Ask the Physicist!

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QUESTION:
When they talk about entangled particles, what types of particles can get entangled? And If 1 of 2 entangled particles experienced time dilation, (Through trips around a particle accelerator) with the affect of time dilation split the difference between the 2 particles and 1 is having the affect of "observer" speed applied while the other has "slower" speed applied? Would there be any measurable affect? Would "Spooky action" occur at time-since entanglement? (ie, If "Normal" particle was impacted x seconds after entanglement and time dilation, would the affect apply in the observer time "instant" or delay until "Slow" particle caught up to x seconds since entanglement and time dilation)

ANSWER:
The easiest way to think about entangled particles is that they are both parts of the same system. The particle states are correlated such that the system all changes together. That, of course, is Einstein's "spooky action at a distance". So rather than thinking of the second particle changing its state when the other is measured because it somehow detects this change, it is simply doing what it is required to do because the two particles are really one system. Because the changes are apparently instantaneously happening, there is no need to worry about the clocks of either of them. This is why they are called entangled—they are really a single quantum system. Indeed, two entangled particles are entangled regardless of their positions and frames of reference. An observer in any frame of reference will see the change of quantum states to be instantaneous and if that is true, clocks are irrelevant—instaneous is instantaneous.

QUESTION:
Context: you're an astronaut in geostationary orbit around the earth. How much smaller in % would the earth look if there was no gravitational lensing? Or in other words how much bigger does gravitational lensing make the earth appear (if at all)?

ANSWER:
The equation to calculate the angle of deflection θ is θ=4GM/(rc²). I get θ=(1.7x10^-9)°. Something tells me that you will have no chance of observing a nanodegree! Don't even bother messing with the percentage.

QUESTION:
I am a retired UMASS Medical School associate professor in molecular neuroscience who has a longtime interest in the quantum realm but unfortunately never pursued it when getting my chemistry degree from UCSD. I have a question that relates to the quantum nature of electron energy levels in atoms. Electron energy levels in atoms relate to the nuclear structure of the atom itself. Is it possible to know anything about these orbitals in the absence of electrons? In other words, when electrons are stripped from an atom, do these orbitals remain of the same nature in the absence of electrons, or is it only possible to truly speak of these energy levels when an electron is present or nearby; ie, they only make sense when the interactions take place between the electron and the nucleus? Maybe in the absence of an electron these orbitals exist in a virtual sense only.

ANSWER:
What matters in quantum mechanics is what the 'system' is. Once you have specified the system, you need to solve Schrödinger's equation for the system. The wave function for an oxygen with 5 electrons stripped off is different than the wave function for an un-ionized oxygen atom; therefore, the answer to your question is no, changing what the 'system' is will change what the wave function is. Along these same lines, the energy spectrum of the O⁵⁺ will be different than un-ionized oxygen.

QUESTION:
So, we can determine the earth's movement in relation to other astronomical objects, but we have no idea of our absolute movement, correct? The earth is spinning and revolving around the sun which is revolving around a galaxy that is expanding from other galaxies in the universe. But wouldn't we need to know our velocity in absolute terms to know our relative time? For arguments sake, the universe could be spinning or moving at a million miles/hr. Our time is dependent on our velocity. If light speed is a 'constant'… doesn't velocity depend on time (m/s) whereas time depends on velocity? Wouldn't that be circular logic?

ANSWER:
The time you are interested in is the time in your frame of reference. In your example of the "universe moving at a million mph", it has no relevance to whether the laws of physics are true in your frame. (Actually making that statement is not meaningful because what is the whole universe moving relative to?) You seem to think that in your moving frame something would change if you accelerated to a higher speed. Once you got to that speed and turned off your engines, there would be no experiment which you could perform to detect that change. There is no such thing as "your absolute velocity". All times in frames relative to yours would depend on their velocities relative to you; your time would, for you just tick away like it always has. I guess you have not studied special relativity which reveals to us that there is no such thing as "absolute time", there is no god somewhere with the master clock which everyone compares to.

QUESTION:
If a system is spinning at 2% of the speed of light or more, it starts leaking energy in the form of photons. Could you remind me what that is called?

ANSWER:
I don't know about your 2% number but I do know that an accelerating charge radiates photons; this is called bremsstrahlung.

QUESTION:
I have made one of these for use in the kitchen for measuring small quantities but I cant figure out the scale graduations because its non-linear. If I can get the formula I can use a spreadsheet to display the scale graduations and then do a screenprint. The beam I used was 3mm steel rod. I can send a photo of the device if that would be useful. This type of scale is useful because its very cheap and easy to construct (and no batteries needed ! )

ANSWER:
I have redrawn the picture you sent, replacing your thread by a fulcrum which does not matter. My notation:

M is the mass being measured.
g is the acceleration due to gravity.
D is the length or the beam.
f is the fraction of the length of the beam on the left of the fulcrum.
μ is the mass density of the beam, mass per unit length, so μD is the mass of the beam.
θ is the angle which the beam makes with the horizontal.

Let's sum all torques about the fulcrum,

MgfDcosθ+(μgfD²/2)cosθ-(μg(1-f)D²/2)cosθ=0.

This is a scale, so we want M as a function of θ. Solving the above equation for M I find M=μD[1-(1/(2f))]. Note that M is not a function of the angle, so your scale will not work. Once you have found the right f for the given M, it will be in equilibrium regardless of the angle. If you want to learn more about balances and their designs, there is an earlier answer to a question which you might like to link to.

QUESTION:
I work in the medical imaging field and once had demonstrated to me how much resistance to mechanical motion eddy currents can have. The MRI field service engineer placed a non-ferrous quarter in the bore of the MRI. In one orientation it fell over normally. Rotated 90 degrees if fell as if moving through molasses. He told me it was due to eddy currents. I also play electric guitar. Electric guitar strings are both conductive and ferrous materials. I tested that with my Ohm meter and a magnet to double verity. The electric guitar has pickups to pick up the vibrations of the strings and amplify them. I do not understand how they work. The common explanation is it's due to the magnetic force pulling the strings down, changing the moving strings path in some way. I am also wondering if eddy currents may be playing a role. Granted the pole pieces are not particularly strong, especially compared to an MRI.
Question 1) I assume eddy currents occur in ferrous and non-ferrous materials both so long as they are conductive, correct or incorrect?
Question 2) Is it feasible that eddy currents in the strings resist motion in combination with a magnetic pull even though the magnetic strength is nowhere near 1T?

ANSWER:
I edited your question to take out a lot of extraneous rambling. You clearly do not understand how the pickup works, but you are in good company—when I was googling I found that easily half the explanations about the physics were wrong or so vague as to provide no explanation. To understand what is going on, you need to understand magnetic induction of electric current, first discovered by Michael Faraday. In a nutshell, if you have a changing magnetic field you will induce electric currents in nearby conductors; also, because of relativity, if the nearby conductors are moving and the magnetic field is constant, you will still find that currents are induced in the conductor (as what is happening with your quarter in the MRI). Of course, if you have currents they will have their own magnetic fields; this plays an important role in the guitar pickup too. So let's look at what is going on:

There is a magnet for each string and each magnet is a small bar magnet;
there is a coil of wire (hundreds or thousands of turns) and the two ends of the wire will be accesible to be the output of the coil to amplify.
of course, there is a string, which is which is fabricated from a conducting metal;
if the string is not vibrating there are no currents anywhere because the magnetic field is static, not changing.
now the string is plucked so the string vibrates so it is moving through the unmoving field of its magnet;
because the string is a conductor there will be an electric current induced in it and it will be varying with the same frequency as the vibration;
now we know that electric currents give rise to magnetic fields so the vibrating string will have its own magnetic field;
this induced field will be seen by the coil;
since the current will be constantly changing, its magnetic field will be changing;
this changing field will now be passing through the coil and there will therefore be an induced current, same frequency as the string;
the current will be passed through a load causing a voltage which can be amplified electronically and sent to the speakers.

QUESTION:
I’m not well schooled and I’m well up in age, but I’ve been wondering if the Earth rotates at 1000 mph and a passenger airplane can only fly at a top speed of 500 mph. How can it fly from California to New York without the Earth staying ahead of it is it gravity a magnetic pull ?

ANSWER:
The atmosphere moves, for the most part with the rotating earth. The airplane flies relative to the air, so it's like paddling a canoe downstream in a river; if you don't paddle, you move along with the river but if you do paddle you move faster than the water. Also, the air does move at different speeds, it's not just a static blanket of air; winds blow depending on the conditions of the air (weather). The most permanent high-speed (more than 100 mph) winds are called the jet stream. Flying across North America or to/from Europe to North America, because jet planes fly at the altitudes (30,000-40,000 feet) where the jet stream is, the flying time is almost always shorter going W to E than E to W because the jet stream mainly goes to the east.

QUESTION:
I was wondering what would happen if 2 10g ball bearings travelling close to the speed of light collide in the LHC. Obviously I'm assuming a lot here. Such as The LHC wouldn't break by putting non-particles in there. The structural integrity of the ball bearings would remain stable for the duration of the experiment The balls could even reach that speed in the first place There was enough energy to do this. I think it would be fascinating but I'm not a physicist. I hope I'm not wasting your time.

ANSWER:
Most people, like you, have no concept about how huge the energy of macroscopic objects going close to light speed are. I have answered questions like this before which might interest you (See faq.html#faq game scifi guns). Apart from the vacuum inside the LHC, there is no reason for it to be in this question. Let's just say that two bearings in a vacuum, each of mass m=10^-2 kg and moving with a speed of 90% the speed of light. Each has energy E=[10^-2/√(1-0.9²)](3x10⁸)²=2x10¹⁵ J. The Nagasaki bomb had a total energy of about 10¹⁴ J, so each of these bearings would have about 20 WWII era atomic bombs. If you took 2x10¹⁵ J and delivered it over the period of one day, it would be equivalent to a power of P=E/t=2x10¹⁵/(3600x24)=23x10⁹ W=23 GigaWatts; that's about 6 times larger than the biggest power station in the US. You may assume that the "structural integrity of the ball bearings would remain stable for the duration of the experiment" but there is no way that could possibly happen if there is this much energy. We are talking about 4x10¹⁵ Joules of energy which is the energy of a typical H bomb.

QUESTION:
I'm trying to measure a slit width by passing a laser through it of known wavelength and distance the slit is from screen. Unfortunately, that requires close to absolute darkness as the resultant beam, after passing through the slit, is like a super fine thread and very hard to measure. My task is to determine how fine the adjustable slit can go down to, in order to maximize the resolution of a home brew spectrophotometer . My problem is that I need to get my eyes "dark adapted" to see this very fine line of light on my white graph screen. Unfortunately, the laser is very bright, which messes up my dark adaptation and makes the projected laser image very hard to see and all the more so, measure on the screen.

ANSWER:
So, by best resolution you probably mean the narrowest width possible. And, it seems that you are having trouble even seeing the light on the screen because the light line is so narrow that it has insufficient light to be seen well. You are barking up the wrong tree, I believe. There are actually two things you need to understand:

The width for geometric optics. If the slit is much bigger than the wavelength of the light, the image is simply identical to the slit as shown in the figure. The angle can be written as θ₁=s/R=d/R

Eventually the slit will become comparable to the wavelength. Significant broadening of the light passing through the slit will occur due to diffraction. The single slit diffraction pattern looks like the next figure:
Now the width of the image, s, will be the distance between the two zeros of intensity of the central peak; this can be shown to be θ₂=λ/d. where λ is the wavelength of the light. Finally I show a graph of θ₂ (red), θ₁ (black), and their sum (blue). The narrowest possible image can be shown, by finding the minimum of the blue curve, to be θ_best=√(λR).

If you are interested you can read a beautiful example of an application of all this, the eye of a bee, in Section 36-4 of the Feynman Lectures. This will also give you more detail filling in my "it can be shown"s.

I suspect that you will not be able, with your apparatus, to get out of the geometrical optics region. But from your question I suspect that you will need to make a compromise because you probably don't have enough intensity to do your measurements.

QUESTION:
How much energy would it take to launch a man 3,000 kilometers away in an instant? And let's say for this hypothetical, the man is somehow durable enough to withstand this.

ANSWER:
You don't define what an "instant" is. It usually means "in zero time" as in instantly. The answer, then, is infinite energy. It is physically impossible for anything to move through some distance instantly.

QUESTION:
Does the speed of light change when it is passed through a prism and is broken up into the various color bands? Do the different colors of light, in the light spectrum, travel at different speed when project into an environment void of light? If, say, red light is projected into a completely red light environment, is the speed of light measurable?

ANSWER:
The speed of light in a vacuum has the same regardless of what its wavelength is. Most experiments would measure the speed in air, not a vacuum, but air has an index of refraction extremely close to 1; therefore the answer to your first two questions is, for all intents and purposes, no. Your third question has no meaning ("…completely red light environment…")

QUESTION:
I know the electric potential energy is negative for opposite charges. But initially when we move the charges (consider bar magnets) from infinity to say 20cm apart the force is applied by my hand since they aren't in each others electric fields. The force is in the same direction as displacement and work is positive. I get that EP becomes negative when they are in the influence of each others fields but why is EP positive intially?

ANSWER:
"…consider bar magnets…"? Bar magnets are not electric charges*. You need to consider two electric charges, q and Q. Imagine Q to be held fixed. The work done to move q from r₀ to r is W=kqQ[(1/r₀)-(1/r)]; if r₀=∞, W=-kqQ/r. The potential energy U is defined as U(r)=-W=kqQ/r. Note that I have chosen U(r₀=∞)=0. So, if q and Q have the same sign, U(r)>0 for all r and if q and Q have the different signs, U(r)<0 for all r. I have left off all the steps leading to W=… and U=-W; you can find these in any physics text or on the web, e.g. this link.

*Because magnetic forces are velocity-dependent, they are not conservative forces so you cannot introduce a potential energy for them.

QUESTION:
I'm currently in the process of Worldbuilding my own character's story. [I have no idea what this means!] If the universe, instead of a vacuum, were an infinite "block" of matter, expanding forever in every direction, how would gravity behave inside a spherical hole (at the theoretical center of this infinite matter) within this infinite block of matter? So far, I have only ever found mathematical formulas to determine gravity inside an empty sphere but I am unsure if that behaves different if there is a vacuum outside the sphere (which, in my case, there is not). [This question was originally way too long and I have edited it because the answer is simple.]

ANSWER:
I assume that the mass of your block is uniform and that it has spherical symmetry; to assure spherical symmetry, the block should be a sphere. Physicists like to avoid infinities so just make it extremely large (like billions of light years). The gravitational field inside a thin uniform spherical shell is zero everywhere. This was proven by Newton, in fact it is supposed that this question was the main reason why he invented calculus. There are several consequences of what is called the shell problem. First, the field inside a thin spherical shell is zero. Second, a thin spherical shell will give rise to a field outside which is the same as if all the mass of the shell outside were concentrated at the center of the hole. Third, the shell need not be thin since you can imagine adding many thin shells like an onion making a thicker and thicker shell, each having the shell theorem applied to it. The gravitational force in the matter will linearly increase the farther you get from the center (another reason for not being infinite). If it is expanding, is mass being created somehow? Or, if mass is conserved, it must be getting less dense. And if it is getting less dense, it must be compressable which means that it is getting more compressed at larger distances than closer to the hole; this would still result in zero field inside the hole but the field in the matter would not increase linearly anymore. If you added any more holes, all bets are off because that would destroy spherical symmetry; however, if the holes were much farther apart than their sizes it would be an excellent approximation that the fields would be zero.

QUESTION:
When I try to calculate the power draw from the recirculator pump on my furnace, I get two different answers.

When I multiply the volts x Amps, I get 0.71 Amps 115Vx0.71Amps=81.7 W=0.109 hp.
The hp rating of the pump is 0.04 hp=29.8 W.

Why the discrepancy? Do I have to use the RMS value of the AC line voltage to get the right value?

ANSWER:
I actually had no idea because I did not know how water pumps are rated. They are not rated by power but rather by head (H) times flow (F); on a site I found the statement is made that "Unlike motors, pumps are rated by head and flow, not by power. There’s no such thing as a 50-horsepower pump". The relation between H, F, and P is (P=HxF)/e where e is the efficiency of the pump. I believe that where you went wrong is to compare the input power to the output power which will not be equal unless the pump is 100% efficient which it never will be. If I understand all your numbers, the pump efficiency will be e=1-(29.8/81.7)=63.5%. The site I found most helpful is here.

QUESTION:
Why do the moons of other planets have names, while the name of our moon is simply moon.

ANSWER:
This can be confusing! The word moon means any natural satellite orbiting a planet; here natural means it is not man-made. The space station is a satellite but not a moon. "The Moon" refers only to the moon which orbits earth. Note the capitalization of the Moon which sets it apart from all other moons; the Moon is a moon! Astronomers often refer to the Moon as Luna, the Latin word for Moon. Io, Europa, Ganymede, and Callisto are moons of Jupiter and were the first moons of another planet to be discovered (by Galileo in 1610). Hundreds of exoplanets (planets outside our solar system) have been observed but no undisputed evidence for exomoons has been found.

QUESTION:
I have a masters degree in Electrical Engineering. We are mainly schooled in classical electrodynamics (Maxwell and Hertz). The more I read about quantum mechanics, the more I think that electromagnetic waves (as described by Maxwell) really don’t exist and is purely a mathematical contrivance (although Maxwell’s equations do great job predicting what happens at a macro level). I am aware that photons contain discrete amounts of energy. If we look at an antenna and why the electrons start moving back and forth and produce a small voltage when radio waves hit it is because the waves have oscillating magnetic and electrical fields. Because electrons have electric and magnetic moments, these fields “push” the electrons around (pretty classical description of a force). The quantum mechanics description of what is happening seems very different. My crude interpretation seems to be that a quantum particle called a photon binds to an electron and gives information to the electron (hey, Mr. Electron, that transmitter says you move that way). Some of the literature seems to imply that energy is just information and the electrons move because that’s what they were told to do and that’s what they do.

ANSWER:

Classical Electricity and Magnetism (E&M) is what mathematicians call a classical field theory. It is probably the most fully understood theory in all of physics. It is beautifully expressed and, in its most complete form, fits in perfectly with special relativity (which most physicists consider to be a classical theory replacing Newtonian mechanics). But in the early years of the 20th century a couple of experiments seemed to not be in agreement with E&M. First was the photoelectric effect and the other is Compton scattering. I will leave it to you to learn about these results in detail if you already haven't. In a nutshell, they provided evidence for photons. The photoelectric effect: Shining ultraviolet light on the surface of a metal results in ejection of electrons which have energies consistent with the existence of photons which have given all their energy to an electron in the metal. Compton effect: Photons can also be observed to interact with with an electron by simply scattering from it, thereby losing a some of its energy, and that is also observed. I don't know where you got "…a photon binds to an electron and gives information to the electron…", but it is not correct on any level. Photons interact with electrons. Finding the constituents of a field which are the smallest possible amount of energy that can still be identified as being in that field is indicative that the classical field can be "quantized", expanding it to a quantum field. This was not achieved mathematically until the late 40s, early 50s. In a quantized field the quanta (photons in your case) of the field may be thought of as the "messengers" of the field, the way that the field operates mechanically by communicating forces between electric charges. With quantization of the field, E&M is physics' most well-understood and complete theory.

Just because quantized fields indicate that you understand E&M more deeply does not mean that classical E&M should now be discarded. It is still appropriate to use classical physics to describe classical problems. If you want to describe antennae, you would be foolish to use photons to do so. Your conjecture "…that electromagnetic waves …really don’t exist…" is wrong and I hope that I have convinced you that the inclusion of photons is simply the completion of the theory of electromagnetism.

QUESTION:
Since the the 1970’s air conditioning has become standard on all new vehicles and in all buildings. In my car the AC is set on auto and my windows are closed so anytime the temp is warm the AC cools the interior. I know for every action there is an equal opposite reaction. Cool air inside equals hot air expelled outside. So are all these air conditioners all over the world contributing to global warming by emitting heat into the atmosphere?

ANSWER:

What an air conditioner does is move heat from inside to outside. There has been no net change in the overall energy of the atmosphere. (Heat is the flow of a certain amount of energy from one place to another. However, this doesn't just happen, it must be made to happen. You use electrical energy to cause this zero net change in energy, to move that energy. That energy is likely to come from an electrical power plant pouring tons of carbon into the atmosphere.

QUESTION:
how many H2O molecules does it take to make water ... with the characteristics of water, i.e., solvent, liquid, etc.?

ANSWER:

One of the main concerns of nanotechnology is to answer this question. Of course, there is no clear answer since there will be a continuous transition to the bulk value of each property being measured as more molecules are added. A very important consideration is the the ratio of the surface area of the cluster to the volume; this is because the molecules on the surface are interacting with fewer neighbors than those inside are.

QUESTION:
I am NOT a scientist, just a curious layman. I’m 74 years old; this isn't homework. On the question of singularities— doesn't quantum theory pretty much disprove them? I'm guessing that if energy (and pretty much everything) can be considered a wave function, then doesn't any wave have to have some amount of volume in order to exist? Or is this question just highlighting the well known problem with the existence of singularities.

ANSWER:
Nearly all physicists, like me, are uncomfortable by any discontinuity. Black holes as described by general relativity (GR) are singularities but that just indicates that GR, while a beautiful and successful theory in describing gravity in general, isn't totally right because there aren't really any infinities in nature. You are on the right track by thinking of quantum mechanics because the uncertainty principle basically implies the impossibility of a point particle (infinite density) at rest; on the other hand, there is no guarantee that the laws of physics are the same "inside" a black hole, in fact very likely not. Another example of an infinity in an otherwise successful theory in condensed matter physics; when phase changes of materials occur (e.g. freezing of water) in certain observable properties or their slopes may instaneously change, an artifact of the theory.

QUESTION:
I am a freshman in high school and in science class we were learning about planets, moons, and stars, and their sizes compared to the size of earth. I was wondering how we can determine the sizes of different planets, moons, and stars with them all being at varying distances away and all looking relatively the same size in the sky?

ANSWER:
They only appear to be the same size in the sky because your naked eye sees all distant objects as more or less point light sources. To be able to estimate their sizes you need a telescope to magnify what you see. So, suppose that you have a telescope which allows you see Jupiter, for example. Now, that telescope has nice crosshairs in it so you can first aim it at one side and then rotate it to the opposite side. You measure the angle (A in the figure). Now, the angle is related to the diameter D and the distance to Jupiter R by A=D/R. If you are a freshman you may not be familiar with the units A has in this equation. The units are called radians and there are 2π radians (π=3.1416 as you probably know) in a whole circle just as there are 360 degrees in a whole circle. So let's see what A is for Jupiter: D=1.428x10⁸ meters=88,732 miles and R=7.14x10¹¹ meters=444 million miles. So now we can calculate the angle, A=1.428x10⁸/7.14x10¹¹=0.0002 radians=0.0115 degrees. This is a really tiny angle and you can understand how you would see it as only a dot with your naked eye. We say that Jupiter subtends and angle from earth of about 0.0115 degrees.

Of course, this gets really hard to do for smaller objects like moons or very distant objects. Today, these are determined by getting up close to those objects with probes which we have sent to space for the past 70 or so years. But once you get closer you still use the subtended angle method.

Stars are harder still to measure. If the star is big enough and/or close enough, you can still do the subtended angle direct measurement method; this is done with a really powerful telescope where you can take a photo and then just measure the diameter of the image. There are other methods. Like if you find a binary star pair (two stars rotating about each other) you can measure the time it takes for an eclipse to occur and calculate the size of the star. If you can see a planet in orbit around a star you can infer the time it takes the planet to cross in front of the star.

I could go on, but maybe this is a good place to stop. Say hello to Ms. Welch from me!

QUESTION:
In Einstein's book The Theory of Relativity, I read that gravity is not a force but rather an acceleration or curvature of spacetime that pulls objects toward regions with high mass or energy density. If gravity is not a force but a field curvature, then why are physicists trying to find a theory that unites gravity with the other three forces? After all, if gravity were a force, black holes wouldn’t be able to attract massless photons. Why are we trying to unify something that is not a force with forces?

ANSWER:
I have answered this question several times before. I think that this earlier answer is the first time I answered it (just the paragraph starting with "…While researching this question…" is relevant for you). Unfortunately the link to the essay is no longer on line, but you can get the gist of it from my summary. The bottom line is simply that a mathematical theory, general relativity, can be interpreted in more than one way; in this case, GR could be interpreted as geometry or as a quantum field theory.

