the following scenario?
A 5kg bowling ball is dropped from a height of 2m above the surface of the Earth.
Use 10m/s^2 for g for the ball
And 6x10^24kg for the mass of the Earth
Thanks, Bob.
I enjoyed the pendulum video.
Relative (apparent?) motion is still a bit of a head scratcher.
But I guess that's physics for you
Thanks, Bob.
“From the plane you will see me apparently moving backwards along with the ground, but I think I'm stationary.”
Interesting that you use ‘apparently’ in that sentence. Why ‘apparently’? I was getting the impression that it’s more than just apparently, it was simply the reality of the situation; that in my reference frame you ARE moving backwards. And in your reference frame I AM moving forwards. Have I got that wrong? Is it only 'apparently'? Or are you using 'apparently' for 'relatively'?
“So "The ball starts at position a, and ends up at position b. Absolutely." is the correct version.”
Yet when I join up the ab dots I get a straight line; you get a parabola (?).