Ok lets dive into this so with no air resistance things accelerate until they reach the ground.

Lets have to objects with mass m and M.

The force of gravity F=m×g

So m×a=m×g or M×a=m×g.

Mass factors out a=g in most cases. Cool that was demonstrated on the Moon with a hammer and a feather.

For Newtonian gravity F=G×M×m×1/r² where M is the mass of the planet.

So lets use m for our object and see if it drops out.

ma=GM/r² × m

a=GM/r² Great.

Now lets look into air resistance and terminal velocity. An object reaches terminal velocity when the sum of Fg and Fr (for resistance) is 0. So the forces are in equilibrium.

Fg=Fr.

Fr=½×q×C×A×v² where A is the surface area of the object C is a factor for shape q is the density of the medium and v is the velocity. We can say taht q and C are constant so lets combine the into a factor b. (With the ½.) We will use A and get a formula for v the terminal velocity.

This all leads to the square-cube law if we assume our different objects have the same density.

Fr=bv²A

Lets look at a solid ball with density q. Its volume is V=4/3×pi×r³. So from q=m/V we get

m=qV=q×4/3pi×r³.

Now A=4pi×r². And with that lets plug that into Fr=Fg=mg.

q×4/3×pi×r³×g=bv²×4pi×r²

lets rearrange

qrg=bv²×3

So now we get

⅓×qg/b × r=v² that first bit is a constant so lets call it k

k×r=v²

(k×r)^½ = v

So with same density balls the larger falls faster. If you increase the radius by a factor of 4, v increases by a factor of 2.

The m(r) function is simple we already have that

m=qV=q×4/3×pi×r³

So now for r(m)

r = (3/4×m/(q×pi))^⅓

So now lets plug it into the v² formula.

v²=k×(3/4×m/(q×pi))^⅓

v²=k × (3/4)^⅓ × (1/q×pi)^⅓ × m^⅓. Bunch of constants and m so lets combine them into K giving us.

v²=K × m^⅓

v=K^½ × m^⅕ Lets call K^½ = C

v=C×m^⅕

So as mass increases so does v but that was obvious from the v(r) formula.

To know the value of C you need the shape factor for our sphere its 0.47 the density of the fluid (air) g~9.81m/s² pi is 5 of course and the density of our material. Of course for different shapes the resulting functions can look different but the square-cube law will apply. More mass height terminal velocity. But the results may be very similar as any shape is appropriately a sphere.

So flat does not equal infinite. One option is an infinite universe but there are finite options that are flat in their fundamental domain. The universe has a topology. The part we care about is curvature. Lets drop one spacial dimension. So now the universe could be something like the surface of a sphere it has positive curvature. The defining property of positive curvature is that initially parallel lines converge and the opposite for negative curvature. Now flat topology is where parallel lines remain parallel.

Lets look at an example you got a 2D universe with flat geometry. Its a sheat of paper. One option is that the paper is infinite. Parallel lines on that paper remain parallel. This is the geometry in the fundamental domain (2D) of this universe. Now lets give it a rule, lines going far enough to the left emerge on the right coming back to their start. If you want to embed this topology in 3D you just connect the left and right edge. You got a cylinder. It has flat geometry in its fundamental domain and is almost finite. To make it finite we connect the upper and lower edge to get a torus a donut. It has a flat geometry and a finite size, you go far enough in any direction you get back where you started. As it turns out you can embed a this flat geometry and get a torus without distortions. So add an extra dimension with the same rules and our 3D universe could be though of as the surface of a 4D torus. But there are other options that also have a flat geometry in their fundamental domain but give us a finite universe.

For an infinite universe it has always been infinite. The big bag happened everywhere. You being able to trace back every path to a single point only means that the universe is scale invariant so you can resale it all you want. Lets look at density if you look at the universe now but zoom way out because volume essentially means nothing you see a really high matter density. So your scale for volume is arbitrarily. If you have a collection of points on a grid you can zoom in and conclude that the points have a low density but zoom out and see that they have a high density. If the universe is infinite you can pick an arbitrarily large scale and there is always a scale where the universe looks the same. Infinite means that you have no reference points for scale, there is no true scale to the universe. Well the only problem is matter being finitely divisible so something like the size of an atom kinda gives a scale to the universe. But the thing is an infinite universe is consistent with the data we have.