QUESTION:
I've never realised it before but the solar system is constantly moving. I saw an animation that almost scared me. It's the sun travelling through space and the planets following slightly behind spiraling around it. I found it amazing. My question is if we’re constantly moving, are there areas in space that could be warmer, cooler or have more radiation in it. I’m not sure how to explain it. Almost like an atlantic fish showing up in the pacific and we wouldn’t be able to toleratee it basically.

ANSWER:
First of all, whenever you talk about something moving you must specify "with respect to what?" I found the animation you saw, I think. [There seem to be several of them.] The video I watched was very "pseudoscience". For example, saying that the usual picture of the solar system is incorrect is, well, just incorrect. The usual picture shows how the solar system looks when viewed from the framework of the sun. If you now view from the framework of, say, the center of the galaxy, and look at the paths of the planets, you see those very pretty helixes; now the narrator waxes eloquently about the profound meaning of his "vortexes", calling what he sees the helical model of our solar system, as if this were some meaningful new law of physics. In fact if you just didn't draw the paths you would simply see the usual picture of the solar system moving past you. Also, since the solar system is rotating in an orbit relative to the galaxy center, most of the stars and other nearby stuff is also in about that orbit so that our neighborhood moves with us keeping our region of space more or less constant. And space in our region of the galaxy is more or less empty on scales much bigger than the solar system, so don't spend your time worrying about suddenly being a "fish out of water" (to use your metaphor). If you want to see a very good and scientifically accurate video about the path of the solar system throught space, I would recommend this PBS video.

QUESTION:
What is the optimal size (circumference) and weight of a ball such that it could be thrown faster than any ball currently thrown by a human. I'm assuming that using physics there is a way to determine the amount of force imparted on the baseball thrown by Aroldis Chapman, and then extrapolate that to optimize a ball size/weight such that his throw could exceed 105.8 mph with perhaps this optimized ball. (I have edited this question to make it more concise and well-focused, but it doesn't matter since there is no possible answer.)

ANSWER:
Your question is based on the assumption that if you know the force exerted on the ball you can calculate the speed that the ball would have. In fact that is false. Please go to the this link to see this discussed in some detail; you will see that the mass plays an important role in calculating the speed reached but the size is irrelevant. One thing which would make a larger ball matter would be how awkward it would be to throw it. Also, the human arm doesn't apply a constant force over the distance the ball moves while the throw is in progress and this would likely vary.

QUESTION:
How would I calculate for the μ of kinetic friction with the force of kinetic friction and the angle on which a wood slider starts sliding down an angled wood board.

ANSWER:
Actually you can't. Let me explain why. If you have an object on the board and lift the board so it makes a small angle with the horizontal, the object remains at rest. Why? Because static friction is holding it there. Static friction is what keeps a heavy box from budging when you haven't been able to get it moving. Imagine you have a heavy box on a level floor: If you push with 5 lb of force it doesn't move because the static friction is 5 lb in the other direction; if you push with 10 lb you maybe still woulb be unable to move it. However, at some point the friction is just not big enough stop you and the box starts moving and you can push it across the room against the kinetic friction force. If you have ever done this you know that the box initially "pops" away and starts speeding up. You can infer from this is that the coefficient of kinetic friction is smaller than the coefficient of static friction.

f_s≤μ_sN, f_k=μ_kN, and μ_k<μ_s.

In the figure there are three forces on the block if it is not slipping: its weight W, the static friction coefficient f_s, and the normal force N which the board exerts on the block. Since it is not slipping we can use Newton's first law to determine N=W_y=Wcosθ and f_s=W_x=Wsinθ. Now keep slowly increasing the angle until at some angle θ₀ the block starts sliding, accelerating down the incline. At this angle, f_s=μ_sN=μ_sWcosθ=Wsinθ, so μ_s=tanθ. So you can measure the coeficient of static friction this way, but not the coeficient of kinetic friction.

QUESTION:
I think you can make a microscope using a regular 3,000X microscope and two drops of water in a tube of fluorinated oil. Fluorinated can be used as a lens. They can adjusted the density, to be very near the density of water. You can search the internet: Can you make a lens with a drop of water in fluorinated oil? It says it is potentially possible. So I think you could put two sphere drops of water in a tube of fluorinated oil. Then you could look into the tube with a 3,000X microscope. And, you could put a layer of water in the tube, below the two drop lenses. Each drop could be 100X. So all together, you could magnify, 30 million. You could see viruses while they are alive, and through them. The big scopes, don't do both of those.

ANSWER:
I don't know much about microscopes. But doing a simple search I find that optical microscopes have maximum magnification of 500-1500x but that the usual max is about 1000x because at 1500x the image is enlarged but reveals no more detail than at 1000x. It is simple physics that geometric optics stops working when the object is the same order of magnitude as the wavelength of the light you are using. One rule of thumb is half the wave length of the light is the smallest object you can observe. For example, using 340 nm light (shortest wavelength visible light, violet) you could not get a well-focused image for an object smaller than about 170 nm and viruses are typically about 100 nm. That's the best you can do with an optical microscope, nothing gained by adding magnification.

QUESTION:
It is said that acceleration is equal to meters per seconds squared. a = m/s^2. I agree that the explanation is that it is the rate of change of a velocity. But why is the rate of change based on the time parameter - seconds. Why can I not say that the rate of change is per meter travelled ? What if I say: a = m^2/s. Meters squared per second, so the travelled distance increases per second. (?) So to use a counter-intuition argument: this is what we observe, time ticks the same amount of ticks (not 1/squared), and speed/distance increases

ANSWER:
You can certainly define a quantity which is the displacement rate of change of velocity, but it isn't acceleration which is universally accepted to be time rate of change of velocity.

QUESTION:
My brother was taught in a driving class that if two identical cars collide head-on at 50 mph, that the collision would be equivalent to a 100 mph crash. Now, I assume the instructor meant that the occupant would receive the same damage/injuries as if they had a 100 mph collision. I disagree, I believe the damage would be the same as if either car had hit an immovable wall at 50 mph.

ANSWER:
I think you misunderstand the instructor's meaning. The 100 mph collision is if the collision is still head-on but one car is at rest. I will assume the collisions are all perfectly inelastic, i.e., all cars and walls are stuck together after the collision. Linear momentum of an object is MV, mass times velocity; keep in mind that velocity is a vector, so if a car is going south with velocity 50 mph, a car going north with the same speed will have a velocity -50 mph. Also, if a system has zero net force acting on it, its momentum remains constant (momentum conservation). If you look at one of the colliding cars, its momentum is not conserved because during the collision whatever it is colliding with exerts a force on it which slows it down. But if you look at both the cars momentum is conserved because they experience equal and opposite forces.

The initial case the two identical cars going 50 mph in opposite directions. The situation:

Their total linear momentum is zero, the magnitude of each car's momentum is 50M
Since they are identical, they will crush at the same amount in exactly the same time.
Average force is the average time rate of change of linear momentum. Since magnitude of the time rate of change of momentum for both cars is 50M, the forces are all equal in magnitude comparing the two cars and their occupants.

Next, compare the case of one car going 100 mph and the other at rest:

The total linear momentum is 100M.
After the collision the momentum is 2MV where V iw the mass of both cars, stuck together now.
V may be determined using momentum conservation, 100M=2MV, so V=50 mph. So the at-rest car and the 100 mph car both have a magnitude of momentum change of 50M, the same as the 50 mph in opposite direction cars, so same forces. So you and your brother misunderstood the instructor's lesson.

Finally, what if a car hits an immovable wall?

The car has an inital momentum of MV which could be anything. If it hits a wall its final momentum is zero. So the change of momentum is MV. If V=50, the change of momentum is MV as you correctly deduce.

QUESTION:
I have a question regarding quantum tunnelling and trying to understand how a particle can cross a barrier and appear on the other side. Let’s take a helium nucleus for example that is fired at an energy barrier. From my understanding, the particle’s wave function provides for a non-zero probability of it appearing on the other side if it is fired at a barrier instead of just bouncing back as is most probable. I am having trouble conceptually understanding the mechanics of this. Does the particle’s ability to pass through the barrier have anything to do with the frequencies, ’spin’, coherence or otherwise of the subatomic constituents being in phase to allow it to traverse the barrier? If not please explain how it can pop out on the other side.

ANSWER:
I think it important that you understand the classical problem where a particle encounters a force which is represented by its corresponding potential energy. The figure plots kinetic (green), potential (red), and total (black) energies for a particle moving, initially, in the positive x direction. to remind you of what's happening, I have drawn a cartoon of the particle below but it doesn't belong on this graph because this is an energy vs. position graph. So the particle moves in from the left with a constant kinetic energy ½mv² until it gets to x=0 where it sees a linearly increasing potential energy which corresponds to a force in the negative x direction; therefore it begins to slow down and comes to rest momentarily. Now the force starts pushing the particle back, speeding it up until it is now moving in the negative direction with the same speed it came in with. For obvious reason the point where the total energy is equal to the potential energy is called the turning point; it is classically impossible for the party to penetrate beyond there.

Now, let's examine same problem but using quantum mechanics instead of classical mechanics. Again we imagine an incident beam moving toward a triangular wedge potential. Now we cannot just think about a particle because all particles are also waves and to understand how tunneling works we must calculate the particle's wave function using the Schrödinger equation. The particle can exist only if there is a solution to to the equation; in the classical case the particle could not exist below the potential energy because it would imply that the particle had an imaginary velocity. However, the Schrödinger does have have a solution in the classically forbidon region, it just doesn't look like a wave; it can therefore re-emerge as a wave beyond the barrier, hence tunneling. Because of nuances of quantum mechanics, you cannot actually directly observe the particle when it is in the "forbidden zone". When you are solving these equations you must pay particular attention to the boundaries where the wave function and its slope must be continuous.

FOLLOWUP QUESTION:
I’m looking over the diagrams you provided in your explanation, which are very helpful. Can you tell me if I’m getting it right: Regarding the quantum mechanics diagram it tells us that:

the particle’s kinetic energy has to be greater than the potential energy of the barrier/force, and when the particle traverses the barrier it loses some of its energy in the process and appears on the other side with a reduced kinetic energy.
The shroedinger equation solutions would show us that that the reason not all the particles cross the barrier is that the odds of crossing it are so small Would it also follow then that the greater the difference in the energy of the particle and the barrier, the more likely the odds of the particle(s) appearing on the other side and/or in greater numbers?
If a particle does actually appear on the other side of the barrier - will it have left a hole in the barrier (if it were a physical object) where it passed through?

ANSWER:

Note that the vertical axis is the energy axis, just as it was for the classical case. The horizontal line represents the total energy of the particle. The wave function, the solution to Schrodinger equation in each of the three regions, is drawn on this graph even though its amplitude is not an energy; rather you may think of the amplitude squared as the probability of finding the particle left, right, or inside the barrier. The particle has the same energy everywhere. The wave function of a particle is smeared all over this one-dimensional space, it is both untunnelled and tunneled but it is liklier to be untunneled. It may be more usefull for you to imagine ithe wave function as being a flux of many particles. then the wave function gives you the probability that an incident particle tunneled out.
The width of the barrier is what determines how likely tunneling is. So, for my triangular barrier, tunneling is less likely near the bottom than the top of the barrier. I would rather not get into the energy difference dependence since it gets too technical.
Think of it as a flux of particles. Now it gets a littll tricky because we have only been thinking of the particles coming in from the left; but particles which do not tunnel will be reflected and move to the left. So if you add the tunneled flux to the reflected flux you should get the incoming flux.

I’m looking over the diagrams you provided in your explanation, which are very helpful. Can you tell me if I’m getting it right:

QUESTION:
I am a curious 49 year old who has become rather obsessed with the nature of reality. My question, I think, concerns what counts as an observer. For example, if we consider a single calcium atom in the bone of my arm. It is bonded with other atoms.Are the 20 electrons in a superposition? Or, because they are bonded or close to or in a relationship with another atom(aka "observed), they are in a definitive position/velocity/etc...? I think where I am eventually going is an attempt to picture vibrations in fields moving through the universe. As I wave my arm in front of my face, should I imagine vibrations in superpositions moving through a field? Or should I imagine wavicles that have relationships to other wavicles moving through a field?

ANSWER:
You play fast and loose with the term superposition. Regarding your example of the calcium atom, we talk about the superposition of all the electrons. First let's talk about a single atom. Because the electrons are identical fermions in a bound system, the wave function of the atom must be antisymmetric*; for your purposes, this means that the electrons have lost their identities, you cannot imagine a bunch of electrons which you could label 1, 2, 3, …40. Every electron is a mixture of all the electrons in the atom. (This is the same as your superposition.) Now if you add more atoms to your arm bone, they bond to make the bone and it is the electrons which are responsible for holding the atoms together—all the electrons' interactions form a solid bone. In principle, the bone is one system which must be antisymmetrized and every electron has a "piece" of each other electron in your bone. Of course that would be impossible to compute and normally only "nearest neighbors" are considered which surely have the most significant contributions. And, indeed, the electrons in both of two atoms have pieces of all the electrons in the pair. I do not understand what your statement about "observed" means and they are certainly not "…in a definitive position/velocity/etc…" I am afraid everything after this makes no sense to me.

*An antisymmetric wave function is one in which you interchange the position coordinates of any pair of particles, the wave function remains unchanged.

QUESTION:
We pump very large amounts of water through kevlar hose and I'm having trouble grasping a concept. Our MAWP (Max Allowable Working Pressure) for 10" hose is 200psi. This is the maximum pressure we can safely operate at legally, our hose is rated for higher pressure but due to safety/legal concerns we are not allowed to pass that threshold For 12" hose our MAWP is 150psi. My question is Why with a larger diameter of hose does the pressure rating decrease? Keep in mind that our 10" and 12" hose walls are the same thickness.

ANSWER:
Poiseuille's law describes the flow of a viscous fluid through a hose of length L, radius R, viscosity μ, and volumetric flow rate F: P=8μLF/(πR⁴) where P is the pressure drop over the length. Everything else being equal, P∝R^-4 or P₁₂/P₁₀=(R₁₀/R₁₂)⁴=0.48, So P₁₂=0.48P₁₀=96 PSI. This demonstrates that it should be less. Since I do not know how MAWP is defined, I don't know if "…Everything else being equal…" is a correct assumption. If, perhaps it is assumed that the velocity v of the fluid is the same in each, then the flow rate would be F=vπR² and Poiseuille's law would be written P=8μLv/R². In this case you would calculate that P₁₂=0.69P₁₀=139 PSI which is essentially what your numbers are.

QUESTION:
Can two objects of infinitesimal size have different infinitesimal sizes from each other?

ANSWER:
The volume element in cylindrical coordinates is dV=ρ·dρ·dz·dφ. Volume elements having different distances from the z-axis, ρ, have different sizes.

QUESTION:
Clearly I am not a phycist but I am an archmchair enthusiast and tinker with theories. My first question to you is that if the universe started as a big bang, we image that there was a zero or one dimensional point which incredible pressure and heat. So, it’s degrese of freedom are pressure/girth, expansion, possibly rotation and heat. As it expands in the first second, before the devolution into the four primary forces, it naturally loses heat and pressure. If heat requires friction between atomic particles or the oscillation of electrons into higher magnetic shells, or in some way generates kinetic energy; How would this be possible if there were no forms before 1 x 10 (-17). Is it actually heat, or an allegory? I’ve heard it described as frenetic radiation, but at this early-stage all we had were photons which are massless mass. Therefore, how was heat generated or is it a placeholder for something that eventually became heat?

ANSWER:
As stated on the site, I do not do cosmology.

FOLLOWUP QUESTION:
Well that was an enormous waste of 5 minutes. It is not really a cosmological question. In the absence of particles, can there still be heat?

ANSWER:
I am not sure which of us wasted 5 minutes; must be you because I am wasting a whole lot more (just kidding). When I look at your question, I see a whole lot about the big bang and early times of the existence of the universe, it looks like cosmology to me. Now, I will answer your most recent question which need not be labeled cosmology.

The main problem is that you do not understand what heat is. Heat is not energy, heat is the flow of thermal energy usually from high temperature to low temperature. If you look at your question you will see that you talk about heat as being thermal energy. This is an important distinction. The flow of thermal energy is accomplished three ways.

Solids can conduct energy because thermal energy (in the simplest sense) is the kinetic energy of the atoms/molecules. So at the hot end the particles have oscillations larger than their neighbors which the transfer to their neighbors causing thermal energy flow (heat).
Usually gases and fluids move macroscopically, currents, winds, etc. This is called convection. And, yes, without particles there is no convention, hence no thermal energy flow (heat).
Finally, all objects are constantly emitting and absorbing electromagnetic radiation. Classically, EM radiation is waves, not particles. In that sense thermal energy flow (heat) can occur without the presence of particles—EM waves carry energy. Of course you could split hairs (and you seem like the kind of guy who would!), and say that EM waves are composed of photons which we usually call particles, but they have no mass.

QUESTION:
We are going to build a mobile chicken coop so they can graze over different parts around the farm. This is what we are looking at: We plan on having solar panels on the roof near the back(above the wheels). We also plan on having a battery bank to help heat the water in the coop or run fans depending on season. The question I have is would the weight of the batteries be better situated higher or lower in the back of the coop to make it easier to pick up by the doors?

ANSWER:
I will assume that you have some familiarity with elementary Newtonian statics. If you don't, you can just jump to the end for the answer; I like to always try to show how my answers are arrived at. The first figure shows your coop schematically before you try to lift it. The battery has a weight w which is a distance H above the ground and acting straight down. The center of gravity ⊗ is where all the weight W (except w) may be thought of as acting, and is a distance C from the axle of the wheel. The ground exerts upward forces of N on the wheel and F at the doors; the doors are a distance D from the wheel. We can find what the unknown forces F and N are:

Sum of forces is zero, F+N-W-w=0.
Sum of torques is zero (I choose wheel axle as axis of rotation), FD-WC=0. (Note that w and N exert no torque.)
Solve: F=W(C/D) and N=w+W[1-(C/D)].
Note that N and F do not depend on H.

Next you lift at the door end until the floor makes an angle of θ relative th the horizontal. All forces remain vertical (assuming you lift vertically). F now is the force you exert to lift the coop rather than the force the ground exerts at the doors. Now proceed in the same manner as before:

Sum of forces is zero, F+N-W-w=0.
Sum of torques is zero (I choose wheel axle as axis of rotation),
FDcosθ-WCcosθ+wHsinθ=0=F-W(C/D)+w(H/D)tanθ
Solve: F=W(C/D)-w(H/D)tanθ. Here is your answer: as H gets larger, F gets smaller. Therefore the higher up you put the battery, the easier it will be to lift the coop. It is easy to see this intuitively because note that you and the battery are both tending to rotate the coop counterclockwise, i.e., the torque from the battery is tending to "help you out".
I have one proviso here though: the angle you will need to lift it is likely small and the weight of the battery is very small compared to the whole coop, so in the real world the difference between low and high mounting is likely to be negligibly small.

QUESTION:
I have more of an abstract mind as opposed to a more mathematical sort, so I'm sure you can help me here. Regarding Hawking radiation, virtual pairs of particles and anti-particles manifest and annihilate. I wonder if that's due to the anomalous nature of 'zero'. The formula is there 1-1=0, particle plus anti particle equals nothing. However, by definition, nothing can't exist. Space is constantly trying to resolve an equation that can't be balanced, thus it forever 'boils' no matter how much or how little energy there may be. As a consequence to this, I imagine going well beyond the universe and Into the vast empty space. A place so cold that only these virtual parings can exist. In this space no force is greater than the other, and the 'tug' on the fabric of space is uniform. This makes me think about a sort of 'tug of war' where both sides are of equal strength. In that scenario it isn't either side that will give in, instead, the rope will eventually break. This 'cosmic 'tug' would eventually result in a massive reaction of chaotic expansion we think of as 'the big bang'. So, I believe there are multiple expansions allover the infinite void. This might explain why the cosmic background radiation image was so asymmetrical. I don't have any of the math to really back this concept up, but I'd be deeply grateful if you consider it.

ANSWER:
Your question makes no sense to me. But it is not too hard to understand Hawking energy without any math at all. You just need to understand what the Heisenber uncertainty princlple (HUP) is. Secondly, if some process can happen (that is, it is not forbidden by any physical law), it will happen although perhaps with an extremely low probability. Consider a particle-antiparticle pair suddenly pop into existence; are any physical laws broken? Yes, perhaps the most inviolable of all, energy conservation. The pair will have energy which was not there before they appeared, so they have violated energy conservation. But the HUP says that you cannot simultaneously know the energy of the pair and the time it exists; i.e., you can violate energy conservation as long as it exists only for the short time which the HUP says it can exist. When you understand these two, go to this link to see how Hawking radiation works.

QUESTION:
I have a question in the realm of mechanical physics (I think?). I'm a quad skater, old school style with four wheels aligned in square pattern rather than inline. Let's say I'm going down a straight flat road, build up speed and then coast on that speed without propelling myself forward. If I want to try and maximize my forward force/distance and reduce the effect of friction slowing me down, would adjusting the amount of wheels in contact with the ground make a difference? Part of me feels that if I can balance in two of the 8 wheels, there's less contact with the surface of my wheels and the ground, meaning less friction. But, the distribution of my weight across that surface of contact is far less, meaning there would be greater friction anyway. Which feels like a conservation of energy sort of thing but it's been a while since I've been in a physics class. So, my question is ultimately, is there a technique where I could reduce friction and maximize distance by adjusting the amount of wheel on the ground, or is physics laughing at me for trying to outsmart it?

ANSWER:
It is a fact that the frictional force is proportional only to how hard two surfaces are pressed together and depends on the nature of the surfaces (e.g., there is a lot less friction from steel sliding on teflon than steel sliding on rubber.) The force pressing your wheels to the ground is equal to your weight and it makes no difference the area over which you cause that force to be distributed. I would guess that the air drag would more important in slowing you down than the rolling friction of your wheels; the drag is proportional to the square of your speed and the area you present to the onrushing air, so you could minimize the area by squatting down, thereby maximizing the distance.

QUESTION:
Say you have 2 masses separated by a distance of 1 lightsecond. Mass A begins moving away from B (relative to their common fixed inertial frame). Will B feel a weakening gravitational force from A immediately or starting 1 second later? A from B?

ANSWER:
Although the speed of gravity has not been measured precisely, everybody assumes it is the same as the speed of light and many measurements indicate that it is so. Therefore, B will not observe any change in the field for a second, the same time he would have to wait to see A depart. A will observe the same sequence of events.

QUESTION:
If a rocket were to take off vertically from earth and travel in a straight line away from earth at 10000 mph continuously for an indefinite amount of time with unlimited fuel what would happen? Escape velocity is 24000ish mph so it shouldn’t get into outer space, but it is always maintaining a tremendous forward motion. Would it hit an imaginary wall and just hover at the same point in the sky as the earth rotates under it, would it maintain the same height but orbit with the earth or would something else happen all together? This is assuming that the nose of the ship is pointed into outer space and its trajectory never changes. For directional sake say it’s taking off from the very center of the North Pole or whatever the highest point on earth longitudinally is and it keeps flying away from the earth where would it go?

ANSWER:
You have evidently never had a physics course. I am going to assume there is nothing we need to focus on in the whole universe other than your rocket and the earth. (I ignore any forces from the sun, any planets you might encounter, any stars you might encounter, etc.) The escape velocity is the speed which the rocket must start with at the surface of the earth if your engines never fire again; this rocket would immediately start slowing down because of the gravitational pull of the earth on it and continue slowing down until at an infinite distance where it would stop. Now, your rocket starts at some constant value (it would not have to be large, maybe just 100 mph) because the force the engines give it are always adjusted so that it is exactly the same as the gravitational force trying to slow it down. Eventually you would get so far from the earth that the force trying to slow it down is maybe one ounce. Except for an occasional tiny pulse from your engine you could coast almost forever and use almost no fuel.

QUESTION:
Many experiments have been done bombarding protons with energetic electrons. Most results appear to be focused on how the electrons bounce or recoil off the protons or their imputed constituents, perchance to illuminate what's in the proton. Have any experiments been done focusing on how many (what %) of bombarding electrons pass directly through the proton without bounce or recoil? Or is it impossible to determine whether any do? I would appreciate any light you can shed on this. Perhaps how narrowly the beam is focused and the density of protons in the target mean one can't distinguish between electrons that missed all the protons in the target entirely, and those that might have passed through the protons.

