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From a physics website;
"Every time the surface of a solid collides with an air molecule, there is force acting on the solid slowing it down"
My question;
What if the air molecule collides with the BACK of the solid? What if a man walking down the street on a windy day is struck on the back by molecules of air; wouldn't this, if anything, speed him up?
Thanks, Bob.
I thought I'd replied to this already. Must have dreamt it
Anyway. I look forward to checking out the MIF page, although those pages, for me at this stage, tend to go beyond GCSE level quite quickly, I think (?).
In the meantime, some basics.
With GCSE physics questions (such as, A man with a mass of 90kg is travelling at 3m/s; what is his momentum?) do we assume that the 3m/s is his instantaneous velocity? And by instantaneous momentum I meant; as opposed to the average momentum that we would get if the 3m/s was his avg v.
I have this theory about collisions, but it's not gaining momentum.
Momentum is instantaneous, yeah?
I’m 90kg, walking east at 3m/s, my momentum (p) is;
p=mass x instantaneous velocity
= 90kg x 3m/s
=270kg m/s
Yeah?
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What’s the significance of momentum?
I’m guessing Force is part of it.
If I bump into you while I’m walking at 3m/s, with my mass of 90kg, I’ll hit you (push you?) with quite a force
But I’d hit you with a greater force if I was travelling at 6m/s (p=540kg m/s), or I had a mass of 180kg (also p=540kg m/s)
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Also, how do I work out what that force is?
F=ma
F=90kg x a
What is a?
a=v-u/(t)
Answer; we don’t have enough information? We only have instantanous v, we don’t have final v and intial v. And what about t? This isn’t happening over x seconds, it’s happening instantaneously, yeah?
How reliable do you think the search result that this provides is?
So far I've been impressed. When I used to look through the first handful of search results there was sometimes one or more that wasn't quite right. Now, my first result is the AI one and so far it's always seemed correct.
Is the AI result taken from the other ones, a distillation of these? If so, how does the AI know which of the other ones are valid, which are not?
Thanks, guys.
And Lol!, Bob.
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The ground pushing me up is a curious one, for me. When I push on the ground it's conscious effort, my muscles do their thing, I use up energy, I might sweat, etc, etc. Does the ground 'do' anything? Or does it push up on me merely by virtue of its being there?
I do a push-up. I push down on the ground; the ground pushes up on me.
I’m trying to relate this to water bottle rockets, where you part fill a plastic bottle with water, cork it, then pump air through a hole in the cork until the air pressure inside is great enough to push the water out and the bottle takes off like a rocket.
Relating this to me doing a push-up; am I the air and the water is the ground?
The air pushes down on the water like me pushing down on the ground (until the water forces the cork out and the water empties from the bottle) and the water pushes back up on the air like the ground pushing back up on me (causing it -and the bottle that contains it - to take off in the opposite direction)?
What do you call a cat who just ate ten ducks?
A duck-filled fatty puss
BBC Bitesize introduced me to Newton’s 2nd Law of motion by way of F=ma. After a few examples of that they introduced the topic of intertial mass by way of a rearrangement of the above formula (F=ma becoming m=F/a). They said the mass in m=F/a is inertial mass, i.e, a measure of how difficult it is to change the velocity of an object.
This seemed to me to imply, that the m in m=F/a was somehow different from the m in F=ma. But if the formula has simply been rearranged it would seem that the m is the same in both versions?
Is the m in F=ma also inertial mass?
Thanks, everyone.
I'm going to let all of this brew for a while then come back to it
Thanks.
What edit did you make?
From the (newtonian) equations of motion for a object falling from rest with an acceleration g (ok to take this as 10m/s/s) the final velocity is given by v = 0.5gt^2
So when t = 0.1 v = 0.05
t = 0.2 v = 0.2
t = 0.3 v = 0.45................................................
t = 1 v = 5
The v values aren't in arithmetic progression so that formula won't work.
Bob
Thanks, Bob.
Are you sure of that equation?
I thought it was v=gt.
E.g, v=10x1
v=10m/s
I.e, an object in freefall reaches 10m/s after 1 second (with an average of 5m/s for the period 0-1s)
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Your equation seems to give that average.
Forgive me if I'm wrong, my brain is starting to hurt after an hour on this
(Just got this from the Google AI overview; To find the final velocity of a free falling object, you can use the formula v = gt)
Write each number in scientific notation as a decimal.