And yes infinite energy is a consequence. And infinite density only means that the volume you pick is arbitrarily. Scale it up and density grows approaching infinite scale it down and it approaches 0.

And flattenes from the CMB is basically just draw a triangle as big as you can, so the two other points on the CMB is the largest we can make. Add up the angles, if its <180° thats negative curvature if ist >180° positive curvature and =180° means flat geometry.

So all in all you can think like this the global properties of the universe don't change. It always has an infinite density and an infinite size, but local properties can. You have a numberline with all the natural numbers. You can stretch the nubers and create larger gaps, the length of the numberline and the amount of stuff it contains remains and you can zoom out to get back the original "number density".

1-2-3-4-... early universe

1---2---3---4---... current universe zoom out and you see

1-2-3-4-... again.

Zoom out even further and everything overlaps.

● - you pretty much see a point.

Calories is a measure of energy. Your body uses energy to do anything. If you do more work you use more energy. So the chemically stored energy is used to do work it can power your muscles to exert a force.

So you care about how much energy you use per unit of time. Its power. Running at a certain speed will require E amount of energy per t unit of time so you need P amount of power to run. So P×T where T is the duration of your run equals E_tot the total energy you used.

We can get energy into our body by eating. The food contains E_in amount of energy and you burn E_out amount of energy if E_in/E_out = 1 your weight doesn't change if its >1 you take in more than you use so you gain weight, if its <1 you lose more than take in so you lose weight.

Of course measure how much power is required for an exercise isn't easy but there are estimates that can be looked up. The amount of energy you need to run a distance depends on your mass and the time it took. More mass more energy less time more energy. So work would look like this W ~ m/t × d if you needed less time you had to apply a larger force and for more mass you also had to apply a larger force and the longer distance you run you need more energy. The fine details of energy consumption is in your muscles.

The point is if you eat less you will only loose weight if you use more energy a healthy adult need roughly 1500-2000 kcal a day if you exercise more than the average with that kind of consumption you will loose weight. Of course the fastest is to eat little and exercise a lot. The difference of E_out-E_in=DE and thid DE will come from your storage, fat mainly.

Simply because that is how it remains conserved.

Lets look at the conservation of momentum first. Its a vector quantity. We define momentum p=m×v that way because that is how it remains conserved. Now not only the direction but the "lenght" of that vector is conserved. We can create a scalar quantity based of of that fact. So just a number and call it kinetic energy.

One of the important things about these scattering experiments is the thickness of your target object. If the target is thin the interaction can be treated as individual scatterings on the atoms of your target object. If the object is thick the interaction at the front of the object effect the interaction at the back, so your scattering is no longer the sum of individual scatterings.

Knowing the crossection of the experiment you can calculate a distance that the particles of the oncoming beam can freely travel in your target. Which means that a collision isn't guaranteed. A thin target would let a lot of the particles through without interaction but some of them would end up colliding and those can be treated as independent collisions. If the target is thicker than that distance almost all particles of the beam would collide statistically even those that scattered at the front of the object scattering again in the target.

So you want your target to be as thin as possible and gold foil can be really thin like a few atoms thin. Which is perfect for scattering experiments but of course there are other options too.

Algorithms, I saw other methods in the commets, but I like this one:

The square root of a number like 9 is 3, 9 is called the radicand. Now if you have x^½ = y then x/y = y. So the radicand decided by the results is the result. 9/3 = 3.

So let's try the following for say 16.

Guess: for example 6.

16/6 = 2.6667

Lets take the average of your guess and this number:

(2.6667+6)/2 = 4.3334

And try this number:

16/4.3334 = 3.692

Take the average:

(3.692+4.3334)/2 = 4.01

16/4.01 = 3.99

(4.01+3.99)/2 = 4

16/4 = 4

So 4²=16, the square root of 16 is 4. And you find a number that is pretty close really quickly.