ANSWER:
This is not a collision of classical billiard balls, it is an electrically charged electron interacting with a charge distribution. "Bounce or recoil" really has no meaning in this context because the electron has no option other than to interact with the field. What can happen is that the electron might exit without changing the the state of the nucleus; this is called elastic scattering and the electron could be scattered at any angle including 0°. In fact this is, by far, the likeliest thing to happen.

QUESTION:
I am a newbie to electricity and magnetism. I am confused about 1 electron charge is 1.64x10^-19 Coulombs and 1Coulmbs = 6.24x10^18 Elementary charge Unit. First of all what does it mean by elementary charge unit and what is the difference between two values 1.64x10^-19 and 6.24x10^18.

ANSWER:
First of all, one of your numbers is wrong, 1.64x10^-19 should be 1.602x10^-19. All electrons have an electric charge or e=1.602x10^-19 C/electron where C is the unit of electric charge, the Coulomb. Until recently the Coulomb was defined in terms of an Ampere (A) which is the unit of electric current; 1 C=1 A/s (s is second). An Ampere was defined in terms of the force which two long parallel current carying wires exerted on each other. This was always difficult because it was just too convoluted. Recently the SI unit of electric unit was redefined as the magnitude of the charge of something which contained as N electrons of charge e such that Ne would be equal to a Coulomb: 1 C/electron=Nx1.602x10^-19 C/electron. Solving, N=1/1.602x10^-19 =6.242x10¹⁸ electrons/Coulomb. (Coulombs per electron is the reciprocal of electrons per Coulomb.) This is now how the Coulomb is defined, the magnitude of the charge of 6.242x10¹⁸ electrons.

QUESTION:
Years ago, I used to listen to the NPR Saturday radio program "Car Talk" They always had a puzzler question that they asked (new puzzle) ant then answered the prior week's puzzel later in the program. A puzzler was asked that I never heard the answer to. I'll try to repeat as much of the details as I remember. A guy is floating in outter space and is "buck naked". He is within reach of two iron bars. One is magnitized and the other is not. How can he tell which bar is magnitized? They didn't say it, but I presume that the non-magnitized bar would be attracted to the magnitized bar, and vice versa. That's as much as I remember. I have wondered for years what the answer was but searches of NPR "Car Talk" website only turn up re-plays of the radio programs and I don't know the dates of broadcast where I might be able to listen to the next-week's program to hear the answer.

ANSWER:
Wow, I loved Car Talk. I really miss those two guys. I have had a couple other questions involving that show*. Now, the iron bars puzzler, two identical iron bars but one is magnetized, the other is not. How to find which is which. I must admit that this took a little thought to come up with the answer. Look at the figure below. If you grab the magnetized bar it exerts the same large force, always attractive, wherever you bring it up to the other bar. The pictures all look the same if the south pole is up. Now, if you pick up the unmagnitized bar, the same force will be found at the ends; but at the center there will be only a very small force.

Other mentions of car talk are at carstuck, pickup, saltwater.

QUESTION:
I would like know what affect solar flares have on the earth's orbit. Since we talk about solar sails as a means of propulsion, wouldn't that mean the solar wind pushes against the earth as well?

ANSWER:
In principle, yes. In the real world it is negligible because the earth is so massive; there would be no observable effect on the earth's orbit. A more interesting question is why aren't we getting mortally fried by all the radiation we should be exposed to? Actually, the earth's magnetic field serves like a shield to protect us from the solar wind. So most of the solar wind which would have hit us is deflected by the magnetic field and ends by just flying by. Still, some gets through and is responsible for the northern and southern lights. The intensity of the wind fluctuates a lot. I have read that, because the orbits of satellites which GPS uses must be known to extraordinary accuracy, the orbital calculations are crucial; the biggest errors in GPS results come from when the solar wind is big and it does affect these orbits and it is not as not possible to know when the satellites will be affected.

QUESTION:
I recently thought of this question that I could not confidently answer myself, and I have googled as well to no avail. I apologise if this question seems trivial, but I am hoping to be blessed with some of your knowledge. The question is: How does visual and tactile nerve input match in our perception? For example; if I tap the back of my hand, I see it happening instantly, and feel it happening at the same time. The pathway of course as you know is light hitting the back of my retina, then having that nerve impulse travel through the optic nerve towards the brain, while the nerve input from the back of my hand does the same as well. And apparently, nerve impulse conduction velocity does not even come close to the speed of light. Is it because of the differing lengths of the nerves involved allowing them to somehow match each impulse upon arrival in the brain? I thought of this, but it feels so simple and incomplete.

ANSWER:
This is really neuroscience, but I can give you the physics slant on the question. I looked up what the speed of nerve impulses: the speeds range from about 0.5 to 120 m/s. There are two kinds of nerves, myelinated and unmyelinated. The myelinated are in the peripheral nervous system and have generally much bigger speeds than the unmyelinated. So, let's just suppose the speed of the impulse carrying the information of feeling the tap is 120 m/s. The distance from your hand to your brain is about one meter, so the time is about 0.0083 s. I am sure that the time until your brain sees the tap is about the same. But I don't think that your brain can comprehend a time interval that small.

QUESTION:
I was watching a film starring Arnold Schwarznegger and when explaining the futuristic weapons (90s film), an actress said the bad guys were using railguns as rifles. When I looked it up on google, it said that the concept was not viable. If it can be done and built as a rifle; what challenges do we need to solve and if not what are the unsolvable problems ?

ANSWER:
I have already answered a question almost identical to yours. You can see that answer here. There are also links to several more questions "about sci-fi and video games guns" on the faq page.

QUESTION:
The fabric of spacetime itself is expanding exponentially. This would imply that eventually there will be an amount of spacetime with absolutely nothing within its volume. Eventually this area would be a complete void being too distant from any interaction with anything whatsoever. Not a neutron, not a photon or anything. Absolutely nothing. What kind of events would occur in such an environment?

ANSWER:
Actually, most of the universe already satisfies your criterion: intergalactic space contains approximately one particle per cubic meter. So you would have no trouble finding some cubic meter where there is nothing inside its volume, a perfect vacuum. Since there is no mass and no field inside your volume, the total energy in that volume is zero. In classical physics you would say that the total energy remains zero forever, energy conservation (no external forces doing work). But there is something which says that energy conservation can be violated as long as that violation exists for a short-enough time—the Heisenberg uncertainty principle which can be applied to energy and time, ΔEΔt≤&hbar; for example, if an M=5 kg bowling ball popped into existence that would be ok as long as it disappeared before a time Δt=&hbar;/ΔE where ΔE=Mc²=5x(3x10⁸)=4.5x10¹⁷ J and &hbar;=1.05x10^-34 J·s, so Δt=4.7x10^-51 s. So a big mass would exist an impossibly short time. What normally happens is that a particle-antiparticle pair, say an electron-positron pair, pops into and promptly out of existence. This is what is called virtual pair production and you cannot observe it directly. However, if you do some electrostatic experiment in a perfect vacuum you will not get the correct answer unless you account for the virtual pair production processes. This is why this is sometimes called vacuum polariztion.

QUESTION:
Someone originally posited the question: "Is it possible for two different balls of identical mass and velocity to have different kinetic energies?" At first I agreed with most others that it was impossible and used the KE equation as the basis for my opinion. After some thought however, I came up with a scenario that makes me question this as follows: Imagine two spheres of identical masses, one with an internal void space of some amount. Impart to them both a spin and a velocity, both of which being equal. Now if you were to measure the KE of the homogenous spere as it moves, it would be a constant as one would expect. With the spere with an internal void spinning as it moves along its path, would this not create a kinetic energy level whose vale plotted on a graph would look like a sin wave whose frequency would match the spheres rotation and the amplitude would be a relationship between the total volume of the sphere and the amount in the void space. I think I belive that thte kinetic energy problems I have done all assumed the center of the item I was calculating for as an easement of solution. I think that if the ball is psinning under these circumstanes you would have t consider the sphee having a 'moent' that you have to calcullate its position for insead of the center and thus the sin wave effect. Am I close to the solution or is it truly just the KE equation regardless of what is going on iwthin the mass?

ANSWER:
You have the answer, but it is not necessary to go through all the stuff about voids, sine waves, etc. You simply say that one sphere is spinning as it also moves with speed v. Even a ball not having a translational speed at all but spinning has kinetic energy. The total kinetic energy of rotation is K_rotation=½Iω² where I is the moment of inertia of the rotating object and ω is the angular velocity. So if one of the balls is spinning it has more kinetic energy than the other. If both balls are spinning with the same angular velocity, the kinetic energies could still be different because of the moment of inertia. For example, the moment of inertia of a uniform solid sphere is 2MR²/5 and of a hollow spherical shell 2MR²/3.

QUESTION:
My question is about how torque and power each contributes to the performance, and acceleration of a car. The acceleration should be equal to the force the wheels pushes the road with, devided by the mass of the car. The force is determined by the torque of the engine if i’m not mistaken. On the other hand, acceleration of a car can be described as the accumulation of kinetic energy, given constant mass of course. The rate of that is determined by the power delivered to the wheels. My hope is that you can give me a better understanding of «torque vs power».

ANSWER:
You are right, the force which accelerates the car is the frictional force between the road and tires. But, performance-wise, what matters is how quickly the engine can transmit energy to the wheels. In physics, think first of force: Force is what you need to do in order to give some mass an acceleration. Torque is the rotation analog of force, that which you must do in order to give some object an angular acceleration—for example to increase the angular speed of a gear from 5 rpm to 10 rpm. Torque is defined as product of the force F times the distance from r from the axis of rotation, T=rF; if you double the torque in a particular situation, you will double the rate at which the rotational speed (RPM) increases. Torque for a combustion engine usually envisions the drive shaft as the agent for delivering torque from the engine. But you really cannot just say that a particular engine has a particular torque because the torque depends on many variables such as the RPM of the engine and also on the gearing between the drive shaft and the wheels. Now, the engine delivers energy to the car at some rate. If something is accelerating in a straight line, the force F times the velocity v is the power P, P=Fv. For example, suppose you want to see what force is required to deliver P=200 horsepower (hp)=1491 watts (W) at the speed v=50 mph=22 m/s; F=P/v=1491/22=68 N=15 lb. Now the same thing can be done for delivering energy to a rotating object with a torque: The torque required to deliver power P to something with a rotational velocity ω is given by the equation P=Tω; in this equation P is in N·m/s, T is in N·m, and ω is in s^-1. I presume you would like to work in RPM and horsepower, so making those conversions,

P=(2π(ω/60)T)/746, with ω in RPM, P in hp, and T in N·m.

T=7114(P/ω).

So if you want to know how many N·m of torque you need to generate 200 hp at 1000 RPM, it is 7114x200/1000=1423 N·m=1050 ft·lb.

QUESTION:
If light slows down ( I appreciate it is by very little ) in water what speed does it travel at after leaving the water and entering a vacuum again? If it returns to c then where does the extra energy come from to speed up again? If it does not return to c then does that not mean that the only light in the universe travelling at c is light that has never been slowed down by anything else? This came up in a conversation with some friends ( we have weird conversations ) I have a sneaky feeling that it will be some special case because of the nature of light ( being massless? ) that means conservation of energy does not apply. Could you explain in terms a 55yo with a degree in maths ( not physics ) can understand?

ANSWER:
Let's start with a fundemental fact about waves: the speed of a wave v equals its wavelength λ times its frequency f, v=λf. I assume you know what wavelength and frequency of a wave are. Now imagine a very long string which we are wiggling at one end with some frequency some frequency f. So there will be waves of frequency f traveling down the string. The speed of waves on a string depends on the mass density of the string (grams per meter, for example) and the tension in the string, how taut it is. Now, suddenly the mass density gets bigger at some particular location on the string, so at that point the wave on the new part of the string, while still vibrating with the same frequency as the lighter part of the string, moves more slowly. As a result, the wavelength gets shorter. Note that no energy leaves or enters the wave when the speed of the wave changes. If the string becomes lighter again, the waves will speed back up again.

Light behaves the same way, slowing down in the medium and speeding back up at the interfaces with the vacuum. The difference is the mechanism for slowing the waves down or speeding them up; v=λf still holds. There is another way you can look at light: You may know that light waves of some given frequency may be thought of as a swarm of photons, each having an energy E=hf where h is Planck's constant. Since f is the same in and out of the medium, the photons always have the same energy whatever the wavelength of the corresponding waves.

QUESTION:
(the video holds my try to solve a textbook question with my intuition, it's not about how to solve a textbook question but rather it is that whyy my intuition is incorrect. Although it's not necessary to watch a video of me struggling to solve a problem and in the end get the wrong answer.) Sir it's not about the textbook question.. but I am asking that why is equivalent mass of a pulley system (the mass of a block by which we can substitute the whole pulley) different than, the (m1+m2) where m1 is mass at the left side and m2 is mass at the right side of the massless, frictionless pulley. Whyy is it not m1+m2 because I only see that much mass there on the pulley And in the equivalent mass formula, 4m1m2/(m1+m2) Where does some extra mass come from??? Well, if we measure the weight of the whole pulley system, it should be (m1+m2)g !! And not 4gm1m2/(m1+m2) right? see some say that the tension in the upward string that is holding down the whole pulley system will depend on the accelerations of masses happening inside the pulley, But Those are internal accelerations, and if one block goes up, the other one goes downand therefore balancing the accelerations. Can you please solve the confusion on the equivalent weight of the whole pulley being different then the actual weight of the pulley??? Why is it different?

ANSWER:
I have not posted the video because it is not informative except to tell me what the problem is. Also, this is not a tutoring site so I have no intention of trying to figure out what the questioner is doing wrong. So I have not even actually read his question in any detail. It is obviously wrong and I will outline how to get it done correctly. We are given that m₁ is at rest. The questioner then was correct in his reasoning that the tension in the thread connected to m₁ must be m₁g and it will be the same on both sides of the upper pulley. He then seems to think that the whole lower pulley sistem can be thought of as an object of mass m₂+m₃. This is only true if those two masses are equal. If they are not, then one accelerates down and the other up and all bets are off, you cannot do the problem that way. (Whoops, I just did what I said I wouldn't—tell him what he was doing wrong!) The first thing to do is to look at the lower pulley. There is an upward force m₁g and two downward forces which must be equal in magnitude which I will call T; the pulley must be in equilibrium, so T=m₁g/2. Next I will look at m₂. I will choose the +y direction to be vertically up and so

m₁g/2-m₂g=m₂a₂

where a₂ is the acceleration of m₂. Next I will look at m₃. I will choose the +y direction to be vertically down and so

-m₁g/2+m₃g=m₃a₃

where a₃ is the acceleration of m₃. But the magnitude of the two accelerations must be the same so a₃=±a₂. Since I have chosen +y to be in opposite directions, a₃=a₂=a. If you now solve the two equations (eliminating a) you will find [4/m₁]=[1/m₂]+[1/m₃].

QUESTION:
I'm a recent physics graduate, and I've realized that most of what I've done while studying physics is very theoretical, with the exception of some basic kinematics, circuits, and optics labs. However, if at all possible, I would like to perform an experiment to show Relativity (or at least the underlying postulates) in the real world for myself. After all, it's different enough from our everyday experience to be difficult to accept on an intuitive level, even if you learn the math and such. I figured I'd ask around and see if you had any ideas for such an experiment using either materials you can find at home or basic undergraduate lab equipment.

ANSWER:
Well, I couldn't think of a table-top experiment which would meet your criteria. So I did a quick Google search and found a question basically the same as yours on Physics Forums. I think you will find the ensuing discussion interesting but probably nothing which would fit your wishes.

QUESTION:
Hello, I'm another 11th grader and I'm trying to understand how EM waves work. The electric and magnetic waves are in the same phase. However, I was told that during induction the changing magnetic field induces the electrice one, and vice versa. From this I'd assume that B is shifted by compared to E, supposedly being its derivative, but that's not the case. I looked some things up on ChatGPT and the internet and found the following Maxwell equation: After getting a rough understanding of gradients and partial differentiation (took a long time), I concluded that the rate of change in the electric field equals the negative rate of change of the magnetic field (divided by the change in time). Which still doesn't make sense! If E and B change alongside each other (beside some constant, or directional difference), then charged objects would gain a constant magnetic field and freshly magnetized pieces of iron would gain an electric field, which they could keep, until their magnetic field declines.

Besides the former, I'm struggling at understanding how energy is stored within photons, if E and B oscillate at the same time. Due to how inductors and capacitors can be charged, I am of the opinion that electric and magnetic fields contain energy, like in this equation about energy density: However, if E and B reach zero at the same time, that would mean the photon has no energy, which it clearly has, when E and B are at their maximum. This goes against the conservation of energy, which means I'm really missing something. Can you help me in this regard?

ANSWER:
[This question had equations and a figure which I have edited out because I do not do LaTex symbol language. Also, because the questioner is not really prepared to understand electromagnetism at this level.]

I admire your quest to understand electromagnetism. You are already in a good place in understanding qualitatively. Until you have studied vector calculus and differential equations, however, be satisfied that you have already achieved way beyond your years. I will briefly give you some more information to help with your pursuit of this difficult subjece. The first equation you show is ∇xE=-∂B/∂t. This is not a gradient of E, rather a curl of E. So you need to understand the curl operator to understand the relative phases of E and B.

The other equation you quote (corrected here) is energy density of an electromagnetic wave,

U=½(ε₀E²+B²/μ₀)

in empty space. What you are missing is that this equation has been time-averaged. And before it is time averaged any spot in space would be oscillating between 0 and some maximum value; there is no violation of energy conservation as you suggest. Regarding the photons, this is not what the energy density is for them. If the frequency of the wave is f then the energy of one photon is hf where h is Planck's constant. Hence, for photons U_photons=Nhf where N is the number of photons per unit volume. In both cases U has units J/m³.

QUESTION:
For the past several months, I (an aerospace engineer with an interest in quantum mechanics) have been trying understand Heisenberg's frame of mind when he first came up with his crucial insights in 1925. Everywhere I look I see essentially the same statement: "His leading idea was that only those quantities that are in principle observable should play a role in the theory, and that all attempts to form a picture of what goes on inside the atom should be avoided." What none of the sources has conveyed to me is the following: granted that at the time Heisenberg developed his theory there was not a microscope powerful enough to see individual particles within the atom, but how did he conclude that seeing the positions and motions of the particles was "in principle" impossible? This was two years before he derived the Uncertainty Principle, which in any case is predicated on his original derivation of quantum mechanics. So without any of that "knowledge," he just arbitrarily decided that some quantities which had not yet been observed were in principle unobservable? I don't understand that. Can you help me?

ANSWER:
Your question is strange. What does "Heisenberg's frame of mind" even mean? Anyhow, that isn't physics, that's history of physics or maybe psychology of physics. In 1925 Heisenberg constructed the first mathematically rigorous theory of quantum mechanics; it was based on matrix algebra. Prior to that there was only the Bohr-Sommerfeld semiclassical description of what I refer to as quantum physics. Within a year an alternate theory, based of differential equations, the root being the Schrödinger equation which is a wave equation. The latter was generally more accessible to physicists and is the favored theory to introduce to students for that reason. Still, Heisenberg's quantum mechanics is very elegant but more difficult to calculate with. But the two were subsequently shown to be 100% identical in rigor and both describe the exact same physics. What is different is the mathematics used by each.

The uncertainly principle does not say some quantities are unobservable. It says that some pairs variables are 'conjugate variables' and therefore you cannot measure both to arbitray good accuracy, you cannot know both of them exactly. It is actually built right into quantum mechanics, so no 'arbitrary' decisions needed to be made.

I think you should give priorty to learning quantum mechanics rather than Heisenberg's state of mind.

QUESTION:
Is it possible for a person to jump/drop onto an impact force measuring device and have the initial impact force be less than body weight? Reference/context: Science world had a kinesiology exhibit, and one of the displays was about impact force measurement. The person gets weighed and then climbs up to a platform approximately 18-24 inches high. The person then jumps/drops down onto an impact force measurement device, and a display shows impact force by body weight. After a couple attempts I achieved an impact force reading of 1.2 times my body weight, and I feel I could have done better. Could that number have been below 1.0?

ANSWER:
I need to know what the device actually measures. "Impact force" is not a quantity in physics. A force is a force, but an impact is the product of a force times a time. For example, if you have a mass m, hit the device with a speed v and come to a stop in a time t, your average impulse over that interval is I=Ft=mv. where F is the average force you feel during the collision time. So you could write F=mv/t. I suspect that F is what this device measures. Assuming m is some fixed number, how could you control how big or small F is? If you increase v or decrease t, F will be bigger; if you decrease v or increase t, F will be smaller. An everyday example is consistent with this: If you fall and hit your head on a cement sidewalk, you might get badly hurt because your head stops in a very short time; if instead you fall on a thick foam mat, you would probably not get hurt because it takes a pretty long time to stop.

So let's see if there is a way to get an F which is less than your weight. Being a scientist, I must use SI units, but I will try to convert to Imperial units when it seems helpful. Mass is measured in kilograms (kg), length in meters (m), and time in seconds (s). So suppose that your mass is 100 kg, the height is 0.5 m, and the time is t. ; acceleration due to gravity is about g=10 m/s² so your weight is W=1000 N (a newton is mass times g and is 1 N=0.224 lb). So, your weight would be about 224 lb. since v/t is your acceleration, I finally come up with F/W=a/g where a is your average acceleration. So you just need to have the rate at which you are slowing down less than the acceleration due to gravity. If you drop from 0.5 m your speed when you hit the device is about v=3.16 m/s. So if you can make v/t=10, F/W=1.0, so t=0.316 s. If you make the time greater than about a third of a second you will get below 1.0; you can do that by watching what parachutists do when they land—go into a crouch as slowly as you can. Of course, I do not know how the measuring device works so my numbers may be off, but anything you can do to increase the time from when you hit and when you stop will lessen the average force you feel.

QUESTION:
When Hiroshima and Nagasaki where bombed with nuclear weapons in World War 2 there was total devastation. However within a relatively short period of time people were able to move back in and re-build the cities. However when there is a nuclear power plant meltdown like Chernoble or Fukushima the area around the power plant remains contaminated with radioactivity for thousands of years. Why is that?

ANSWER:
I have previously answered this question.

QUESTION:
I'm having trouble with Hawking radiation. It seems to me that if you separate a pair produced at the "event" horizon, with one half of the pair falling into the black hole and the other half escaping, you have added mass\energy to the black hole rather than subtracted from it. The problem for me here is the assumption that the mass\energy to produce the pair in the first place is somehow subtracted from the black hole?! To me this seems to presuppose that the mechanism for the pair production is well understood and the reservoir for the mass\energy for this pair production can communicate across the "event" horizon.

ANSWER:
AI have answered this question before.

QUESTION:
Red shift/blue shift—light traveling toward us (Andromeda Galaxy as an example of blue shift) has a shorter wavelength than stars/galaxies that are moving away from us. These light waves are part of the visible part of the electromagnetic spectrum (right?), but what about the parts that are invisible to the eye—ultra-violet, infrared, radio waves, etc? Do these waves change frequency in the above examples? (I know about Doppler radar and how it works, so is the same principle at play here?)

ANSWER:
All frequencies of electromagnetic waves are Doppler shifted.

QUESTION:
Black holes because of their mass and gravitational 'pull' suck things in and there is no escape (at least as I understand it), and this includes light. But light has no mass, so what is it about light that makes it susceptible to the gravity of a black hole?

ANSWER:
Our best theory of gravity is general relativity. In general relativity a photon, the quantum of the electromagnetic field, has energy. (Of course it does because the electromagnetic field itself has energy.) The energy E the photon has is E=hf where h is Planck's constant and f is the frequency of the field. If you throw a ball straight up in the air with some initial speed v, it initially has energy E=½mv², but the higher it goes the smaller v becomes until it is zero; so when it runs out of energy it stops and, like any mass, will fall back to the ground. The photon, however, still loses its energy but in another way: Because of the red shift of light moving away from a heavy object, the photon's frequency will get smaller as it rises. For most objects, a star for example, the photons will end up with a pretty small shift in frequency. For a black hole, however, the photon loses all its energy before it can escape, that is, it simply disappears. If you are wondering where the energy of the photon went, it went to increasing the mass of the black hole by an amount m, mc²=hf or the mass of the black hole has increased by m=hf/c².