A. 9.7 x 10^3
B. 6.453 x 10^(-1)
Question A
Positive exponent 3 tells me to move the decimal point 3 places to the right.
Doing so, I get 9,700.
Question B
Negative exponent 1 tells me to move the decimal point one unit to the left.
Doing so, I get 0.6453.
Is 9700 a decimal?
Should it be 9700.0, to make it a decimal?
I guess you know that x^m/x^n = x^(m-n)
Therefore, x^a/x^a = 1 = x^0, x could be any number.Now we can write:
1/s = s^0/s^1 = s^(0-1) = s^-1 ===> s-¹
Similarly,
1/s^2 = s^0/s^2 = s^(0-2) = s^-2 ===> s-²
Thanks, KerimF
A very helpful explanation. I think I've followed all of that.
But I think my problem is how I see the term, "metres per second".
Although I've always used the shorthand, "m/s", I never saw that forward slash symbol as a division symbol; I never saw "m/s" as a fraction. I just saw it as a short way of writing "metres per second". I'm now wondering if this sounds absurd, that I've not seen it in this way. But, until now, it's not been put to me that it's a fraction.
So, why does, "m/s" mean, "metres DIVIDED BY seconds"?
If I walk at a pace of 2m/s, I cover a distance of 2m for every second that ticks by. Does that mean 2 metres DIVIDED BY 1 second? Which equals 2m (per second).
And if I walked at a pace of 2m/2s, i.e, 2m for every 2s that tick by, the fraction would be; 2m DIVIDED BY 2S, which equals 1m per second?
Apologies in advance if this sounds ridiculously basic.
Velocity or speed is given by m/s (meters per second) and Acceleration is given in Meters per second squared.
.
Thanks, Jai Ganesh.
Why does ms^-2 give velocity?
ms^-2
= m x (1/s^2), yeah?
= m/s^2, yeah?
= m/(s x s), yeah?
If I've got that right, why does m/(s x s) give velocity?
I thought velocity was simply, for example, 10m/s downwards, or 30mph East, etc?
When referring to speed on this page they often say things like,
“The bus has a constant speed of 10ms-¹"
I've not seen this before. I'm used to seeing speed given as simply, for example, 10m/s
Why the -¹ ?
I've noticed also for acceleration instead of saying, 10m/s/s, or 10m/s², they write 10m s -². Why might they do this?
I’m trying to apply Gauss’s formula for adding consecutive integers to the first second of a freefalling object.
The formula is n(n+1)/2
E.g, to add the numbers 1 to 100 we do,
n=100
100(100+1)/2
=100(101)/2
=10100/2
=5050
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During the first second of freefall an object, I’m told, averages a velocity of 5m/s downwards (is this the average of Om/s and 5m/s?) *edit*; (is this the average of 0m/s and 10m/s, not the average of 0m/s and 5m/s)
I’m assuming that one way to get this average would be to divide that first second into tenths, add the ten corresponding velocities, and divide by 10.
So for the first tenth we would reach 1/10m/s, for the second tenth we would reach 2/10m/s, etc, up to the tenth tenth where we reach 10/10m/s.
Using Gauss’s formula for the addition we get;
n(n+1)/2
n=10
=10(10+1)/2
=10(11)/2
=110/2
=55
Which we divide by 10 as we’re talking about tenths, so we get 5.5
But the average velocity for the first second of freefall is 5m/s, not 5.5m/s.
Where am I going wrong
Thanks, KerimF (I've only just seen your comment).
Thanks, Bob.
Filed away
Although displacement is regarded as a vector quantity is there one exception where it is a scalar quantity, i.e, when one undertakes a round trip and the displacment is zero?
(There is no direction to give, just the magnitude, i.e, zero)
7! What the highest grade could be in this case?
In my math exams [French style], I couldn't avoid, once a while, to make silly mistakes due to mistyping or the like.
Fortunately, in real life, I have time to correct such silly mistakes which I still do in every new design. After all, I am just a human, robots only don't do such mistakes
The highest is 9, which is equivalent to the old A star.
Cheers, Bob
I got a grade 7. Google tells me that's equivalent to an A in old money. I'm happy with that.
Thanks to Bob, KerimF, and everybody else for the help and encouragment. Cheers!
Thanks, Bob, thanks, Phil
Thanks, KerimF