Incidentally, There are two kinds of red/blue shifts. One is the one you are probably familiar with, generally called "Doppler shift" and caused by motion of the source or observer. The other, as discussed in this answer, is referred to as gravitational red/blue shift.

QUESTION:
I've just generally been thinking about this for a while, and it's difficult to get an answer from Google. I think if I understand things correctly, it's actually vital for our universe that the speed of light is constant, and this constant effectively creates the electromagnetic spectrum we observe. Light can't go faster. It can only increase in frequency as more energy is added. If there was no constant speed of light, it might be the case that all energy in the universe would evaporate instantly? Maybe? I don't know. I'm just trying to conceptualize what's going on. But what would be the physical mechanism by which the speed of light is constant? It must be a property of space? Do physicists have a solid understanding of what causes this?

ANSWER:
Go to the faq page. You will find "Why is the speed of light constant to all observers?"

QUESTION:
I am an 11th Grade Student, and my teacher currently teaching us newton's laws of motion, in the above image there is a pulley arrangement, with all surfaces smooth! my teacher was discussing about this in the last few minutes of school and i couldnt understand it. He said that the big 10kg block will move towards the left! what? why would it move towards left? it would need a force in the left direction right? but here, i dont see any force on the 10kg block which is on left direction... [followed by 364 unnecessary rambling words!]

ANSWER:
Once again I am faced with a question which violates site ground rules requiring "...single, concise, well-focused questions." Oh well, I have just edited it down and, rather than doing a complete solution, I shall only address the student's primary concern: when released, what happens to the 10 kg block? I always told my students to "choose a body" when starting a Newton's second law problem. I choose the pulley which, although not stated explicitly, must be frictionless and massless. The next step is to find all the forces on that body. Clearly the string is pulling on the pulley to the left by T₁ and down by T₂; because this is a massless pulley we can assume T₁=T₂=T. But, wait, is there anything else touching the pulley which might exert a force on it. Yes! This setup would not work if the pulley were not attached to the 10 kg block. The block must therefore exert some force on the pulley to hold it on the corner of the block, F in my diagram. If the block exerts a force on the pulley, the pulley exerts an equal and opposite force on the block, and that force has components to the left and vertically down. One more thing: The questioner indicates that, when the system is released and the two 5 kg masses start accelerating, the tension is the string is 5g. This is wrong.

QUESTION:
My question is about the way steam and heat works when food is cooking. For instance, if you cook food too long in a steamer, the food generally gets mushy if cooked too long. However if you heat up food in an airtight container in the microwave, the food does not get mushy if cooked for too long. In fact it will burn and get crunchy instead, despite having steam and lots of moisture in the container. Is there a physics-related reason that the two steaming methods vary so drastically in outcome if overcooked?

ANSWER:
In a steamer there is a reservoir of water continually adding stream to the inside of the pot. The cooking happens from the outside in. If you cook too long it will be overcooked. The same happens simply boiling the food. When you microwave you heat up the whole thing uniformly. And any steam that is in there came from the food itself, not nearly as much as in a steamer. But if you do that too long you will boil the water out of the food and it will certainly become "crunchy".

QUESTION:
I need to move an 80 pound speaker 28.5 inches above the floor using a dolly on a 7 foot long plank. How much force will I need to apply? (A very long question heavily edited by me!)

ANSWER:
Before you push there are two forces on the speaker, its 80 lb weight (red force in the figure) which points straight down, and the force which the plank exerts on it to keep it from breaking through the plank and which is perpendicular to the plank (the green force in the figure). There is also a frictional force which is parallel to the plank and pointing up the plane; because you have managed to put the speaker on wheels, this force is negligibly small for our purposes. When doing inclined plane problems it is useful to resolve all forces into components parallel and perpendicular to the plane; these are shown by the dashed forces. I won't go into detail about resolving forces here, but these two components are completely identical to the 80 lb force and show us that there is a 27 lb force trying to push the speaker down the plane and a 75 lb force trying break the plank. The plank is strong enough so the plank doesn't break; but if nothing (this is where you come in) intervenes, the 27 lb force will accelerate the speaker down the plane. If you now push up the plane with just a little more force tha 27 lb, it will go up the plane, as you wanted it to do, instead.

QUESTION:
You have 4 hot hard-boiled eggs you want to cool down quickly. They are in a pot. You throw in a handful of ice and then add cold water. Do you just cover the eggs with the water or fill the pot? Which is faster?

ANSWER:
I want to start out by noting this is an academic question. The way cooks normally cool hot foods quickly to halt the cooking is to plunge the hot food into an ice bath, water to which enough ice has been added to bring the temperature of the water down to 0°C (ice water).

This can get to be a very complicated system. I will simplify it to include only what is going on with the water, ice, and eggs. In other words, imagine that these three are enclosed in an insulating box through which heat cannot flow. This is probably a good approximation because cooling the food happens quickly enough that the interaction with the bowl and air in the room is minimal. What happens is:

ICE: There are two ways ice can absorb heat: (a) if the temperature is less than 0°C, it can warm up until it gets to 0°C, and (b) at 0°C it will absorb heat but not increase in temperature, instead melt. Most home freezers have a temperature of 0°F=-18°C, so initially, the way the ice will absorb the energy is by warming up; when the ice reaches 0°C it will absorb energy by melting.
WATER: The question seems to ask, mainly, how much water should we add. I will look at several amounts of water. I will start the water at 15°C, a typical temperature for tap water.
EGGS: The mass of an egg is about 50 grams and we have four of them. They have just come out of boiling water at 100°C.

I struggled with answering this question. The main problem is that the speed at which the temperatures change depends sensitively on things like how heat diffuses from the centers of the eggs into the water and the size of the pieces of the "handful of ice"; crushed ice will absorb energy much faster than a big block of ice. So I cannot answer the question, which presumes that only the amount of water determines the speed to come to equilibrium—there are many more factors which play. I think it likely that it would take longer times for large amounts of water to come to equilibrium, so I could glibly suggest that the answer is the smaller amount of water is faster; but the problem has too many variables to know for sure. What I have decided to do is demonstrate the physics of how the final temperature T is achieved and work that out for one example: I will find the minimum amount of ice we must add to 1 liter of water so that the equilibrium of the ice/water/eggs system will be T=0°C.

Heat is just the flow of energy. If an object has a mass M and a temperature T₁ and an amount of heat Q is added its temperature increases to T₂. Then the change of temperature is proportional to Q, Q=MC(T₂-T₁), where C is a proportionality constant called the specific heat of the material absorbing energy. If the object is instead having energy taken from it, T₂-T₁<0 so Q<0. If a solid is melting it is absorbing energy but not getting warmer; in that case one needs a constant called latent heat of fusion, L which is the energy absorbed per kilogram of material melted, Q=ML. So, the following is the computation.

ice: C_i=2090 J/(kg·°C),
water: C_w=4182 J/(kg·°C),
eggs: C_e=3180 J/(kg·°C),

and the latent heat of fusion for:

ice: L=3.33x10⁵ J/kg.

I will calculate the amount of ice which is needed for the final temperature to be T=0°C starting with 1 kg (a liter) of water.

M_w=1 kg of water at T_w=15°C,
M_e=0.2 kg of eggs at T_e=100°C, and
an unknown mass M_i of ice at T_i=-18°C.

So, the equation (which is simply energy conservation) to solve is:

C_iM_i(T-T_i)+C_wM_w(T-T_w)+C_eM_e(T-T_e)+M_iL=0
=2090x18xM_i-4182x1x15-3180x0.2x100+3.33x10⁵M_i
=3.71x10⁵M_i-6.27x10⁴-6.36x10⁴, so solving,
M_i=0.34 kg.

Note that this is a linear equation for M_i as a function of M_w.; this is shown in the figure. So, 1 kg will about cover the eggs and about a third of a kilogram of ice will totally melt when T reaches 0°C. If you start with more ice, there will simply be ice still unmelted in the pot when equilibrium is reached.

QUESTION:
I am currently studying for my physics exam. A student in the group chat asked us all a question and it got us stunned. Can you help? Please be aware that this is NOT a homework question, we all just share different opinions on the answer.

Question: A container with very thick/sturdy walls is on a frictionless surface. Inside the container, on the left side, there is a cannon with cannonballs. The cannon fires all its cannonballs against the right wall, where they fall directly to the ground upon contact (they do not bounce back). What is the effect on the container?

a) The container stays in place
b) The container moves to the right
c) The container moves to the left
d) Not enough information

I personally think the answer is A, but a lot of people are saying C, and others also say D. I'm lost.

ANSWER:
See the figure. I assume that the box and cannonball are rigidly attached and have a combined mass of M; the cannonball has mass m. Before the cannonball is fired, the whole system is at rest and so its linear momentum is zero. Afterwards the cannon ball is moving (relative to the ground which is where the momentum was previously zero). But the total momentum must still be zero (conservation of momentum) because there are no horizontal external forces on the system. So, mv-MV=0 or V=mv/M; when the cannonball hits the wall its horizontal linear momentum becomes zero. There are still no horizontal external forces so the motion of the box must also stop. Clearly the answer is c).

There is another way you could look the problem. Since after the the cannon is fired the cannonball, previously at rest, now is moving. So the cannon/box must have exerted a force F on the cannonball. But Newton's third law says that the cannonball exerted an equal and opposite force -F on the cannon/box etc.

QUESTION:
I am a high school senior who writing software for the avionics of our school's model rocket for the TARC rocketry competition. I have taken AP mechanics (Calculus based) and had a difficult time figuring out the projectile path of a rocket. After burnout the rocket will coast to a target apogee (around 800 ft) at subsonic speeds. The rocket has onboard an accelerometer (plus gyro), a really accurate clock, and an altimeter (with barometer). I was wondering if there was a way to predict the exact apogee of the rocket based soley on the accelerometer readings and the clock (while knowing all the constants: mass, drag coefficient, etc.)? My team and I have tried several ways to do this but we get stuck in dependency loops. Drag is based on velocity while velocity is based subsequently on drag. We tried using energy, momentum, even modified kinematics (for changing acceleration), none of which worked. What we need is a single function that takes in current vertical acceleration and the constants and predicts a delta height which we can added to the current height to predict apogee. Also, since air pressure is part of drag and that changes based on height according to a function, it would be nice if we could put that into the predictive function. I really don't want to make this question too long but if I could add one more layer of complexity. We are using air brakes to control the altitude to achieve a target apogee. We planned to run the above mentioned predictive function in a loop and reverse engineering it to create another function to find out how much more drag area we would need to achieve the target altitude since the airbrakes are radial flaps that just increase our cross sectional area.

ANSWER:
First of all, you have not given me nearly enough information about what you are trying to do. Is the time the rocket is burning part of your calculation? It is very different (and harder) than the trajectory after burnout for several reasons. Is the trajectory one-dimensional, i.e. straight up and back down? Or does the motion of the rocket also have a horizontal component? Does the accelerometer give you the direction of the acceleration vector? And, if not, it is not very useful. Do you know the mass of the rocket before and after burnout?

All the above is necessary for me to give you any guidance. But the way you are approaching the problem, common among newcomers to physics, is all wrong. One of the most important things to consider when solving a problem in the real world (not the ideal models of on-the-blackboard problems in a physics classroom) is to determine what is important and what can be approximated without serious error. Except in textbook problems, you should never expect to be able to get the appropriate equation in just one step, you need to make sure you understand things step by step. You should be starting out by imagining you are doing a textbook problem—ignore drag and ignore the change of mass during the burn. Then you might put in drag. Here is a little tip: A fairly good approximation for air drag D is D≈¼Av². Here A is the area presented to onrushing air and v is the speed; the vector D must be opposite the vector v. The catch is that you must use SI units here and D will come out in Newtons. The estimate for D does not look dimensionally correct; that is because lots of stuff is included in that little factor ¼ like the fluid is air, the density is atmospheric pressure at sea level, the shape is a generic sphere, etc. Including D in your calculations will tell you if it is actually very important for your problem.

That's probably all I should do at this point. If you answer my questions about what you are trying to accomplish, maybe I can give you some other instructions.

By the way, if you know the acceleration as a function of time you should be able to determine the position and velocity as functions of time if you know the initial position and velocity. It would require numerical integrations because a will not be constant.

REPLY:

I am only simulating the rocket after burnout. As such, the problem is more like finding the trajectory of a cannonball mid-flight.
Like a cannonball, the projectile will have velocities in the horizontal and vertical direction, however only the vertical component is needed as I am trying to predict apogee.
The problem is that at an arbitrary time during the ascent of the cannonball, we will be given vertical acceleration (drag and gravity with direction), current altitude, and time (if that is needed though it might not be), and we must determine the apogee time and altitude.
Constants like mass (after burnout), cross sectional area, drag coefficient, etc. are known.
Drag is quadratic (v^2) Without drag, the apogee is simply found with kinematics (constant acceleration), but since drag is a changing force that imparts a changing acceleration unlike the constant 9.81 of gravity, no one on the rocket team could figure it out.

Also, I have found other equations online that include drag and give altitude as a function of time, but the equation we need should just give out a number for max altitude based on the given values stated above. That's why we tried using energy formulas to solve the problem since at the top there is no kinetic energy, only potential (again only vertical components).

ANSWER:
There is a good reason why you are having so much trouble: You are high school students who, at most, have had introductory differential and integral calculus. As I will show, this problem will require differential equations. And not just simple linear differential equations, but coupled second-order nonlinear differentials. Such equations can only be solved numerically, almost never analytically. I have given a little thought and have thought of a way that just might work for you. It won't be easy. One thing before we get started: You said "only the vertical component [of the velocity] is needed." This is not correct because to get the behavior of the vertical component you must also know the horizontal velocity.

One thing which I am not clear on is what you are given. It is imperative that you know the initial velocity, both magnitude and direction, to solve this problem. When you say "we will be given vertical acceleration" I think you must mean "we will be given acceleration" because the acceleration is not vertical. When you say "drag and gravity with direction" I assume that the drag force, both vertical and horizontal components are given. The drag force is usually written
    D=-C|v|v, so
    D_x=-C√[v_x²+v_y²]v_xD_y=-C√[v_x²+v_y²]v_y From these equations you can deduce that
    v=√(D/C) and
    v_y/v_x=D_y/D_x=tanθ
where θ=tan^-1(D_y/D_x) is the angle v makes with the horizontal. From the initial drag you can now get the two components of v,
    v_x=√(D/C)cosθ and v_y=√(D/C)sinθ.
So this is how you get the initial values (t=0) values of the components of the velocity. I will choose that the initial values of x and y are both zero, i.e. I choose the origin of my coordinate system to be at the initial position of the rocket.

So now we are ready to do some physics. What's key in classical mechanics? Newton's second law, F=ma.
    m(d²r/dt²)=-C|v|v-mg1_y
where 1_y is the unit vector pointing vertically upward, m is the mass, r is the vector pointing from the origin to the rocket at time t, and g is the acceleration due to gravity. I realize that vector calculus is something you have not yet studied, but it's not so bad because we just write it in component form:
    m(d²x/dt²)=-C|v|v_x=-C√[v_x²+v_y²]v_xm(d²y/dt²)=-C|v|v_y-mg=-C√[v_x²+v_y²]v_y-mg.
These are second-order (second derivative), non-linear (square root), coupled (v_x and v_y both appear in both equations) equations. But we can make them much more tractable by noting that (d²x/dt²)=(dv_x/dt) and (d²y/dt²)=(dv_y/dt). Finally we have
    (dv_x/dt)=-β√[v_x²+v_y²]v_xand
    (dv_y/dt)=-β√[v_x²+v_y²]v_y-g
where β=C/m. Now we can't solve these analytically but we can do it numerically (that's where you come in!). This will just involve arithmetic.
   Δv_x={-β√[v_x²+v_y²]v_x}Δt and
   Δv_y={-β√[v_x²+v_y²]v_x-g}Δt.
So here is your computer programming problem. You know β, g, v_x, and v_y at t=0, they all are just numbers. Choose some Δt, say Δt=1 s and then you can calculate the changes in the velocity components. Now you know the (approximate) values of the component velocities at t=1 s. Now you calculate with those and get the values of the velocities at t=2 s. Continue until you find the time when v_y≈0; going past this t v_y will become negative. You will probably need to try several values of deltat until it is small enough that the final answer doesn't change (you are, after all, simulating what you would get with infinitesimal Δt). After you find the total time to the top, you still haven't answered where it is. What you need to do is to integrate into your program calculation and tabulation of Δx=v_xΔt and Δy=v_yΔt at each interval, and calculate running sums. The final sum of Δy is your answer.

I know that this is a lot for a high schooler to digest. But I have tried to simplify as much as I could. As I warned you at the beginning, when you have air drag to contend with, you have a much more complicated problem than with constant forces. If you are able to pull it off I would love to see your results.

ADDED THOUGHT:
To check that I have not made an algebra error somewhere I have written a PYTHON code to do the calculations I suggested above. I randomly chose reasonable boundary conditions, β=0.001, v_x(t=0)=100 m/s, v_y(t=0)=200 m/s, g=9.8, and Δt=0.25 s.

Shown are the velocity and the position components as functions of time and the path of the rocket, y as a function of x.

QUESTION:
We have read for years that the universe is composed of certain percentages of visible matter, dark matter, and dark energy. I can understand that there are ways of reasonably estimating the amounts of visible and dark matter. They are described for us in published articles. How does anybody estimate the amount of dark energy? And what is it anyway? Is dark energy expected to have all the characteristics types that regular energy has - kinetic, thermal, etc.? All of the articles that I have seen simply present a percentage associated with dark energy. There don't seem even to be some typical characteristics of dark energy that would allow us in the public to form a concept of what it is. Falling back on the Classical Physics model, dark energy has to be the ability to do work on dark matter, if that helps! Regardless of what dark energy actually, is one still wonders: how is it estimated?

ANSWER:
Site ground rules clearly state that I do not do astronomy/astrophysics/cosmology. I can tell you though that it is not clearly understood what either dark matter or dark energy actually are.

FOLLOWUP QUESTION:
But as a last thought, they tell us that dark matter can be evidenced by its observed effect on real matter. And then they quote actual percentage estimates for the amounts of dark matter and energy. And yet dark energy, I think, has not even been detected. There must be a process used to arrive at these numbers. That's what I am really after. Can you direct me to anyone or any organization that I can communicate with to find out just how it is done? My thought is that, if I can understand the process, I can more clearly Understand what "they" think these things are. Maybe not.

ANSWER:
Dark matter is thought by most astrophysicists to be some kind of particle which interacts only via gravity. It is very hard to "observe" something if you cannot interact with it using electromagnetism. There are numerous anomalies in the universe which seem to not be described by our current "best" theory of gravity, general relativity. You may make a model of hard-to-detect particles with mass but which only interact with the rest of the universe via gravity, but if that model works it does not mean that you have "detected" dark matter, you have only fitted the data, not necessarily one which tells you what dark matter is; at this level it is empiricism. It may be that dark matter is not stuff, but an indication that we don't understand gravity as well as we think we do.

Dark energy is a different kind of thing. You say that it "…has not even been detected…" That is not correct, it has been "detected" just as surely as dark matter has. The evidence for it is that very distant galaxies are observed to be not just moving away from us but accelerating. In other words you can assume that there must be some universal repulsive force pushing the universe apart. In my opinion, dark energy is in just as detected as dark matter is.

There is an interesting back story about dark energy. Circa 1918, when Einstein was putting the finishing touches on his theory of general relativity, it had not yet been discovered that the universe is actually expanding; it was assumed to be static. That presents a big problem since gravity is a long-range attractive force which should have been causing the universe to be collapsing. So he added a term to general relativity to hold everything in place and it became called the cosmoligical constant λ. Much later in his life, when it was discovered that the universe is not static, he said that λ was his "biggest blunder". In 1998 when acceleration of distant galaxies was observed, new life was given to λ! If you are interested in delving into this more deeply, check out this link.

QUESTION:
I was wondering if a feather and a bowling ball actually pull towards the earth at the same rate? Why would the gravitational fields produced by the feather and the bowling ball not increase the rate of their individual attraction towards the earth?

ANSWER:
First we need to specify that this notion requires that there be no forces (like air drag in the real world) other than gravity acting, the earth must be assumed to be enormously more massive than the feather and bowling ball (which is true), and the gravitational force which the earth exerts on the objects is uniform (which is true only close to the surface of the earth). Also, you seem to be a bit muddled as to what is going on. The reason that the two masses accelerate is that each feels a force due to the gravitational field of the earth, it has nothing to do with the gravitational fields of the objects themselves. You seem to be thinking that the bowling ball feels a bigger force and that is true; the force anything feels (which is called the weight of that thing) is proportional to (depends on) its mass. But the effectiveness of that force, what acceleration (the rate at which its speed will change while falling) it falls with is inversely proportional to its mass. So, if the mass of the ball is 1000 times bigger than the feather's mass, the force on it will be 1000 times bigger. But, the more massive an object is, the more force required to give it a certain acceleration is. It just so happens (because of Newton's second law that the force necessary is proportional to the mass. Therefore although the ball has 1000 times the mass, it takes 1000 times as much force to accelerate it as it takes to accelerate the feather. The acceleration is g=9.8 m/s². If you want a more mathematical discussion, see an earlier answer.

QUESTION:
One of the issues YouTube physicists discuss regarding interstellar travel is that the high speeds we would achieve would render the spaceship more vulnerable to constant bombardment by particles, gases, and dust, requiring a thicker hull in the direction of travel. Assuming interstellar travel is possible with current or near-future tech (not a given), is it true that more particles and gases would impact in the opposite direction of travel? That would seem to assume that most particles are more or less at rest in the frame of reference we are coming from with Earth. Wouldn't they all be going at random velocities up to near the speed of light relative to us? Within the solar system, are particles mostly at rest relative to Earth's frame of reference? Within the gravitationally-bound Local Group?

ANSWER:
Unfortunately, the space between planets, stars galaxies are not true vacuums as we often think of them. So, inevitably, atoms and molecules will be present in varying densities, sometimes just moving around in relative vacuum, sometimes congregated in clouds. And you are right, they will be moving in random directions with a distribution of speeds, but the speeds are all very small compared to the speed of light, c. But a ship traveling to distant stars would have to have speeds which are not so tiny compared to c. And to someone on that ship, all the particles will be moving mainly at speeds close to the speed which the ship has. When you say "mostly at rest" you do not define what that means. What is true is that their speeds are very slow compared to c; if that's what you mean by "mostly at rest", then yes most particles which you encounter will end up coming toward you with a speed very close to the speed you are traveling.

FOLLOWUP QUESTION:
Thank you for answering my question!! I have refined it to clarify the point you brought up.

We are measuring the velocity of the spacecraft relative to Earth's frame of reference. Therefore, we are measuring the velocities of the particles it might encounter from the same frame of reference - Earth's. You can't compare any particular velocity to the speed of light without a frame of reference. You must have an observer with its own frame of reference to measure the speed. Why should we assume that particles outside the Solar System or Local Group travel at very low speeds compared to the speed of light from our frame of reference? Shouldn't we see the same number of high-speed particles bombard the ship from all directions, regardless of its velocity relative to Earth? Does the idea that the spacecraft will encounter more particles in the direction of relativistic travel rely on the fact that all particles in the Local Group are gravitationally bound together?

ANSWER:
My bad! I should have specified my reference frames. Let's call the earth our "rest" frame. First some numbers:

Typical atom/molecule densities most places in our galaxy is 10-100 particles/cm³
Typical atom/molecule densities in intergalactic space is 1 particle/m³
Typical temperatures of most places in our galaxy is 10-100 K
Typical RMS speed of molecules is slow, e.g., CO₂ at 10 K is about 300 m/s
So if we say that probably particles with speed 10 v_RMS are the fastest with any appreciable number, those have speeds 3000 m/s=10^-5c.

Since the biggest speeds relative to the earth are so tiny compared to the speed of light, they look like they might as well all have been at rest compared with the motion of the ship. Therefore, on the ship with some velocity U, every atom or molecule will be seen moving with a velocity very close to -U.

Suppose that you are driving in a forest at 50 mph. Every tree from your frame of reference is moving backwards at 50 mph, not just trees in front of you.

QUESTION:
Since the discovery of the Higgs Boson, which is said to be what makes matter essentially makes things be physical. Then wouldn't it make sense that there is indeed a graviton? In where the particle is in a wave function until check and collapsed into a particle? Not unlike the double slit. Couldn't this be the case? And it is my opinion that this could fit nicely into an current theory.

ANSWER:
I don't understand hows Higgs or double slits have anything to do with whether "…there is indeed a graviton…". I will assume that the main issue is you want to know is whether "…there is indeed a graviton…". As you may know (see the next question) general relativity is generally depicted as a geometric model, deformation of spacetime by mass. So gravity is not a force, merely geometry? If that is the case, why seek a quantum theory of gravity where the quanta (gravitons) would be "messengers of the force" when there is no force, only geometry? But a few years ago I came across a short essay which made the whole thing clear to me. A theory does not have to have only one framework of characterizing it. For example, when quantum mechanics was in its infancy, two competing mathematical frameworks were used, both of which seemed to come to the same result when calculating: Matrix mechanics was formulated by Heisenberg, Born, and Jordan; differential equation formulation of quantum mechanics was introduced by Schrodinger. The two were eventually shown to be identical. Similarly general relativity has the properties of a field theory and can be interpreted as having forces and being quantized. Hence, there should be a theory of quantum gravity. If you want to read more see my earlier answer and be sure to look at the essay linked there.

QUESTION:
Hi. I’ve been pondering gravity lately. Why is it that -- although Einstein's general relativity best explains what we call "gravity"-- one still sees statements such as "all masses attract each other – you and your pet, the earth and the moon . . ." etc., etc. What magical "attraction" are they talking about? Also, could you recommend a good illustration of what actually happens when an object near the earth "falls"?

ANSWER:
The theory of general relativity (GR) is quite complicated mathematically. To understand a complicated physical theory requires that we have a qualitative explanation of some effects predicted to visualize. GR is a theory which uses advanced geometry formalism, a four-dimensional spacetime (time being the fourth dimension). It therefore becomes a theory we think of as a geometrical theory and envision massive objects to deform the spacetime around them resulting in attraction. But, there's a problem—since we live in a 3-dimensional world we really can't visual a 4-dimensional world. So what is usually done is to imagine a 2-dimensional plane which, if we put a bowling ball in it, gets deformed. If there is a marble nearby it will roll toward the bowling ball; I often refer to this as the trampoline model and it is a cartoon not to be taken too seriously but to provide an example of an object deforming some space around it. You can also give the marble a little shove and it will orbit the bowling ball as shown in the figure. The marble also deforms the space around it but since it has so little mass compared to the bowling ball, it is not apparent in the figure.

QUESTION:
Recently I’ve got interested in one question. Lets assume we have 2 balls with same mass in space on some distance from each other (no other objects around). At the initial moment of time they are staying still (velocities = 0). Then, due to gravitational attraction, they start moving toward each other. At some moment they collide and collision is elastic — so they move away from each other, slow down and then again start moving toward each other (then collide ets.) The question is — will they bounce forever (because of conservation of energy) or they eventually, like, stick to each other?

ANSWER:
The problem with your question is that you ask for an answer in an ideal situation, perfectly elastic collisions and perfectly empty space. Under those conditions, the two balls would bounce forever. The fact is, though, that no macroscopic balls can collide perfectly elastically and no vacuum is perfect. So in the universe we live in, the balls would lose a little energy every time they collided and quickly stop, "…stick[ing] to each other…". And even if you had perfectly elastic collisions, actual "empty space" (intergalactic space) contains about one molecule per cubic meter; it might take eons, but eventually the balls will collide with those molecules giving some of their energy to them and eventually sticking together. And don't forget, there are lots of photons zooming around which could add or subtract energy when they hit the balls.

QUESTION:
A while back I was reading an article about the electromagnetic spectrum, the visible light spectrum specifically. And a question raced through my mind. If I understand correctly the colors we perceive are based on reflective properties of objects in our environment. Certain reflective properties reflect different wavelengths of light and we perceive them as colors. My question is in the absence of all light does anything truly have a color? And if so what do you think that looks like? An infrared photo is the closest that I can imagine.

ANSWER:
If a tree falls in the forest and there is nobody nearby to hear it, does it make any sound? Your question is similar—both are about semantics, not physics. If you say that sound is your perception of it, the tree makes no sound if there is nobody to hear it; on the other hand if you say sound is the presence of compression waves traveling through the air, there is sound. Color is trickier, though, because the color you perceive depends on how it is illuminated. We usually think of white light as a distribution of all the wavelengths of sunlight. So just saying something is red is a grossly undetermined description because something red in one illumination may look pink in another. So physics avoids this confusion and simply uses the word color in a qualitative sense, just like you might describe something as warm to denote temperature. So if you want to know what "color" something is you specify the distribution of wavelengths emitted from an object when it is illuminated by a different distribution of wavelengths.

QUESTION:
I was trying to bounce my Walmart ball at the ground and have it bounce off the ground, hit the ceiling, bounce off the ground and then hit the ceiling again. I noticed that no matter how hard I threw, it would never hit the ceiling twice. I assume it is impossible for a physics reason and would like to know specifically what. The ceiling is about 8 ft high, ball is about 42 inches around.

ANSWER:
The "physics reason" is that your arm is incapable of giving the ball enough energy. Each time the ball has a collision with ceiling or floor it loses a fraction f of its energy, so if its incoming energy is K, its outgoing energy is K(1-f); as it travels from the floor to the ceiling it loses energy V=mgh where m is the mass of the ball, g is the acceleration due to gravity, and h is the distance the center of the ball travels; as it travels from the ceiling to the floor it gains energy V. What I did is to calculate the energy step-by-step to find the minimum initial energy K₀ needed for the final kinetic energy to be zero, just as it reaches the ceiling for the second time. Then I solved the equation I got for K₀. The result was

K₀=V(F²-F+1)/F³ where F=1-f, the fraction of energy retained in a collision.

The energy is just the kinetic energy, K₀=½mv²=mgh(F²-F+1)/F³

so v=√[2gh(F²-F+1)/F³].

So let's look at your case, the radius of the ball is about 6" so h is about 7'=2.1 m, g is 9.8 m/s², and I will guess that F is about 0.5 which means half of the energy is lost in each collision (I dropped a similar ball and it bounced back about halfway from where I dropped it). Therefore, v=15.7 m/s=35.2 mph. I doubt that you could give it that velocity.

QUESTION:
I'm doing my level best to finish my first speculative fiction novel, and this is a question to which I'm unable to have anywhere close to 100% confidence in my answer. I'm hoping your much more vast breadth of knowledge might help give me a bit more certainty with it; even if you have to do some educated guessing - which is more than I'd be able to do (though I've shared my guess below). So here goes: I would like to have as close to an accurate sense of how water would move on other worlds; specifically, were all atmospheric conditions identical to Earth's, how would water (and most other liquids then) move on the Moon ("The Expanse" showed liquids moving much more slowly on the Moon) and on Mars? My guess is, if they were right - I assume Astronomer/producer Naren Shankar has an pretty solid sense of it - and atmospherics are essentially the same, the biggest factor is the strength of the gravity; more than the speed of rotation (but this is a factor I'm uncertain about). Stronger gravity will make liquids move more quickly, I believe, but if they were right on that show, then twice as fast as on the Moon - or if rotation speed is a bigger factor than I'm thinking, that's something I'll have to educate myself about.

ANSWER:
The speed v in deep water is v_deep=√(gL/2π) where L is the wavelength of the wave and g is gravitational acceleration; 'deep' means h≥½L. So the speed is proportional to the square root of g and to the square root of L. For example, a water wave on the moon would have a speed of about v_moon=√(1.6/9.8)v_earth=0.4v_earth. On Mars, v_Mars=√(3.7/9.8)v_Mars=0.61v_earth. For shallow water, h≤0.04L, the speed is approximately v_shallow=√(gh); here the speed is proportional to root h rather than root L. There is even an approximate expression for the intermediate depths, v_inter=v_deeptanh(kh) where k=2π/L. This is probably more than you need to know, but if you want to see the whole reference which I used, go here.

QUESTION:
I understand that time slows down for objects moving at relativistic speeds. However, I don't understand why the slow-down in time directly affects things. Why are biological reactions slowed down (so I age more slowly). Or why are chemical or nuclear reactions slowed down? Even though they call time the 'fourth dimension', it seems to have a much more direct effect of things than space does. If space where increased or decreased by 50% in extent, I wouldn’t become 50% bigger or smaller. If physical objects were directly affected by the growth or reduction of space dimensions, how would you ever be able to measure that change?

ANSWER:
You misunderstand time dilation. If you are moving with some large speed relative to someone else, your clock runs at some rate which is totally the same as you would at any speed. You do not see that clock running at a different rate because you were moving relative to someone else. If your ship had no way to look outside, you would not even know you were moving. What is different is the rate which you clock runs is different than the clock which is at rest relative to you runs. You should read my description of the twin paradox.

QUESTION:
So I am having problem in understanding if the acceleration must always be constant when we think of Free Fall. For example, imagine if I were about 70,000km above Earth’s surface and the only object in my vicinity is Earth, then I’d only feel gravitational force of earth. Suppose I am falling towards Earth, then by definition what I experience is free fall. But when I get closer to Earth, say when I am 1,000km above the surface, the acceleration I’d feel would not be the same as what I felt when I was about 70,000km above the surface. Having said that, can I conclude that the only reason we experience constant acceleration when we experience free fall on Earth is because the height for the motion is significantly less than the radius of Earth and therefore, we can ignore the minute changes in gravitational acceleration?

ANSWER:
The gravitational force F on a mass m at some distance r from the center of a spherically symmetric sphere of radius R and mass M is F=MmG/r²where G is Newton's universal gravitational constant. Because of Newton's second law, a=F/m, so a=MG/r². If you set M and R to be the mass and radius of the earth you will find that a is about 9.8 m/s². That is the only r which has this as the value of a; in other words, the acceleration is never uniform. But, as is often the case in science, a good approximation can yield perfectly adequate approximate results provided that you understand the limitations. It is clear from your question that you already understand that very well. I will give you a quantitative way of estimating how high h in altitude you can go before the approximate acceleration differs from the ground-level acceleration:

First, a(r=R)=g=MG/R², a(r=R+h)=g'=MG/(R+h)²
Next, write (R+h)²=R²[1+(h/R)²] so that we can write g'=g/[1+(h/R)²] or g/g'=1+(h/R)²
Suppose we want there to be only a 1% error at some h; what is h?
(g-g')/g=g(1-g/g')=0.01g=g(h/R)²so h=√(0.01R²)=0.1R

QUESTION:
If the Laws of Thermodynamics and the Conservation of Matter state thate neither energy nor mass can be created nor destroyed, why do theoretical physicists fixate on the big bang as the creation of the universe? To me it seems logical that both energy and mass (being interchangeable) are eternal, and that the big bang that caused the present universe may just be the latest of the likely cycle of eternal big bangs. What do you think, is it logical?

ANSWER:
There is no such thing as "conservation of matter" because matter can be created or destroyed. The total amount of energy in the universe cannot be changed although the amount of mass may change. However, I see no reason why the "eternal" amount of energy in our universe should preclude "fixating" on the big bang.

QUESTION:
But energy cannot exist without matter and matter cannot exist without energy. As both would be stagnant in the absence of the other. The laws state that in an isolated system neither mass nor energy can be created or destroyed. Mass is a body of matter of indefinite shape or size. Therefore mass is matter and cannot be created or destroyed. How then, do you validate your proposition that it can be?

ANSWER:
This is utter nonsense. "The laws" state that in an isolated system the total energy must remain constant. Mass is just a form of energy, E=mc² . You are not alone in believing that conservation of mass is a law of physics; I often get questions based on this assumption. So. where did this "law" come from? Modern chemistry was developed about 400 years ago. Experiments showed that if you measured all the mass which contributed to a chemical reaction and then measured all the mass after the reaction, the two, within the accuracy of measurement, would be equal. The catch is that the accuracy of the measurement is too poor to say that the two are precisely equal. When special relativity was developed, it became clear that mass could be converted into energy because E=mc². In an exotheric reaction, energy is released in the form of heat and the origin of that energy is a tiny decrease in the mass. Indeed, even with the most sensitive instruments it is virtually impossible to measure this; chemistry is a very inefficient source of energy. But if you look at nuclear reactions, the mass change is easy to measure; e.g., the mass of a helium nucleus (two protons and two newtrons) is about 3% lighter than the summed mass of two protons and two newtrons. Stars, like the sun, are using nuclear reactions to release energy from mass. If the sun were burning using chemistry, it would have been totally reduced to ashes millions of years ago.

QUESTION:
I was wondering how the oceans can absorb CO2 from the atmosphere if they have achieved equilibrium.

ANSWER:
Lot's of reasons. First, the CO₂ doesn't just swim around but iteracts chemically so that the water dynamically reduces levels allowing more CO₂ to replace that decrease. If the amount of CO₂ in the atmosphere is increasing which causes the equilibrium to be undone whether the CO₂is decreasing or not. So the real equilibrium is dynamic, not static. By the way, the chemistry causes the ocean to become more acidic, not desireable.

QUESTION:
I have come across this paper written by a person called John Mandlbaur. He claims:

Now I do not believe this because although I am no physicist, I know physics well enough that papers are not rejected without good reason so I am sure his is rejected because the calculation and/or reasoning is fatally flawed but being no physicist myself I am not qualified to comment on the paper. If however you identified the relevant flaws in straightforward English, I could post your reply on Quora and hopefully prevent others from being taken in by these claims of Mr Mandlbaur's. If this is not for you, but you can recommend someone else to explain why the paper is refuted in plain English please let me know.

ANSWER:
I interacted with this fellow about 7 years ago. We had a very lengthy back-and-forth discussion of his "ideas" which eventually became so outrageous that I relegated him to the "Off-the-Wall" Hall of Fame". I don't want to call names, but there has never been a questioner more worthy of off-the-wall status than him. It is very sad that he has held on to these delusions for such a long time. I think it was Bill Murray who said "It’s hard to win an argument with a smart person, but it’s damn near impossible to win an argument with a stupid person." I am sure that if you go over these ramblings you will have plenty of examples to convince yourself that Mr Mandlbaur is a crackpot.

QUESTION:
I have come across this paper written by a person called John Mandlbaur. He claims:

"My papers are rejected because they come to a conclusion which the editor doesn’t believe. That is what rejected without review means. Numerous physicists have described my maths as perfect and nobody can defeat it. Now I do not believe this because although I am no physicist, I know physics well enough that papers are not rejected without good reason so I am sure his is rejected because the calculation and/or reasoning is fatally flawed but being no physicist myself I am not qualified to comment on the paper. If however you identified the relevant flaws in straightforward English, I could post your reply on Quora and hopefully prevent others from being taken in by these claims of Mr Mandlbaur's. If this is not for you, but you can recommend someone else to explain why the paper is refuted in plain English please let me know."

ANSWER:
I interacted with this fellow about 7 years ago. We had a very lengthy back-and-forth discussion of his "ideas" which eventually became so outrageous that I relegated him to the "Off-the-Wall" Hall of Fame. I don't want to call names, but there has never been a questioner more worthy of off-the-wall status than him. It is very sad that he has held on to these delusions for such a long time. I think it was Bill Murray who said "It’s hard to win an argument with a smart person, but it’s damn near impossible to win an argument with a stupid person."

QUESTION:
I am writing a science fiction story, and I’d like the science to be accurate. So: if a space ship starts from Earth and steadily accelerates at 2 G’s, how far will it be from Each in three months, and how fast will it be going? Thank you so much; what I’m writing is a love story intertwined with the difficulties of interstellar commerce.

ANSWER:
Before you do anything you should read the four previous answers which you can find in the faq page. If you don't care to get into the details you can just go to this link which will let you know how I did the calculations for you. You first need to understand that acceleration is not a very useful quantity in physics if velocities are comparable to light speed; this is discussed in the previous answers I refered to above. Since you are interested in high speeds, we need to first determine whether classical Newtonian physics can be used or whether we need to use special relativity. We first need to specify who is measuring the acceleration since it is not the same as seen by everyone with high relative speeds; I assume that you are interested in where the ship is and how fast it is going after three months. The figures show calculations I have done assuming g=10 m/s², for both relativistic (red) and Newtonian (black) mechanics, and an acceleration observed from the rest frame of 2g. and essentially velocity as a function of time. The first figure calculates up to about t=23 months, the second to 3 months. Note that the first calculation has the relativistic calculation approaches c at large t whereas the Newtonian calculation increases linearly forever, already at about twice light speed at about one year. But, as shown in the second figure, the two calculations are approximately equal at 3 months. Therefore at 3 months you could say that the speed of the ship is about half of light speed and the distance traveled was about d=½at² =10x(7.78x10⁶)²=6x10¹⁴ m=0.06 light years. It would be much more complicated to calculate the distance relativistically.

QUESTION:
My question is about spinning black holes. Or rather spinning singularities. Like singularities are one dimensional objects right? A point of infinite density. Anyway. don't you need like...another dimension to be capable of spinning? Like what axis does the point spin around? How can you rotate when there is no geometry? I Thought it was a point. How does a point rotate? It wouldn't make sense though. The thing would suddenly stop spinning because there was no more geometry to rotate. Anyway, everytime I try to think about it I sorta reach this block in my head.

ANSWER:
Not one dimension, zero dimensions. Still your questions are pertinent. The first thing you should accept is that we do not know what the physics of a black hole is. The laws of physics are not known inside the Schwarzschild radius. But reasonable calculations which can be done seem to lead to a true mathematical singularity. I feel that most physicists are uncomfortable with the literal idea of infinities. That said, there is no reason why a point particle cannot have angular momentum. When black holes initially form from the gravitational collapse of a dying star and the star originally had angular momentum, that property will be conserved. From that point of view, one would expect just about every black hole to be "spinning". You are thinking classically, but most elementary particles have intrinsic angular momentum (spin) but if you try to interpret it as the physical spinning of some mass distribution about some axis you will fail. Electrons are thought to be point particles and they certainly have spin.

QUESTION:
I am trying to figure out how quickly (approz) a tangential velocity (I think) something would have to have, rotating around the earth at an altitude of about 400kms, to have a centripetal force of 1g ie 9.82 newtons. I think that's the right terminology? Here's my thinking:

I think escape velocity at that distance is (very) roughly 27,360kph?
So if you're going at just or just under that speed you're in constant freefall round the earth, ie experiencing zero gravity, in a vertically constant orbit. (First hurdle: I don't even know if that's correct!)
FYI, I'm imagining a hypothetical band looping entirely around the earth at ~400kms altitude, for example, creating its own centripetal force.
If you stood on the inside of that band, head towards the Earth, and this is where my thinking gets v v doubtful, would you then have to spin whatever the whole artefact (I think it comes out at something like a tangential velocity of 29,366kph to create a 1g centripetal force?) on top of the originally calculated escape velocity in order to create 1g's reactionary force?
So, is the end tangential speed the band would need to rotate approx 56,726kph?

ANSWER:
I am sorry to tell you that most of the things you say are nonsense. Let me just tell you how things are rather than picking your question apart. First, qualitatively: A circular orbit of radius r (r being the distance from the center of the earth) will have a particular velocity associated with it; you seem to think that you can choose the velocity you would like to have in that orbit. The gravitational force on any orbiting object gets smaller the farther away from the center of the earth you are; as you know, the gravitational force on an object of mass m at r=R_E is

F(R_E)=9.82m

where R_E=6.37x10⁶ m is the radius of the earth. So the acceleration is g(R_E)=9.82 m/s².

But, in general g(r)=M_EG/r²where G=6.67x10^-11 N·m/kg² is Newton's universal gravitational constant and M_E=6x10²⁴ kg. So now you can see that every orbit is in free fall except the acceleration of that fall is dependent on the radius of the orbit and is only approximately equal to what you think of as g at near-earth orbits. We can also calculate the speed at any orbital radius r from Newton's second law

F=mg(r)=mv²/r=mM_EG/r². So, solving, v(r)=√(M_EG/r).

You are interested in the altitude h=4x10⁵ m so r=R_E+h=6.41x10⁶ m. Doing the arithmetic, g(r)=9.74 m/s² and v(r)=7.9x10³ m/s=28,440 km/hr.

ADDED THOUGHT:
Rereading your question it appears that you might want to be making "artificial gravity" like the big donut-shaped space ships which you see sometimes in sci-fi movies like 2001 A Space Odyssey. But I don't know why you would want to do it around the earth where there is already gravity present. If you are interested in rotating space ships to create the appearance of gravity because of the centrifugal force, I have answered questions along those lines in the past. Find links on the faq page.

I think escape velocity at that distance is (very) roughly 27,360kph?
So if you're going at just or just under that speed you're in constant freefall round the earth, ie experiencing zero gravity, in a vertically constant orbit. (First hurdle: I don't even know if that's correct!)
FYI, I'm imagining a hypothetical band looping entirely around the earth at ~400kms altitude, for example, creating its own centripetal force.
If you stood on the inside of that band, head towards the Earth, and this is where my thinking gets v v doubtful, would you then have to spin whatever the whole artefact (I think it comes out at something like a tangential velocity of 29,366kph to create a 1g centripetal force?) on top of the originally calculated escape velocity in order to create 1g's reactionary force?
So, is the end tangential speed the band would need to rotate approx 56,726kph?

F(R_E)=9.82m

where R_E=6.37x10⁶ m is the radius of the earth. So the acceleration is g(R_E)=9.82 m/s².

F=mg(r)=mv²/r=mM_EG/r². So, solving, v(r)=√(M_EG/r).

You are interested in the altitude h=4x10⁵ m so r=R_E+h=6.41x10⁶ m. Doing the arithmetic, g(r)=9.74 m/s² and v(r)=7.9x10³ m/s=28,440 km/hr.

QUESTION:
I suppose this is an off the wall question, but it is quite practical, and I’m not sure where to ask it. A web search turns up nothing of interest. I’ve convinced myself that my coffee cup is interfering with the 2.4GHz signal between my mouse and its receiver, which is plugged into the back of my display. The spatial orientation of the three objects is not ideal, but my desk setup is a bit limited, basically thus:

2.4GHz wireless mouse on pull-out keyboard tray, mouse on the right.
My ceramic coffee mug on the surface of the desk, just above and beyond the mouse, also on the right.
My display is just above and beyond that on the desk, with the mouse receiver plugged into a USB port in the middle of the back of the display.

Ordinarily, these three things tend unconsciously toward an unfortunate syzygy*. I suspect the coffee cup is interfering with the signal, causing severe mouse lag. So far unexceptional. I don’t see why the ceramic mug or its mostly water contents might not be reflecting the signal. But there doesn’t seem to be any interference when the mug is full. Only after drinking half or more do I notice the lag. If I move the mug way over to the left, that seems to eliminate the lag. So my question is, does the water level in the mug change the overall resonant frequency of the mug such that it can interfere with the signal at some levels but not others? Is it like filling a jug to different levels to get different tones when you blow across the opening? And considering that microwave ovens work on a similar frequency by heating water, would water be an especially problematic interloper at 2.4GHz?

ANSWER:
Interference and blockage of electromagnetic waves are tricky to understand and track down. Like sometimes when I am driving and come to a stop there will be a specific spot where a local radio station will suddenly almost be totally blocked. All the things you have tried are reasonable: I salute you because physics is, at its root, an experimental science. If you could tolerate moving your cup to the left side, that is likely your best solution. The 2.4 GHz seems like a likely culprit here, particularly since there seems to be negligible blockage once the water is removed. Cups of different construction would be useful to look at, e.g., a metal cup with and without water. I doubt that it is anything to do with interference like the blown over bottle because sound waves are of the same order-of-magnitude wavelength as the resonating object whereas the GHz waves are nowhere close to the size of the cup. Good luck!

*If you, are puzzled, as I was, by the funny word syzygy, it means when three or more objects line up as in eclipses for example.

QUESTION:
Sir,My doubt is in POTENTIAL ENERGY .I am really struggling in understanding this topic .please help me out.Sir my question is that -"is potential energy just a term to make sums look easier?" Sir, taking an example - if we throw a ball with v velocity perpendicularly upward then at the highest point the velocity becomes zero therefore the kinetic energy becomes zero but at this point we say that the energy has been transferred into potential energy but acutually where did the energy go ? And thus can we say that we defined potential energy so that our famous rule does not gets violated i.e. "energy can neither be created or destroyed it can only be transferred from one form to another " is it so ??? What does it mean that energy is stored???(potential energy)?????

ANSWER:
My, you really are struggling. It is clear that you are just beginning your physics education, so I will keep my answer addressed only at that level. In fact, I will only discuss problem involving a uniform gravitational field. First, suppose we know nothing about potential energy. So, the example you give is taking a ball of mass m and throwing it up vertically with some velocity v. Then it starts with energy E₁=½mv² and rises to some height h where all its energy is gone, E₂=0. So energy was not conserved; why not? Because some force must have done work on the mass to take its energy. Of course, that force is gravity which is mg. Because the force is constant and pointing down and the displacement is positive is positive, the work is negative, W=-mgh. The work done is equal to the change in energy W=E₂-E₁=-mgh=0-½mv². You could use this to find out how high the ball will go, h=v²/(2g); you have probably done such a problem before you knew anything about potential energy.

I next want to generalize this problem. The ball is thrown vertically from a height y₁with a velocity v₁ and at a later time it is at different height y₂and now has a velocity v₂. So now the work done is -mg(y₂-y₁) and the change in energy is ½mv₂²-½mv₁². So, rearranging, I can now write (½mv₂²+mgy₂)-(½mv₁²+mgy₁)=0. If there are no forces other than gravity which are doing work on the ball, the quantity ½mv²+mgy never changes! So now I get the brilliant idea (just one way of looking at this problem) to call K=½mv²the energy due to motion (kinetic energy) and V=mgh the energy due to position (potential energy). So you can call gravitational potential energy a clever trick which allows you to never worry about calculating the work done by gravity. Another great reason to use potential energy is that energy is a scalar quantity whereas force is a vector quantity.

There are some things you have to be careful of. If you are using V=mgy for gravitational potential energy, you must choose a coordinate system where +y is up; e.g., if the top of a cliff is at y=0 m, the bottom of the cliff cannot be at y=100 m. You cannot use potential energy for just any old force, it has to be a conservative force. A conservative force is one which does zero total work over a closed path (ending where you started). E.g., gravity does negative work on the ball and positive work on the way down. If you push a book across a table you must do work against friction; but now if you push it back it will be the same amount of work so you cannot have a potential for friction. Your "famous rule" is not very useful. I think you should think in terms of the energy equation, E_final=E_initial+W_ext, the energy you end up with is the energy you started with plus what you added. W_ext is the work done by all external forces, forces for which you have not introduced a potential energy; forces represented by potential energy functions are not external, you have internalized them.

QUESTION:
I saw an older question on your site regarding how fast Flash could run a quarter mile. Using similar assumptions, in Science Fantasy - verse Warhammer 40000, there are Space Marines who are biologically enhanced super warriors with 19-22 implanted extra organs that make them able to overwhelmingly outperform a regular human. Space Marines are roughly 7-8.5t tall (depending on if it is earlier variant with 19 organs or later one with 22 organs then the height is on the taller end of the spectrum) with a mass of 750 pounds unarmoured. Space Marines have been decribed running 87 km/h taking six meter strides. Seeing that much shorter and less massive Flash could run quarter mile in ten seconds giving a velocity of 144 km/h, using the formula, how fast would you put the maximum velocity a Space Marine could run given their mass of 750 pounds? Can you give one estimate using the rubber boot sole, and one for ceramite boot sole that their armour is made of? And when a Space Marine wears armour, his mass is increased to 500-1000kg, so how does wearing armour affect the maximum velocity when affected by laws of friction, gravity, etc...?

ANSWER:
Oh my, you missed the whole point of that caluclation. The mass does not matter because the frictional force he can exert without slipping is proportional to his mass. And his acceleration is inversely proportional to his mass; just the same reason why the acceleration of a falling mass (in vacuum) is independent of mass. And strength doesn't matter either if he can exert the maximum frictional force without slipping, μ_staticmg.

QUESTION:
How an ellipsoid object, say an American football...ball is thrown through the air. It shows projectile motion, but the other interesting observation is that the front of the ball points in the direction of travel. I have thought that air resistance possibly applies unequal torques on different sides of the ball perhaps, but I read somewhere that it has to do with gyroscopic precession - that this axis is different to the one for angular frequency, and somehow that explains it. I'm not sure though, any simpler way to understand this?

ANSWER:
If we were to pass the football with spiral in a vacuum, the spin axis would remain pointing in the same direction because of conservation of angular momentum; this is the case for the lower trajectory shown in the figure. In air, though, the angular momentum mainly remains pointing in the same direction as the velocity of the ball. I always am surprised to find an everyday phenomenon the physics of which is unexplained or explained only recently. This is such a case and was explained only recently (2020) in a paper in American Journal of Physics. A more accessible description of the phenomenon can be found at this link. Your hunch that the air has something to do with it was spot-on.

QUESTION:
so, tell me if I'm wrong but if an object is energy it doesn't have mass yet it is still affected by gravity, but you need mas to be affected by gravity 1. why is energy still affected by gravity 2. is there any way whatsoever for something without mass that isn't energy be affected by gravity.

ANSWER:
No, you do not "need mass to be affected by gravity". The only object we know which has no mass is the photon (a quantum of light). And because it moves, it has kinetic energy which is well-known; you shouldn't say that a photon is energy though, rather it has energy. When a photon finds itself in a strong gravitational field it does not go in a straight line (as you see things) which you might have expected, but rather it follows the curved lines of the space where the gravity is; in other words, in a curved space a curved line may be the shortest distance between two points. Your second question refers to something which has neither mass nor energy; that isn't really something, is it?

QUESTION:
I hope this question isnt stupid, also English isnt my first language, but here I go: People always say that you can't travel faster than light speed, but what is the point from which you count the speed (0 m/s)? Because the earth is already moving through space, so would we need to accelerate less when accelerating in the same direction and more if we accelerate in the direction it comes from?

Follow up, may be explained by the answer to the first question: does the light emitting from let's say a LED travelling at 60mph reach a target 100 miles away faster than a staionary LED (assuming we turn both on the second they are next to each other, so when the moving one passes it)?

ANSWER:
One of the triumphs of the theory of relativity is that its success validates the principle of relativity: The laws of physics are the same in all frames of reference. From this principle we can deduce that the speed of light in a vacuum is independent of the motions of the observer or the source. So that answers your 'followup': a bystander would see the two light flashes arrive at the same time. Your main question asks which frame of reference may a moving object not exceed the speed of light? Since all observers see the same speed of light, the answer is any frame.

QUESTION:
Dad I have a (possibly) physics question: how come tempered glass just spontaneously combusts without any apparent cause? I remember my old roommate's shower fully falling apart during the pandemic when no one was even there

ANSWER:
Hi Hannah! I don't think you mean "spontaneously combust" because that would mean burning/flaming. Glass does spontaneously shatter or break in unexpected circumstances, however. I have answered variations of this question several times ( e.g., 1, 2, 3 ); 1 is probably the most complete.

QUESTION:
I have been thinking about a question, like how we can write the change in unit vectors over a finite time interval, now it may seem like a homework problem but trust me it's not it's a genuine question that I have been thinking about. I know about the change in unit vectors in an infinitesimally small-time interval 'dt' and that it's given by the magnitude of the really small angle that the unit vector has turned about, but what about a change in the unit vector over a finite time interval how can we write that. I tried to find the answer in many sources, but I was unable to find any.

ANSWER:
It is not really clear to me what you are asking. If you are talking about Cartesian coordinates, they never change. But if you have a coordinate system like spherical polar coordinates, the directions of the unit vectors are very dependent on where a point happens to be. I will use a two dimnensional polar coordinate system where the vectors have both radial and tangential components to keep the answer from being too complicated. The figure shows a red dot moving along some path (red dotted line) which shows the path taken by the particle between two times t₁ and t₂. The unit vectors are shown in green. If you know those two locations you can see that the unit vectors all rotate through an angle θ₂-θ₁ in the clockwise direction. The path taken by the particle will be described by some vector function f which has components f_r and f_θ as shown in the figure. So the average rate of change of direction of unit vectors is ω_avg=Δθ/Δt=(θ₂-θ₁)/(t₂-t₁).

QUESTION:
A ship at the speed of light turns on headlights, understand that a person on that ship sees light go out at the speed of light, relativity. What does a person "see" from outside that area of relativity? Can they see that, is that currently possible, what if, if not current? Do they see a ship with no lights, light but no ship, a spectrum of both that equals the speed of light? I am, and with more questions, being as specific as possible: I find it hard to articulate a concept from the written word, please accept and understand my ignorance/lack of understanding!

ANSWER:
First of all, no ship can travel "…at the speed of light…" and I do not answer questions which stipulate at or faster than the speed of light. So I will answer your question assunming v=0.99c, 99% the speed of light. As you say, an observer on the ship sees the light in front of him having a speed of c regardless of how fast the ship is moving. An observer relative to whom the ship is moving with speed 0.99c also would measure the speed of the light to be c. This is one of the postulates of the theory of special relativity: The speed of light in a vacuum is independent of the motion of the source or the observer. However he would not "see" the same "color" of the light because of the doppler shift. Suppose that the light as seen by the ship had a wavelength of 450 nm=4.5x10^-7 m, blue; for 0.99c speed of the source the doppler shifted wavelength would be about 4.5 nm, low-energy x-rays. Hence the stationary observer would see nothing but if he could measure the speed of the x-rays coming out the front of the ship he would see them as having a speed of c.

QUESTION:
Lenses/Mirrors form real / virtual imgs. Let's consider real images. My query is how can mere interaction of light waves create the exact same picture of the source from which they originated? Does this mean that light carries some information in itself and when it interacts / superimposes, it creates the beautiful picture called an "image". To elaborate, light itself is not coloured, but why can we see coloured images (when the light is reflected from the object) ?

ANSWER:
Imagine an object to the left of a spherical lens as seen in the figure. There is an imaginary line drawn perpendicular to the lens and through its center call ed the optic axis. There is some object consisting of many colors. I have chosen to look at just three points in this object, one red, one blue, and one green. Each rays from these points illuminates exactly one point where the image is formed; although I have only drawn three rays for each point, every ray passes through the the same single point on the image. Apart from size and whether the image is inverted or not, the image is an exact copy of the object. Your statement that "…light itself is not coloured…" is not correct--the color of light is determined by its wave length.

QUESTION:
Assuming no engineering challenges like parasitic losses in materials etc…. Is it theoretically possible to take: Speaker A. Play a tone. 440hz. Digitally. Using EXACTLY 100 watts of power. Speaker B. Play a tone. 440hz. Digitally. Using EXACTLY 100 watts of power. Have speakers A and B facing each other head on. EXACTLY 180 degrees. The 2 speakers are playing the same note. Same signal, timbre, frequency amplitude pitch etc…. Would these two “sounds” cancel out perfectly such that “in between” there would be zero “sound”. I understand that small changes in angles and frequencies will have enormous effects on there and how the sound will dissipate if this system is not perfectly balanced from the get go. Very loud. But if the experiment were set up perfectly, is it theoretically possible to zero out the wave? Full 100% destructive interference? Chaos? Feigenbaum?

ANSWER:
No, the result is a standing wave of the same frequency. In the animation the original waves from the left (blue) and right (green) combine to make the black waves which have the same frequency and wavelength but zero velocity. This is due to the superposition principle which states that if two or more waves are moving in the same medium, the net displacement is equal to the sum of the individual displacements at each time and location. Note that the black curve is sometimes totally zero, when the two traveling waves are out of phase with other. If two identical waves are traveling in the same direction they could completely cancel if they were 180° out of phase. In the second (nonanimated) gif, imagine the green and blue are moving in the same direction. The black would be the resultant wave.

QUESTION:
I am trying to understand heat, that is, mostly the idea of hot and hotter. My understanding is that heat is basically motion, and heating something up (cooking, let's say) is about transferring motion from one place to another. So that the changes that take place in cooking are really caused by the molecules of stuff rubbing against each other. So hotter really means MORE motion, because the actual temperature at which things burn, like natural gas, can't be increased just because you burn more of it. Does that make some sort of sense to you? Clearly my mind wanders in the kitchen, and this has been nagging at me for a long time.

ANSWER:
I believe that your question is answered by an earlier answer. You refer to motion; in my earlier answer the kinetic energy referred to is K=½mv² where m is the mass of some atom or molecule in your flame or pot or food, and v is its speed (motion).

QUESTION:
I have a question about fluid velocity/acceleration and its effects on tubular friction. Is the relationship linear? Meaning , if the fluid velocity is being increased, is there a moment or a few moments where the friction increases more than what it would’ve been if the relationship was linear?

ANSWER:
From what I have been able to learn while researching your question is that "friction" as in two solids sliding on their surfaces is not applicable to fluid inside a pipe. The fluid exactly on the surface is not sliding but rather adheres to the surface and is at rest. The motion of remainder of the fluid has some velocity just outside the surface increases to a maximum at the center but certainly not linearly. Everything about this velocity profile is due to the viscosity of the fluid, sort of the fluid version of friction. More detail can be seen in this link and links which it contains. The figure shows a figure from that link and shows the shape of the velocity profile, v(r). Eventually, at high enough speeds the fluid becomes turbulent and "all bets are off".

QUESTION:
How should I get to calculating the force that needs to be applied to a piston to compress a container of gas? Suppose the temperature is constant, and by default, as the volume decreases the pressure increases. Let's suppose the volume gets halved.

ANSWER:
The force depends on the area A of the piston. And, if the gas cannot be well approximated as an ideal gas, you would need detailed properties about the gas. For an ideal gas, PV/T must remain constant and if T is constant, PV is constant. Now, it seems easy, like if you halve the the volume you need to double the pressure; that would give you the answer for the force to hold the piston there once you got it there, F=P₂A=2P₁A. Be sure that you understand that this is what you mean because if you simply started at the beginning with this force and continued pushing with that force, when you got to the half-volume position you would go right by it and it would start slowing down until finally it would turn around and start coming back.

QUESTION:
For reference I'm 31. Am I closer to the age 60 than I am 20? I think I'm closer to the age 60 because time is moving forward and we cannot move backwards to my age at 20. So wouldn't I be closer to 60 because time can only move in a linear motion towards 60 and we are constantly distancing ourselves from the age of 20?

ANSWER:
You are 11 years older than 20 and 29 years younger than 60. You are closer to 20 by almost a factor of 3. If you had asked which you would be closer to being, you would be closer to being 60 than 20; but you didn't.

QUESTION:
Is a mirror VERY different to how people perceive me in reality? I am bothered by the fact that I never realised before but I guess there's a reason… How impactful is the reversal effect? I always thought that's how I look to others.

ANSWER:
The image you see is not how others see you, but a "mirror image". If you look at the image of yourself in the mirror, the image of your right hand is the left hand of the mirror you. If you want to see an image of yourself which is what others see, you need to use a corner mirror. However, since most of us are pretty much left-right symmetrical, there is seldom much difference.

QUESTION:
My eight year old has a question. Get ready. If two particles are in entangled, and one is on the earth and the other has passed the event horizon of a black hole. Would the entanglement be broken or even if it's unmeasurable still be entangled. Does the medium "the ether" transcend the event horizon? Silas Massicotte (8) and Malachi Massicotte (6).

ANSWER:
Seriously? Your 8-year old understands quantum mechanics and cosmology?! That would be pretty amazing. I usually don't do, as stated on the site, astronomy/astrophysics/cosmology, but I did a little research and think I can give you an answer. When one of the two entangled particles is totally destroyed, the other one will not know at all because there is no communication between the two. Since the wave function of the destroyed one simply disappeared, the wave function of the other will remain in the mixed state it was in originally, but it will no longer be entangled with anything. If a measurement is made on the survivor, there will be no change in whatever remains of the other particle. There is a pretty good explanation at this link. There is no such thing as the ether.

QUESTION:
how can electromagnetic waves travel in vacuum? how can something without any physical body and dimensions travel in nothing? it should need something to mediate its propagation like sound waves require air molecules. if its pure energy then what is pure energy?

ANSWER:
This is exactly the way physicists were thinking at the end of the 19th century. Serious searches for the medium required for light to be able to propogate were being made. The best known search was the Michaelson-Morley experiment which failed to find it. It was called the luminferous æther. The punch line is that it doesn't exist at all. Light can travel through empty space. One way to understand this is through electro-magnetic theory, quite well understood at the begining of the 20th century. James Clerk Maxwell unified all that was known about electromagnetism into four equations, today known as Maxwell's equations. One of those equations says that time-varying magnetic fields cause electric fields; another says that time-varying electric fields cause magnetic fields. Using all four you can show that one possible solution is a wave equation for both electric and magnetic fields. Since waves are time-varying, these waves will "keep each other going" which has nothing to do with any medium. If you go to the link above, you can follow other links telling you more.

QUESTION:
I have a question about a calculation in the USPSA (United States Practical Shooting Association) rulebook. The power factor of ammunition is measured by the weight of the bullet in grains times the velocity in feet per second, divided by 1000. A 165 grain bullet at 1000 fps would be 165 power factor. This is intended to keep recoil the same between competitors. Felt recoil is subject to a lot of variables that have more to do with the firearm than anything else. But the rule is to ensure that everyone has a minimum level of recoil to manage. Does a 110 grain bullet at 1500 fps, or 165 power factor exert the same force backwards as a 250 grain bullet at 660 fps, or 165 power factor? All else being equal, is power factor an accurate measure of recoil? The muzzle energy is much higher for the lighter bullet.

ANSWER:
The quantity power factor is nothing more than linear momentum (mass times velocity) in physics but in unusual units, milligrain feet per second: mgr·ft/s); physicists usually express it in kg·m/s. But the units you use don't affect the physics so I will use yours. The thing about linear momentum is that if there are no external forces on a system, it never changes its magnitude; this is called conservation of linear momentum. So, if the rifle has a mass of M=3 lb=21,000 gr=21,000,000 mgr and the two bullets both have equal power factors of 165 you can calculate the speed V of the rifle after the gun is fired: Before the gun is fired momentum is zero, so it must be zero afterwards: 0=165-21,000·V or V=0.00786 ft/s=0.094 in/s. Now, you asked for the force on the shooter. There is not enough information to calculate this because that depends on how quickly the rifle takes to stop—if you have a bony shoulder it would hurt a lot less if you stopped it more slowly by having a squishy pad between you and the rifle butt. Suppose it takes 1 ms to stop; then the average force on you over that time would be [3 lb x 0.00786 ft/s / 0.001 s] = 24 lb. But you said "all else being equal" so I presume that the recoil force would be the same for both scenarios because the masses of the guns and the recoil velocities are the same.

QUESTION:
When text book gets to quanta, it would say E=hf. It would say f = n * fundamental frequency. Ok, I can understand there are 1Hz and 2Hz and so on. But couldn’t one build a circuit that makes 1.5Hz? At the end, it is a voltage completing a cycle within some time. And if f can be decimal, then it is continuous. So E is continuous. I must be missing some concept here.

ANSWER:
Not everything is quantized, but even if everything were in some way quantized, you could only detect with your senses in some situations. Usually we think of quantum physics as what happens in very small systems, systems on the general order of atomic sizes. The thing which is bothering you, I think, is that something which is macroscopic which shows no signs of being quantized means that an a microscopic version of that system would not be quantized either. Take, for instance, a mass m hanging on a spring. So if you pull it down to some amplitude A, say 1 cm and let it go, it will oscillate up and down with some frequency f determined by the stiffenss of the spring k and its energy will be E=½kA² and its frequency will be f=2π√(k/m). You could quadruple the energy by doubling the amplitude. But, of course, you would be able choose any amplitude you liked? Like 1.2379 cm? Of course, assuming you could actually measure that accurately. So quantum mechanics is nonsense, right? No, I could contend that the macroscopic oscillator is also quantized (theoretically at least). What is the spacing of the energy levels of this quantum system? hf, of course. what is hf if f=1 cycle/second? ΔE=6.6x10^-34 J! Suppose that the spring in this oscillator had a spring constant k=1 N/m; then, since E=½kA², going to the next allowed level would require a change of amplitude of ΔA=√[2x6.6x10^-34/1]=3.6x10^-17 m. That is about 1/10 the size of an atomic nucleus! I don't think you would notice this constraint!

Your intuition tells you that quantum mechanics can't be right because you have never noticed its effects. What is your intuition based on? Your experience! Do you have any experience regarding such tiny sized things?

QUESTION:
I am working on a piece of science-fiction centred around the mid – career physicist, upon whose desk objects appear for significantly timed short periods, a significant pattern. At some point in the peace, the working assumption becomes That these objects are being transmitted from "Parallel universe"/other reality in the Multiverse. This may seem to be an odd question if such objects were assumed to be capable of being transmitted between parallel universes or other reality planes in a Multiverse, do you think they would be capable of being photographed and or scanned in some other forms such as x-ray?

ANSWER:
Of course, this is all speculative, but I would say that if your eye can see them a camera could see them because the eye is, essentially a camera. But you could also say that maybe the eye is not really seeing them but rather only your brain was seeing them, like a halucination. I remind you that I do not usually do cosmology, the realm in which multiverse is included.

QUESTION:
I have a theory (or just an idea) concerning faster than light travel that I would like you to answer. My idea is that if FTL travel is possible, then the energy required to make a warp drive, wormhole, etc. work is exactly the same amount of energy as would be required based on Newtonian physics if there were no law of relativity. For example, suppose you have a 1000 metric ton FTL spaceship that can travel four light years in four days. Please calculate how much energy it would take to accomplish this if only Newtonian physics were involved. Then look up some FTL scheme. Here’s one you can use https://newatlas.com/physics/ftl-warp-drive-no-negative-energy/ Or just pick another than you prefer. Then calculate how much energy it would take to move the 1000 metric ton spaceship four light years in four days. My idea is that perhaps the universe is designed in such a way that the energy required is the same. In other words even if FTL travel is possible, you don’t save any energy by using it. In other words, there is no “free lunch.” I donated $20 for your answer, since I suspect this will take a lot of work.

REPLY: (Other readers note: I am only answering this question because the questioner has sent a donation.)
It would been good if you had read the site ground rules before you had sent me a donation since I have no way to refund it. No faster than light questions is one of the ground rules. You can also be sure that anybody designing a "warp drive" is liable to be a crackpot and the amount of work I would have to do to evaluate any such design would be unreasonably large. I am willing to compute the energy using Newtonian mechanics and also calulate the energy to do your four-day trip relativistically. I will assume negiligable time to get up to speed.

ANSWER:
One light year is 365 light days. Using Newtonian mechanics, the speed you would have to go 4 ly in 1 day is 365 times the speed of light, 365x3x10⁸=1.1x10¹¹ m/s. The energy which a your spaceship would have at that speed is
E=½mv²=½x10⁶x(1.1x10¹¹)²=6.2x10²⁷ J=1.7x10²¹ kW·h.

Relativistically, the distance 1 light year (ly) would have to be length-contracted to 1 light day (ld) would be 1=365√(1-(v²/c²)) or v=0.999992494c or β²=0.999984988. Now, total energy is E=γmc² where γ=1/√(1-β²)=258. But we don't want the total energy, we want the kinetic which is K=E-mc²=(γ-1)mc²=257x1.x10⁶x(3x10⁸)²=2.31x10²⁵ J=6.42x10¹⁸ kW·h.

So you see, contrary to your idea, it takes less energy to go relativistically than classically.

QUESTION:
I am a new and upcoming math and physics teacher at a local high school. A more senior physics teacher posed this question to me before he retired, and I am rather embarrassed that I cannot solve it. Apparently, he said this system is accelerating and that acceleration varies with the angle.

[COMMENT:
The questioner shared a copy of a textbook problem with m=50 kg, M=100 kg, θ=13°, μ_s=0.41, and μ_k=0.35. I chose to do it in general and we can stick those numbers in later.]

ANSWER:
I would first choose the knot (massless) as the body; this is a good idea because we don't know whether the system is accelerating or not yet, but it doesn't matter because all the forces on the knot must equal zero because it is massless. Because of the symmetry of the system, the tensions in the strings attached to m and m must have the same magnitudes, t₁=t₂=t, so we have

ΣF_y=0=-T+2tsinθ.

The x-equation just tells us what we already know, that the x-components cancel each other out. We now have one equation with two unknowns, T and t, so we need to choose another body. I choose m on the left.

ΣF_y=0=N-mg-tsinθ

ΣF_x=ma=tcosθ-f

f=-μ_kN if m is moving or f=-tcosθ if tcosθ<μ_sN

Now we have five unknowns, T, t, N, a, and f but only 4 equations. The other m is not going to give us any new information because of the symmetry, so we are left with only M. This gives another equation,

ΣF_y=MA=Mg-T.

Note that I have chosen +y to be down so that m and M are moving in their positive directions. The sixth equation will be the relationship between a and A,

a=A·tanθ.

I also calculated that the net force on m goes to zero at 55° so the static coeffecient of friction would kick in and the system would stop.

t=Mg/(2sinθ)

f=μ_k(mg+Mg/2)

tcosθ-f=0

and therefore,

θ=tan^-1(500/350)=55°.

I leave it to the reader to do the algebra above.

When I did the calculation using WolframAlpha it gave me confidence that the equations of motion are all correct since A=0 at 55°.

So I now had 4 equations with 4 unknowns (I incorporated the a-A and the f-N relations to reduce the number of equations and unknowns to 4.) I put in the value of θ=13° and the solution is shown below, calculated in WolframAlpha. I should note that I checked that the system is actually accelerating and not stuck by static friction at θ=13°.

The results of calculations over the whole range 0-55 are shown in the graph below.

ADDED THOUGHTS:
Just a few comments. I have to admit that I have not rigorously shown that a=Atanθ although I did enough qualitative thought to the relation that I feel pretty confident. Also, the behaviors of all the variables are just what I expect them to be using that relation. It is interesting that at θ=0, because when we usually think if a weight hanging on a horizontal string, that the string would have to have infinite tension to not deviate from the horizontal; that situation, though, requires the (massless) string to be tied between two immovable walls. The equations here have a kinetic frictional force acting on the "walls" so they are not immovable. Since the string at 0° has no vertical component, T must be 0 N, as the calculation shows; M must therefore be in free fall for the instant that θ=0°. Since the frictional force is 0.35x490=-171.5 N, the solution finds that t=171.5 N also; I guess technically t could be anything and still have a=0 because of the tangent factor, but small angle nonzero θ will have whatever t is required. At θ=55° everything has stopped and must be in equilibrium: both N and T are at 980 N and t=980x0.35/cos(55)=598 N. Because all these accurately describe what we know they should actually be, I am pretty confident that a=Atanθ is correct.

QUESTION:
My understanding is that atomic clocks circling the earth run 1) slower because they are moving fast but 2) faster because they 'feel' less of earth's gravity than clocks on the surface of the earth. Which wins out, speed or gravity? How much is the time dilation/contraction due to gravity and how much due to speed? Could there be a planet of some mass that you could put a satellite in orbit around at a certain speed so that the two would balance out so that the orbiting clock would measure time at the SAME rate as one on the planet's surface? And, how much math would I need to figure out the answer to that last question?

ANSWER:
I started trying to use the expressions for the two and doing the algebra demanding that their magnitudes be equal. But then I stumbled on this figure which addresses exactly what you were asking. The red curve is the kinematic time dilation (slowdown) for circular orbits. The green curve is the speedup due to gravitational time dilation. Also shown, in blue, is the sum of the two, the function which you wanted to be zero somewhere. Lo and behold! There is a spot, looks like it is about one third of an earth radius above the surface. Note that where GPS satellites are, the gravitational time dilation is much more important. If your GPS didn't have general relativity corrections built into it, it would not work. The math would not be very demanding, but the physics might be. You need to have an analytic expression for the blue curve and set it equal to zero and solve for r.

Everything below in this answer is incorrect. I was misled by an AI error, several incorrect internet posts, undue confidence in myself to be able to do the GR calculation! It is an excellent example of how incorrect an internet post can be and how AI probably learns from its memorizing everything on the web. I did more research because it seemed extremely odd that the geosynchronous orbit and the zero dilation orbit would be the same and the fact that my incorrect answer was about 10% smaller than the actual geosynchronous orbit.

I am leaving it posted as an example of how wrong the internet can be!

I really liked this question because it was one of those questions which gave me the opportunity to learn something new. I wanted to understand everything which went on in making the graph above. Well, I worked on it for a number of hours and finally got where I understood what should be the functions plotted. I got this graph from the Wikepedia article on gravitational time dilation; I would recommend it because it gives the form of the gravitational dilation (I leave it to the reader to understand the notation here. t₀ is at the surface of the earth and t_fis at some distance r from the center of the earth):

t₀=t_f√[1-(2GM/(rc²))]

What needs to be plotted is (t₀-t_f)/t_f which must be zero on the surface. So I conclude that

(t₀-t_f)/t_f=√[1-(2GM/(rc²))]-√[1-(2GM/(Rc²))]

The kinematical time dilation turned out to be easier for a circular orbit of radius r. Without any details except to note that GM/r=v²

(t₀-t_f)/t_f=-1/√[1-(GM/(rc²))]+1, ensuring that an orbit at r=∞ have zero speed.

So, here's my calculation. The kinetic calculation agrees very well with the Wikepedia graph but the gravitational one dies not. The result is that the zero net dilation is at about 6 earth radii, but the Wikepedia calculation was at about 1.3 earth radii, a really big discrepancy. I stared and stared and could find an error. So I decided that I would ask AI about this; here is my interaction:

Six earth radii is approximately where the geosynchronous orbit, where all the communications satellites are. It would appear that my calculations are right. I did check by googling and the usual answer was the same as ChatGPT's.

I think AI got it backward, though: kinematic dilation is slower, gravitational dilation is faster.

QUESTION:
Picture an attic that covers a house and its driveway. The area of the driveway and the house are the same, so they are each covered by 50% of the attic. In the center of both the house and the driveway are 2 identical circular holes into the attic above. The temperature in the house is 68°F, attic is 86°F and driveway 104°F. Since warmer air flows upwards, will the driveway air get into the attic, or will the colder air in the house work as a barrier to keep the attic air from coming down?

ANSWER:
There are a few things we need to understand before we get to the answer:

The zeroth law of thermodynamics is that heat always flows from a higher temperature to a lower temperature.
The zeroth law is true for an isolated system. If there is some other agent which acts to add, subtract, or redistribute the heat energy, it may not be true.
Particularly for gases, the notion of "warm air rises" is a result of the fact that if you increase the temperature the volume increases and so the gas becomes less dense; it rises because of gravity so the warm "floats" on the cold just like wood floats on water.
There are three ways that heat is transferred—radiation, conduction, and convection.

First, lets consider these systems are totally isolated—no gravity, perfect insulation all over the outsides, no furnaces, ACs, etc. Looking at all three at once would muddy the water, so I will imagine there is a perfect insulating wall in the middle of the attic. Heat would flow from the hot to the cold by convection through the holes and conduction through the ceilings. Eventually all the air would be at the same temperature, somewhere between the two starting temperatures, which would depend on the amount of air in each of the two volumes. The same thing would happen for the garage/attic except heat would flow up, not down. Next, turn on gravity; then there would be a continuous increasing temperature from bottom to top. The ceiling would have the same temperature top and bottom. Similarly for the garage/attic half. You might want to imagine that the ceilings are perfectly insulating; the net result would be the same but only convection through the holes would transfer the heat.

Finally, your scenario, all three volumes connected through the holes and no wall in the middle of the attic. With no gravity everything would end up with the same temperature. With gravity added, there would then be a continuous increase from floors to the roof of the attic. Don't forget that this is the real world and any sources or sinks of heat will change the details.

QUESTION:
I have searched the internet and not found an answer. If there are 2 identical blocks of ice. One is at -1 Degrees C One is at -50 Degrees C Do they melt at different rates? Will the colder piece take longer to melt? I’m assuming the colder piece will take longer, but I don’t actually know.

ANSWER:
I am assuming that each piece is in the same environment, say a room at constant temperature T somewhere above the freezing point or below the boiling point of water. Heat flows from a higher temperature environment to a lower temperature environment. When the heat flows into the lower tempeature environment (the ices in your case) it is absorbed and can do two things: it increases the temperature or it can cause a phase change (melting, freezing, condensing, or evaporating) without a change of temperature. (You might want to read a recent answer to a question similar to yours.) At the beginning of the experiment each piece of ice will heat by increasing its temperature. The 1°C piece will get to 0°C in some short time, t₁, then energy coming in is used to melt the ice until it is fully melted in a time t₂. As the ice is melting, the water from the melting will start warming while the remaining ice will continue to melt without warming. Eventually all the water will be heated to the temperature of the room; call the time from when all the ice was melted until all the water was at T to be t₃. It will take the first piece t₁+t₂+t₃ to get to the end of the experiment. The only difference between the two pieces is that it will take the colder piece fifty times longer to get to 0°C. So it gets to the end of the experiment in a time 50t₁+t₂+t₃. The times for the two pieces to completely melt will be the same, but it will take the colder one longer before it starts melting.

QUESTION:
In the known universe, is there anything else in existence besides matter, waves, forces, movement and change? In other words what are the very basic essential physical things and the effects, excluding any measures that we seem to give a reality to quite often? Maybe waves could exclude matter or maybe waves are just matter in motion? Space itself seems not to be truly part of existence? Time to me is a measure of forced motion and change and the same with energy. What are the bare essentials of the known universe that I can detect through my senses, not just concepts alone? Would you include fields? What are the indisputables?

ANSWER:
This is a strange a very strange question. There is only one way I know to encompus everything which exists. Mass is energy, light is energy, electric and magnetic fields are energy, motion of a mass is energy, etc.. The answer is that the universe is ENERGY.

REVISED ANSWER:
It occurs to me that time also exists. Time is not energy, so the revised answer is ENERGY and TIME.

FOLLOWUP QUESTION:
What is energy and where has it come from? Does the ability to do work come from energy or the forces or both? It seems to me that of all the forms, shapes, colours, sizes, amounts of energy that can be stored or transferred but not created or destroyed.. energy is a quantitative measure of the movement that forces create on an object or particle. Could energy be dispersing waves of just motion through fields since the big bang. Anyway, if not a measure, what is it? Please not the ability to do work.

ANSWER:
You seem to have a talent for submitting questions which, contrary to site ground rules, are not "…single, concise, well-focused questions…" You are expecting me to give you a complete tutorial on energy, something which takes a couple of years of classes to fully understand. Essentially we invent a quantity which, for any isolated system, is conserved, i.e. remains constant no matter what happens inside it. By being isolated we mean that there are no external forces doing work on the system. [Sorry, there is no way to talk about energy without mentioning work.] Conceptually, isolated means any system to which energy is neither being added to nor taken from the system. Assuming that the entire universe is all that there is (or not interacting with other systems outside the universe), the total energy of the universe never changes.

QUESTION:
While water is freezing, energy is being released in the form of heat- how does this jive with a temperature plateau during the freezing phase change?

ANSWER:
Suppose that we have a glass of water which has been put into a freezer which is at -10°C. Heat will flow from the relatively warmer glass of water to the cold freezer and will and will be removed from the freezer into the room so as to keep the freezer at the set temperature. After a couple of hours all the water is frozen at a temperature of -10°C. Now there is a power faliure and, since no freezer is perfectly insulated, heat will leak in from the room so the temperature will will start warming. When the temperature has risen to 0°C heat will continue to leak in and the air will continue heating and the ice will start melting. The air is using the heat to increase its temperature but the ice will use the heat to melt the water, not to increase its temperature. When all the ice has melted, it water will start increasing in temperature.

Well, I see that my little story has not answered your question, it is the reversal. So my story continues: The water and the air are now at some temperature, say +10°C. Now the power comes back on and the water and air start cooling, the freezer removing heat from both. When the water gets to 0°C it starts freezing and the heat being removed results in freezing, not cooling; but the air continues to cool. Finally when all the water is frozen the ice now cools because heat is being removed by the freezer. Eventually everything is at -10°C again.

QUESTION:
I have a question about time. I am a nursing student, taking anatomy courses and learning the atomic physiology of bodily processes. For example, how ions flow in and out of heart cells to make it contract. The whole process takes so much time and thought to track it all out, when in reality, it takes .8 seconds to actually complete a cycle. My question is, how fast are these atoms moving in order to move at the speed to complete our bodily processes in the small fraction of a time we experience as humans? Or, is it that atoms are so small, they experience time different than we do? or is that not an accurate thought? thanks for the help!

ANSWER:
do not know much about electric currents in the heart, but I do understand electric currents in a conducting wire like copper. In a wire, when you have a voltage between two ends of a wire, electrons will flow in the wire with some average speed v. It is a common exercise in an introductory physics class to have the students estimate the speed with which those electrons move in the wire; a typical answer for household wiring is about ½ inches per minute! So, if the switch for a lamp is 10 feet from the lamp, it would take an electron starting at the switch (10 ft)x[(12 in)/(1 ft)]/(0.5 minutes)=240 minutes=4 hours to get to the light. This is, of course, nonsense. What actually happens is that when the switch is turned on there is a voltage across the ends of the wire which establishes an electric field* in the wire; the other end receives the information that the field is there at the speed of light, almost instantly at 10 feet. So all the electrons in the whole wire are moving, however slowly, almost immediately.

Hearts are not conducting wires but they can conduct charge carriers because flesh is an electrolyte which plays the roll of conductor and the charge carriers are ions (atoms or molecules either missing an electron or carrying an extra electron). So when some action in the heart requires a message to be sent from point A to point B, a voltage between A and B is turned on at A resulting almost immediately in the pulse arriving at B; an ion created at A does not get sent to B. I believe that calcium ions are important charge carriers in the heart; I take a drug which is called a Ca channel blocker but I couldn't tell you in any detail how it works.

So the answer to your question is that the ions move very slowly but the information which they carry arrives very, very quickly.

*An electric field is the thing which causes there to be a force on the charge carriers which drives the electric current.

QUESTION:
When light goes from, say, air to glass or glass to air at a non-90° angle, it's refracted. Does that use energy? Are the air and the glass heated? And if so, is that what's happening when a microwave oven heats things the most at the interface between different things (like the ceramic plate and the ham).

ANSWER:
Read the Q&A right after yours. It should be clear that the photon emerging from the first medium has the same energy as when it entered that medium. Therefore no net energy has been lost at boundaries. However, real materials also absorb some of the photons so a net loss of energy occurs when an ensemble of photons traverse the medium and it shows up as thermal energy.

QUESTION:
Concerns the speed of light in a medium. I understand the speed of light varies slightly with hot and cold earth atmospheres, primarily (apparently) with density, as it does in other mediums. Intergalactic space is a medium, however tenuous, and not a true vacuum. It would therefore seem that the speed of light would vary slightly even in it. The standard speed of light is calculated as in a true vacuum. Is this an answerable question?

ANSWER:
To understand refraction in classical E&M is difficult; essentially the alternating electric fields induce vibrations in the atoms in the medium and the result of complicated calculations is that the speed of the wave changes. But it is easier to understand, at least qualitatively, in terms of photons; the photons are absorbed by atoms and promptly re-emitted. This takes a tiny amount of time and is repeated enough times to cause a change in the average speed of the photons. In intergalactic space the density of hydrogen is about one atom per m³. The likelihood of there being an interaction which would measurably change the average speed of a single photon are about as close to infinitesimal as we could imagine.

QUESTION:
Air is moved downwards, how can momentum be conserved if the helicopter remains stationary? Should there not be an equal and opposite momentum upwards?

ANSWER:
Momentum isn't always conserved, only if there are no external forces acting on system (or if all external forces add to zero). The rotating blades are pushing down on the air which causes the air to accelerate downward. But if the blades push down on the air the air must exert an equal but opposite force on the helicopter (Newton's third law). So, if we ask if there are any external forces forces on the helicopter, the answer is yes— the earth exerts a downward force W called the weight and the air is exerting an upward force L often called the lift; we can therefore conclude that momentum is not conserved for the helicopter. In the special case L=-W, and the helicopter is either at rest or moving with constant speed down or up, the momentum is conserved. For the air it really makes no sense to talk about momentum unless you look at it microscopically; fluid dynamics is very complicated and performed by computers. There is no doubt, though, that the helicopter and air exert forces on each other.

QUESTION:
My question is with regard to an apparent mathematical disagreement with inverse square law. Let's say you have point source in 3D space positioned at (x_0, y_0, z_0). It emits photons in uniform 3D directions. --- I just want to exclaim that properly picking uniform 3D directions is not obvious! See Wolfram's Sphere Point Picking page ---. Anyway, you have an infinite length/height vertical y/z plane detector at x=1. One can derive the governing distribution of hits on this detector and it turns out to be the multi-variate Cauchy distribution given by: p(y,z) = 1/(2*pi) * (1-x_0)/sqrt((1-x_0)^2 + (y-y_0)^2 + (z-z_0)^2)^3 This density does not seem to follow the inverse square law. It has a 3/2 power in the denominator! It would need to be 2/2. I have derived densities for other 3D geometries (e.g. cylinder) that are not exactly inverse square law either. I am not a physicist but I am struggling with a physicist who claims it must be inverse square law. It is not. It is similar, but agrees less than the Cauchy density when compared to Monte Carlo simulation. What am I missing?

ANSWER:
Why would an infinitely large detector give you information on the r dependance of what ever the "source" is emitting? By definition, a point source of a vector force field creates a field pointing radially and having a magnitude which decreases like 1/r². I don't know what a multi-variate Cauchy distribution is, or if I ever did I don't remember. You seem to be talking mathematics here and not physics.

REPLY:
The Cauchy distribution (maybe you know it as Lorentz?) is the distribution of where rays hit a vertical or horizontal line when emitted in uniform directions from a 2D point source (x_0, y_0). The bivariate one I wrote is the distribution of where rays hit a y/z plane (at x=1 in this case) when rays are emitted in uniform directions from a 3D point source (x_0, y_0, z_0).

ANSWER:
I have absolutely no idea what you are talking about! But whatever it is, it is not a means of observing or proving an inverse square law. If you want to find out what the field is, since whatever is coming from the source (I will call it "stuff") has to be isotropic, your detector should be either a sphere with the source at the center or a segment of that sphere. Since the area of a sphere increases like r² and the total amount of stuff hitting the surface of the sphere does not change (because there are no other sources or sinks), it follows that the density of the stuff must decrease like 1/r².

REPLY:
I think I have an answer for you! Though it has stumped me for more than a month now. If I look at the behavior of probabilities on a small area of the y/z detector plane and I allow the plane to move... say x=1,2,3,etc... Then the ratio of probabilities of detection in this small area will approximately follow the inverse square law. A PDF is not a probability itself. You have to integrate over an area of the detector plane to calculate probability of detection within that area. So if I calculate the probability of detection in, say, [-0.1,0.1]^2 for the y/z detector plane positioned at x=2, then it's approximately 1/4th the probability of detection in [-0.1,0.1]^2 for x=1. If I do the same for the y/z plane positioned at x=3, then it's approximately 1/9th the probability of detection relative to the x=1 scenario. And so forth and so forth. I think this other physicist assumed you could just arbitrarily define a density function with inverse square law. This seems to only be true for histograms! The bins of a histogram can be thought of as the integral of some PDF over a small area. Still, that Cauchy/Lorentz density is an inverse cube where the calculated probabilities behave like inverse square law. That's pretty weird and unintuitive to me!

OK. I'm a poor communicator at this stuff, I'm sorry. So the way this physicist thinks is that if a point source (x_0,y_0,z_0) emits a number of photons in 3D uniform directions, then the number of photons that should hit a specific location on a surface that is a distance r away from the source should be proportional to 1/r^2. So he asserts that you can just define a probability density of photons hitting a detector surface directly using the inverse square law. For example, if it's a y/z plane at x=1, then r=sqrt((1-x_0)^2 + (y-y_0)^2 + (z-z_0)^2) and you get your inverse square law probability density function of p(y,z) = C*1/r^2 where C is a normalizing constant. That seems reasonable to me! And it's quite elegant, right? If you know how to calculate the distance to the detector surface, you can trivially describe the probability density of photons hitting any location of the detector surface. Well... but there's this Cauchy distribution that defines a distribution of photons hitting a planar surface when emitted from the point source. That one is proportional to 1/r^3... and it lines up almost perfectly with Monte Carlo simulation. So how can that be 1/r^3? The physicist asserts inverse square law must be true. It's well-tested and known to be true. I understand that. But the Cauchy distribution describes the simulation better than the inverse square law-based density. From a math point-of-view, it's so straightforward to derive the Cauchy distribution that it would be baffling if it wasn't the correct probability density. But the darn thing really has 1/r^3 in it!

ANSWER:
Since I can not understand your Cauchy approach or what it tells you about about the point "source", I thought I would start at the beginning with a point "source", in particular a point charge, and ask what can be learned by asking about that field in the vicinity of an infinite plane. I believe I can square the confusion between my physics and your mathematics; as you will see, your 1/r³ function appears quite naturally. I will look at an infinite sheet occupying the xy plane and a point charge Q, a distance d from the sheet. The electric field has a magnitude E=kQ/r² and points radially away from Q. Although your calculations were in Cartesian coordinates, cylindrical coordinates are much more natural. At the surface of the sheet the electric field has no azimuthal component and so the area element to consider is dA=2πρdρ. My aim here is to get the total flux through the infinite sheet, but I will look at other things as I go along. A few things we need are:
    r=√(ρ²+d²)
    E=kQ/r²=kQ/(ρ²+d²)
    cosθ=d/√(ρ²+d²)
    E_z=Ecosθ=kQd/(ρ²+d²)^3/2    sinθ=ρ/√(ρ²+d²)
    E_ρ=Esinθ=kQρ/(ρ²+d²)
    dA=2πρdρ

The first thing to note is that E_z is where your 1/r³ is; just because you happen to get a 1/r³ behavior on the x,y plane does not contradict what the behavior of the field is. E_z^*is what is important here if we are interested in something spread over the entire sheet because for every E_ρ on the area element, there is an oppositely-pointing E_ρ 180° away which cancels it. dA·Ecosθ is what physicists call electric flux and, taken over the area dA, the flux is dΦ=2πkQρdρ/(ρ²+d²)^3/2. Finally we can calculate the total flux through the sheet:

Φ=2π ₀∫^∞E_zρdρ=2πkQ

* The graph shows E_z normalized to E_z(ρ=0, z=0) as a function of ρ in units of d.

QUESTION:
I recently learned that at CERN they produced streams of protons by applying large magnetic fields to hydrogen gas. This removes the electrons and these only the nucleus specifically protons. This made me wonder as alpha particles are merely the nucleus of the helium gas, were you to apply a large magnetic field to helium gas would you merely be left with alpha particles?

ANSWER:
I don't where you "learned" this, but it is almost 100% wrong. In a proton accelerator you need to do three things:

Ionize hydrogen gas to make a plasma. This is done with a radio frequency electric field.
You then extract the protons from this plasma using electric fields and inject those into an accelerator which speeds them up.
Then you repeatedly speed them up, again using electric fields and magnetic fields to steer and focus the protons being accelerated.

Most accelerators, including for alpha particles, work the same way. In fact, magnetic fields cannot ever exert the forces necessary to accelerate charged particles. The reason is that the force which a magnetic field B exerts on a charged particle with a velocity v is perpendicular to v.

QUESTION:
So I just read an article about virtual particles and Hawkins radiation, and how a black hole can take virtual particles, separate them, and take one of them while the other Hawkins radiation is ejected out and that's how a black hole can evaporate, which I still find hard to believe. If that is the case, and a virtual particle pops existence and one is taken away and the other is ejected what happens with that energy? So, if there's energy between the virtual particles and that energy can't be destroyed only transferred could that energy end up as dark energy? I'll explain it again in the hopes of clarifying, near a black hole, virtual particle, can pop into existence, be separated, with one particle, being devoured by the black hole, and the other one being ejected out into space, if there is energy between the two particles at the moment of them being separated could that energy simply turn into dark energy or would it stay with the Hawkins radiation being pushed into outer space?

ANSWER:
If a particle/antiparticle pair is created near the Schwarzschild radius, and one of them is inside and one outside, the one inside cannot escape but the one outside can. If this had happened either inside or outside the black hole, they would quickly have recombined to conserve energy consistent with the uncertainty principle. But, now, because the particle outside has energy, energy would appear to be not conserved. But that does happen, so that energy excess has to be provided by the black hole which means that it must lose a bit of mass. The details doesn't matter of how this happen, it has to happen to conserve total energy.

QUESTION:
When a golf putt hits the edge of the hole, sometimes the ball will "lip out," whipping around the edge and heading off in a new direction. Some lip outs can look quite violent and travel surprisingly far from the hole. My question: can hitting the edge actually increase the distance to the hole? Formally, if point H the center of the hole, is it possible for a lip out to finish further from H than an identical putt would if there were no hole punched in the ground? Or does conservation of momentum make this impossible, and this is all just an optical illusion caused by the brain misinterpreting the abrupt change in direction?

ANSWER:
Two golf questions in a row! In an earlier question I discussed the motion of a putted ball in great detail. If you want to understand my answer to your question, you must read this Q&A first to see how I use the frictional force which is what I used there. I will also assume that the parameter μ_R=0.093 and the mass of a golf ball is m=0.046 kg. So the frictional force of a rolling golf ball f=-mgμ_R=-0.045x9.8x0.093=0.042 N if the green is level which I will assume to be the case. The acceleration of the ball along a straight-line path will be a=f/m=-0.91 m/s². The figure shows the two paths you stipulate, one rolling a distance D₁; the second, rolling a distance D₂+D₃+D₄. D₃, not labeled in the figure, is the path during which the ball is "lipping out", moving on the rim around an arc of angle θ; so D₄=Rθ where θ is in radians and R=0.054 m is the radius of the hole. I will assume* that the acceleration along the curved path is the same as along the straight paths. Now, we can find D₁ because, from the earlier answer, D₁=0.55v₀²=where v₀ is the velocity which the ball has as it leaves the putter. I will choose v₀=1.5 m/s which yields D₁=2.05 m. Since I have assumed all paths have the same acceleration along all paths, each of the two putts must travel the same distance, so
D₂+D₃+D₄=D₁=2.05 m D₄=2.05-D₂-0.054θ

Since we are trying to find out whether or not it is likely that the ball which lipped out is farther from the hole or not, knowing D₄will likely tell us. D₂ is just the distance from the hole center where the lie was; θ will probably not be bigger than 90°=1.6 radians and, besides, it is very small in the scheme of things. If we choose D₂=1 m and θ=90°, D₄=2.05-1-0.085=0.965 m. It makes more sense to look at a straight shot which just missed getting lipped with the one that lipped by 1.6 radians. The second figure shows how far each are from the center after stopping. The lipped one is closer by about 8%.

*The question about what the rolling friction should be in the arc was what worried me most when I started analyzing this problem. But the normal force on the ball by the ground has to be bigger than it is when going straight because of a centrifugal force acting outward. But the conclusion would be the same that the lipped ball will end closer to the center of the hole would would not be changed; if the normal force is bigger, the friction is bigger, so it loses more speed and would be even closer.

QUESTION:
I was trying to measure the bounce of a golf ball to determine whether its lifespan was up. In order to control the ball's bounce as to not get away I thought bouncing it in a tube with viewing hole would be a good environment. Unfortunately I noticed the balls bounce approx 50% less in the tube than outside of it. I would like to know why the golf ball bounces higher outside than inside a 2 inch PVC tube from 5 feet off the ground? Is there anything I can do to alter the tube to get the same results (ie drill holes in it, cut out sections in the bottom to release air)?

ANSWER:
This is so far from how the freely falling ball falls, that I won't even discuss air drag which is the main source of friction for the ball which is not in the pipe. But something else is going on here. Imagine that the diameter of the ball and the inner diameter of the pipe were exactly the same and that there was no friction between the two; also imagine the bottom is sealed. You release the ball and all the forces on it are its weight down, the force of the atmospheric pressure on the top which is also down, and the force of the pressure of the air (which starts as atmospheric pressure) which is up. So the net force on the ball is initially only its weight down; so it starts out just like the ball without the pipe. But, as soon as its starts falling the pressure of the air below it begins increasing. The farther it falls, the higher the pressure gets, so eventually the net force from the pressures is up and has a value exactly equal to the weight. But, when it gets to that point the ball which has been accelerating downward begins to be more up than down so the ball starts slowing down eventually coming to a momentary stop. Now the ball has a net force up so it starts accelerating upward and eventually goes back to where it started. It keeps oscillating up and down just as if it were on a spring.

Now your situation is not like this but it is very similar except it is leaky; the air will leak out as the ball falls, but the pressure in the air below the ball will still increase. All the air which the ball is pushing through has to squeeze through a much smaller space than if the pipe weren't there. That is going to considerably impede the fall of the ball. If you used a pipe much larger to keep from having to chase the ball it would probably better match the freely falling ball.

QUESTION:
This is the 90 year old man in a nursing home and would like to know if it is possible to have a total vacuum in a closed space.

ANSWER:
In a very excellent vacuum system there are about 20,000,000 molecules/cm³. In intergalactic space there is typically 1 molecule per m³. We now understand that even if we had a volume with absolutely no molecules in it, it would still not be truly empty because there are constantly particle-antiparticle pairs which pop into existence and pop back out, called virtual particles; this is called vacuum polarization.

QUESTION:
I am in a state-run nursing home that has toilet facilities in a small enclosure with two doors measuring 5ft by 8ft and height 10ft. The temp rises to 6 degrees above the adjacent rooms. Is this due to the heating of the particles in the atoms creating energy in the form of heat?

ANSWER:
If the toilet room has its own heat register or radiator and this room is much smaller than the adjacent rooms, the heat will increase the temperature more compared to larger rooms; and if the doors are closed for most of the time or even less, it will naturally be hotter than a larger room on the same thermostat. This could be corrected by partially closing the register or partially reducing the water/steam flow to the radiator.

QUESTION:
I just really confused myself with a strange hypothetical situation during the impulse/momentum unit at my school. If a kid is riding a sled on completely frictionless ice with negligible air resistance, and then separates himself from the side of the sled in a way such that both the sled and the kid are moving in the same direction, would the sled still speed up? I know the sled's mass would seem to decrease from the sled's perspective, so it should speed up to conserve momentum, but if the kid is still moving at the same velocity as before, wouldn't any increase in speed for either thing not conserve momentum for the system? I tried to explain this to my teacher, and we clearly didn't understand each other.

ANSWER:
This depends on how the kid separates from the sled. There are two things you need to be aware in questions like this: Is linear momentum conserved and is kinetic energy conserved. If there are no external forces acting on a system the linear momentum will be conserved; most problems you are likely to encounter have momentum conservation. If there are no external forces doing work on a system the kinetic energy would be conserved; if it is (which it usually isn't), it is called an elastic event. I shall look at two scenarios:

What is the relation between the two final velocities if we demand that linear momentum is conserved? If the subscripts labeled 1 and 2 are for the boy and the sled, then
(m₁+m₂)u=m₁v₁+m₂v₂
or v₁=u[1+(m₁/m₂)]-(m₁/m₂)v₂where u is the speed before separation and v_i are the velocities after.For example, suppose that v₂=u/2, then v₁=u[1+(½m₁/m₂)]. As you can see, there are an infinite number of possible velocities after the collisions depending just on how the kid got off the sled. Of course, that is not surprising because we have only one equation for two unknowns.
Now, suppose that we impose the condition that energy must also be conserved. The derivation of the solutions for a perfectly elastic event can be found in any introductory physics textbook or go to the Wikepedia article on elastic scattering in one-dimension. Now we have two equations for two unknowns which means of course there will only be one solution. If both objects have the same speed u before the collisions, then these equations become
v₁=u and v₂=u.
So, you see, the boy and the sled have the same speed afterwards as they did before the boy got off but only if he got off in such a way that no energy was added nor taken away from the system.

It looks to me that you did not give me all the information about the problem. If it had been something like "Suppose that after the separation the sled had a speed of 3u/2, how fast is the boy moving?" you could find out that the boy was moving more slowly, v₁=u[1-(½m₁/m₂)].

QUESTION:
I am a new adjunct instructor for a introductory physical science online class. My background is in chemistry. Anyway, I ran across a conceptual problem in OpenStax Physics 2e, that I am confused about. In the chapter reading, Coulomb’s law is introduced followed by electric fields which are derived from Coulomb’s law. Neither of which I find confusing. Here is the problem I am confused about: Compare and contrast the Coulomb field and the electric field.

ANSWER:
Electric field is a general term, Coulomb field is presumably referring to a field which falls off like 1/r². The prototypical Coulomb field originates in an isolated point charge q: E=r_okq/r² where r_o is a dimensionless vector of magnitude 1 which points directly away from the point charge, called a unit vector; the constant k is sometimes written as 1/(4πε_o). This Coulomb field plays a crucial role in the determination of the field of any charge distribution which is not a point charge. If you have a chemistry background, you surely know calculus, so I will give you the whole general picture so you can understand what goes on. If the course you are teaching is "physical science" you surely will not teach it at this level, probably the electric field due to 2 or more point charges. In the figure a tiny (infinetesimal, really) piece of the blob of charge has been focused on. The field at the point labeled P due to the tiny (point) charge labeled q_i is a Coulomb field; now you have to integrate over the whole volume for every point in 3-D space. Except for simple shapes like a sphere, this is an extremely difficult calculation to do analytically. That is why your course will probably not talk about anything much beyond point charges. You might also touch on the field of a large uniformly charged plane which is pretty easy to conceptionalize and introduces a uniform electric field.

QUESTION:
Einstein, in his theory of General Relativity, stated that Gravity is not a force. So, why the phisicists are still trying to develop a theory to explain Gravity as a force next to the other three fundamental forces of nature?

ANSWER:
I doubt very much that he did said that gravity is not a force. Although thinking of space-time warping is the most popular way to visualize the results of his theory, it is not a unique way. General relativity (GR) is also a field theory and interactions in field theory are manifestations of forces. Einstein was aware of this alternate view, in fact he embraced it—because GR is deterministic, he believed in predestination. Also, the search for grand unification including gravity is rooted in the hoped-for quantization of GR; field theories are quantizable. See earlier Q&As on this topic; be sure to see links in that Q&A.

QUESTION:
I am writing you to seek your expertise in the field of physics. As you know, Newton's law of universal gravitation demonstrates that the gravitational force between two objects is proportional to the product of their masses and inversely proportional to the square of the distance between them. The use of r² instead of r arouses my curiosity. Any further explanations would be highly appreciated.

ANSWER:
At the Newton was proposing his three laws there was a lot known about the solar system. Most importantly, Tycho Brahe had made thousands of precise measurements of planetary orbits, their shapes and periods. His assistant Johannes Kepler continued after Brahe's death but he also used those data to formulate the properties, the three Kepler's laws. Keep in mind that Kepler's laws were purely empirical, nothing we would call a theory. It was now known that the orbit of a planet was an elipse one focus being at the sun, that the quantity T²/a³ was the same for all the known planets (T is the period and a is the semimajor axis of the elipse); also, the line from the center of the sun to the orbit swept out equal areas in equal times. Newton's laws introduced the notion of force and its relation to motion. He realized that the planets moving around the sun implies that some force was acting between the sun and the planets. And surely he realized that the force would not be constant nor would it get bigger as the objects got farther apart. Maybe he did try a force which went like 1/r, but if he did it would not explain the data—Kepler's three laws were the data. But it would not work. Nor would 1/√r, 1/r³, etc., but a 1/r² force did work.

There is another way you can look at forces. A force usually may be thought of as a field where we think of lines of force directions in the direction a mass would be attracted; the closer the lines are together, the stronger the field; and total number of lines is proportional to the mass. The red lines in the figure show the earth's gravitational field but only in one plane. The blue lines show spheres again only in two dimensions. You should imagine the lines coming in from all directions and many imaginary spheres. The strength of the field would be the number N of lines per unit area A. But every sphere is punctured by exactly the same number of lines as all the others. The area of a sphere is 4πr² so the number of lines per unit area, strength of field, is N/(4πr²). Eureka! There's your 1/r².

QUESTION:
I am a student in Belgium. Near my school, there is a wind turbine that has caught my interest. I have been wondering if it's possible to determine the wind speed by measuring the time it takes for the turbine blades to complete one full rotation, given the radius of the wind turbine. I am assuming a constant wind speed for this calculation. I believe that the wind speed (V_w) can be expressed as V_w = (3.6 * 2 * π * R_turbine) / T (in kilometers per hour), where R_turbine is the radius of the wind turbine, and T is the time it takes for the blades to complete one rotation. However, this formula does not seem to yield realistic results. Despite conducting some research on this matter, I haven't been able to find a satisfactory answer.

ANSWER:
You would have been right that the rotational speed of the blades depends on the speed of the wind if you were looking at a simple pinwheel. There are no other significant torques on this machine than that caused by the wind on the blades so, to a pretty good approximation, the only thing affecting the rotational speed is the wind. The wind turbine, however, is a very complicated machine. Here are some things which are interesting:

The blades must move slowly, usually around 10-20 rpm, because of structural limits and noise polution.
So there are mechanisms which keep the rotational speed within those limits.
Because this is too slow for the generator to work there is a gear box to increase rotation to the generator by a factor of about 100.
The blades do not work like paddles or sails, rather like the wings of an airplane or rotor of a helicopter.
The speed of the tips of the blades is more important than the rotational speed of the blades themselves for maximizing efficiency.
Don't forget that, unlike the pinwheel, energy is being constantly taken away from the blades and converted into electrical energy which will be sent out to the world; so there must some energy balance where the energy taken out is equal to the energy being supplied by the wind minus energy lost to friction, inefficiency, etc.

There is a very good webpage going into the details of wind turbines; be sure to look at the video too. There is another video which is also pretty detailed.

QUESTION:
if pi measures a circle and and it never repeats it self, but its used to measure a circle doesn't that mean the numbers would eventually go around the circle and repeat.

ANSWER:
This isn't physics. π does not "go around the circle", whatever that means; it is a measure of the ratio of two lengths, the radius R and circumference C of a circle, π=C/(2R).

QUESTION:
For matter that is falling into a black hole, even before it gets to the event horizon, it would be impacted by both gravitational and velocity time dilation in accordance with standard and generally relativity. This time dilation can be very extreme at these extremes. So would this mean that through the perspective of the matter falling in, it's still hasn't reached the center or even the event horizon? Because from our perspective outside the black hole it is going to take them billions or even trillions of years to get there. So while for them it's a quick decent, for us it's a very long time. And since we are here now could that possibly mean that all that matter in black holes has yet to reach the center because they are nearly frozen in time dilation? If I'm wrong what aspect is not working the way I think?

ANSWER:
Your question is not really a "single, concise, well-focused" question as required by site ground rules. I suggest this link to get a tutorial about falling into a black hole.

QUESTION:
Is infinite energy possible in a frictionless surface

ANSWER:
There is not an infinite amount of energy in the whole universe. The only way it could get infinite energy is if it goes with a velocity equal to the speed of light and that is impossible exactly because you cannot get an infinite amount to give to it.

QUESTION:
I was learning about a STEAM activity for my children. This activity involved cutting up pool noodles and putting toothpicks in them to stick the noodle pieces together. I was wondering, why doesn't the toothpick fall out? What force(s) are acting on it to keep the two together? It seems strange to me, because the pool noodle is porous. I want to know so I can teach my children the physics behind the activity.

ANSWER:
I have raised 4 children, and I never heard of this—must be pretty new. It is friction between the toothpick and the noodle piece that keeps it in. The frictional force can be felt because if it takes a force to push it into the noodle (or pull it out). An interesting thing is that the frictional force between two surfaces depends on how hard the surfaces are pushed together—when you push the toothpick in it pushes the noodle over to make room for itself. Since the noodle stuff is elastic, (if you squeeze it, it springs back), the two surfaces are pushed together pretty hard which gives you a lot of friction. If you were to drill a hole exactly the same size as the toothpick is thick, there would be much less friction. If you think about it, that makes sense because you know it would much easier to push the toothpick into that hole than to push it into the noodle without a hole.

QUESTION:
If sound like light travels on earth, does the sound like light just continue on continuously.

ANSWER:
Sound does not travel anything like light does. The only thing they have in common is both are waves. The most important difference is that sound needs a medium through which to travel, air in the case you are thinking of I presume. Light does not need a medium, it can travel through a perfect vacuum. So light keeps right on going when the atmosphere disappears but sound does not because there is no air to travel in.

QUESTION:
As a solid object (ex a box) travels in a circular path, the side of the box closest to the inside of the curve moves a shorter distance than the side of the box towards to the outside of the curve. Yet, the sides of the box remain in the same relative position to each other............ How can one side of the box travel a greater distance than the other and the box still remain intact?

ANSWER:
That's just the way it is in Euclidian geometry—every part of a rigid body which is rotating about some axis remains locked to the where it is in the rigid body. In fact, that is what a rigid body is defined to be. Think of a CD rotating about its central axis. Any point a distance r from the axis has a speed v=rω where ω is the angular velocity in radians per second. So the farther out any point is the faster its speed is. But if the spinning object is not a rigid body it will not remain intact. If you start with a spinning pancake made of soft putty it will spread out. If the box you were thinking about was made of rubber it would not retain the same shape if you started to spin it.

FOLLOWUP QUESTION:
Thank you so much for replying to my question, but.... I'm sorry, even tho your answer makes perfect sense, it did not answer my question . How can one side of the box travel a greater distance than the other and the box still remain intact? Imagine the box is going around a circular race track (like a race car). The track has a diameter of 1000' at the center of the track - the circumference of the center of the track is 3141.59' . Say the box is 4' long and 2' wide. In one trip around the track, the center of the box travels 3141.59'. The outside of the box (at a diam of 1001') travels 3144.73'. The inside of the box (at a diam of 999') travels 3138.44'. So, the outside of the box has travelled 6.29' farther than the inside of the box.... How can one side of the box travel a greater distance than the other side and the box still remain intact?. I have wondered about this for many years of my life and have never been able to find someone to explain it to me. (straight-line physics is so much easier than curved-line physics....) Thank you so much for your attention to this question.

ANSWER:
The answer is actually simple, if one part of the box is farther from the center it moves with a larger speed than another part of the box closer to the center. What I think the problem is that you do not know the basics of rotational motion. Central to rotational kinematics is what the linear velocity v of any object rotating about some axis with angular velocity ω and at some point a distance d from the axis: v=ωd. The angular velocity must be measured in radians/second (s^-1); since there are 2π radians in a circle, an angular velocity of 1 rotation per second has ω=2πd. I have drawn a diagram of a stick rotating around one end with angular velocity ω showing one end having a speed v but the middle having a speed of v/2 and the end it is rotating around is at rest; v=Lω. Your instinct that if it wasn't strong enough to have those differing speeds, remain a rigid body, it will break. A good example of that is when a very tall chimney falls, it breaks before hits the ground.

QUESTION:
I suspect that Elon is not producing trucks because of the way electric vehicles brake.. Am I correct in saying that there is a transfer or energy to the road surface on EVs, not on normal vehicles that rely on pads.a big rig, heavy.. those wheels, the weight, the method of slowing, I believe that e trucks will not suit our roads. Do you get me?

ANSWER:
Yes, you are wrong that there is "…a transfer of energy to the road surface…" when EVs brake. Let's discuss what braking does. When a vehicle is moving it has energy by virtue of its motion called kinetic energy. In order to stop the vehicle from moving, you must get rid of that kinetic energy. The traditional way to do that is to use friction of brake pads (or shoes) rubbing on a metal disc (or drum). Where does the kinetic energy of the vehicle go? Into heat energy because the brakes get very hot. Wouldn't it be nice if you could capture some of that energy and store it to use to change back to kinetic energy later rather than get the energy from the burning fuel. There is a way other than friction that could be used; that method is called regenerative braking (regen) and involves using electromagnetism to create an electric current which could be stored by sending the current to a battery (not to the road surface!) Before recent times there were no batteries to store all this energy except the usual 12 V battery which gets recharged but not by a braking system but by the alternator. But with hybrids or EVs, there are big batteries which are hungry to grab any energy they can to lengthen the time before they next need recharging. But no vehicle relies solely on regen because regen might not slow the vehicle down fast enough or it might fail altogether; all hybrids and EVs have both conventional brakes and regen brakes and usually you can adjust what fraction of the braking is done by each. Furthermore, if the vehicle is not braking fast enough with just regen, conventional brakes would jump in. Therefore, as you can see, braking is not an issue at all in whether the vehicle is feasible. But adding regen to any vehicle is to your advantage if you have the ability to store the electric energy.

QUESTION:
I read that quantum spin is intrinsic angular momentum. I do not understand what intrinsic angular momentum is. Could you please tell me? In ignorance, I guess that it is some number which quantum physicists derive from the positions, movements and interactions of electrons, other particles and photons, and which predicts their positions, movements and interactions, but does not refer to classical rotation because the number is so high that under the equation E = MC2 the electron, other particle or photon would have infinite mass, which is wrong.

ANSWER:
I believe that it is always good to have a clear picture first of what a quantity is in classical before trying to understand corresponding quantities in quantum physics. We often break angular momentum (AM) into two pieces, orbital angular momentum (OAM) and intrinsic angular momentum (IAM). IAM corresponds to the AM an object has by virtue of spinning about an axis in the object itself. OAM corresponds to AM which an object has by virtue of its moving relative to some other point. An example is the earth which has OAM by virtue of its motion around the sun and IAM by virtue by virtue of its rotation about its axis. A rough, but often successful model of the atoms can be understood by thinking of electrons being in orbits around the nucleus (OAM) but also you must assume that the electrons have IOM to understand what is going on in detail. Electrons have IOM which never changes. In general, however, in quantum mechanics the angular momentum of a system cannot be just anything you like (for example, if you spin a ball there is no limit on how fast it spins, 1, 23, 32.5, 0, etc. RPM); instead they can only have certain discrete values which is called quantization of AM. Elementary particles have one fixed AM but the total angular momentum of a quantum system in some particular state which has quantum numbers either integers n=0, 1, 2, 3… or half-odd half integers, n=1/2, 3/2, 5/2… The first set are called bosons and the second set are called fermions. The angular momentum J of a system with quantum number n is J is given by J=&hbar;√(n(n+1)) where &hbar; is the rationalized Planck's constant. IOM is often referred to as spin. But if you try to make a semiclassical model of an electron as a little spinning ball you get impossible results, like the surface of the ball traveling faster than the speed of light. So we can think of it like a spinning ball but have to keep in mind that it is not a classical object and just has a property that behaves sort of like classical objects do.

QUESTION:
I am trying to understand something about gravity. To it seems that the gravitational pull equations I was taught is flawed. It is stated that it is directly proportional to the mass of the object and inverse to the distance from the object. The distance makes sense. The mass does not because gravity itself is needed to make the mass. Without it, mass would not exist. It is like trying to describe the color red using crimson.

ANSWER:
You have it backwards, gravity is the result of the presence of mass (or any energy density for that matter), not vice versa. Also, gravitational force is inversely proportional to the square of the distance.

QUESTION:
Light is able to be bent by both gravity and it easy bent when traveling through clear objects like a prism. Could it be that the secrets to gravity and anti-gravity might be found in light?

ANSWER:
These two bendings are for two entirely different reasons. So the answers are no and no.

QUESTION:
I'm not sure if this counts as astrophysics, but if an object was falling towards something, like the earth (no air resistance), and then the earth disappeared, would it keep the inertia from the fall? Because if considering gravity as just space-time curvature, it disappearing seems like it would leave the object motionless, as it always was motionless in freefall, because standing on the earth is the same as accelerating up?

ANSWER:
I seem to always be saying...if you ask about velocity you have to specify velocity relative to what. If you are watching this scenario in a frame at rest relative to the earth, the object will move with the velocity it had at the instant earth disappeared.

QUESTION:
My question is does it matter if we have two Dynamos which produce the same voltage but different frequency does the frequency matter and how?

ANSWER:
It depends on what you are powering with the dynamos. But most devices are designed to operate at a particular frequency.

QUESTION:
Is it theoretically possible for a black hole to contain other smaller black holes?

ANSWER:
If you mean by contain inside the Schwarzschild radius, the answer is yes, although it wouldn't last for very long. If you mean inside the black hole itself, what does it mean for a point to "contain" anything? Keep in mind that I, as stated on the site, do not normally do astronomy/astrophysics/cosmology.

QUESTION:
How fast would a missile be flying if it was fired from an SR-71 blackbird flying mach 3.2?

ANSWER:
As always with questions about velocity, you must specify velocity with respect to what. If the blackbird has a velocity v with respect to the ground and the speed of this particular missile has a velocity u with respect to the ground if fired from the ground, and the velocities are in the same direction, then the missile has the speed u+v with respect to the ground if fired from the blackbird.

QUESTION:
So let’s say my buddy and I are driving along the highway in our Pontiac Trans Am’s at 99% the speed of light maybe a couple hundred yards apart. I’d like to chat with him on the CB radio. Are we traveling faster than the radio waves can be received?

ANSWER:
It sounds like you and your buddy are traveling in the same direction with the same speeds. In that case, according to the principle of relativity, there is no difference from your both sitting at rest separated by "a couple of hundred yards apart."