Yahoo Answers is shutting down on May 4th, 2021 (Eastern Time) and beginning April 20th, 2021 (Eastern Time) the Yahoo Answers website will be in read-only mode. There will be no changes to other Yahoo properties or services, or your Yahoo account. You can find more information about the Yahoo Answers shutdown and how to download your data on this help page.

Lv 725,734 points

morningstar

Favorite Answers46%
Answers3,431
  • Can you see Polaris from Chimborazo?

    If you were at sea level, you couldn't, but maybe the horizon is expanded enough by the elevation.

    3 AnswersAstronomy & Space7 years ago
  • Why do supernovas in other galaxies appear as big dots?

    For example in these images.

    http://www.eso.org/public/archives/images/screen/e...

    http://www.dailygalaxy.com/.a/6a00d8341bf7f753ef01...

    I understand how they can briefly outshine their entire parent galaxy, but shouldn't they still be a small, incredibly bright dot? How do they span a significant angular area in the sky as compared to other stars in their galaxy? I don't think the expanding gas can spread out that quickly after the event. Is it some artifact of the method of photography, rather than the object actually spanning that much space?

    3 AnswersAstronomy & Space7 years ago
  • Is ice stable in space?

    So I know that liquid at any temperature evaporates in a vacuum, but can't solids also sublimate? Does ice have a vapor pressure at any temperature, no matter how low, or is there some temperature below which sublimation stops entirely or for all practical purposes?

    Is stability of an icy body in space based on the sublimation of enough vapor, that then creates a thin atmosphere with barely enough pressure to equal the vapor pressure? If so then how to icy bodies coalesce in the first place? How could solid grains form if they evaporate immediately and have negligible gravity to form gas pressure?

    7 AnswersAstronomy & Space7 years ago
  • What's so great about fusion?

    If we think about the kind of fusion the sun does, it's great. It just consumes plentifully available hydrogen and produces harmless helium and almost undetectable neutrinos. However, from what I've heard, the likely candidates for fusion power generation are hydrogen-1 to helium fusion.

    They use more rare isotopes like deuterium or tritium, or even elements other than hydrogen. Finding fuel won't be quite as easy. They might produce more hazardous by-products, such as neutrons, that would have to be shielded, and by absorbing them the shielding might become radioactive waste. Or they might produce light radioactive isotopes as a direct fusion product.

    So does fusion have any advantage over fission? Fission also produces a large amount of energy from a small quantity of fuel, and produces no greenhouse gases. It has disadvantages like hazardous, difficult to clean up waste products, and the possibility of accidents, but would fusion be just as dangerous?

    2 AnswersPhysics7 years ago
  • Why didn't Arabs discover America?

    Arabs were excellent sailors, making regular voyages around the Indian Ocean. They ruled in western north Africa, a closer starting point than Columbus. Was it purely lack of necessity? Europe needed an unobstructed route to the orient; the Arabs already controlled the land route.

    4 AnswersHistory8 years ago
  • How much helium will be produced as a by-product of world fusion energy generation?

    I was just wondering. Suppose technology to make net energy gain from fusion is discovered. How much helium will be produced? Enough to be useful? So much that it's a pollutant?

    I could probably take a guess myself, but I want to see what other people suggest. Possible approaches to an answer:

    1) Zero. The fusion reaction that will be successful is not one that produces helium as a product.

    2) Estimate world electrical energy use. Calculate the energy produced by one fusion reaction and the amount of helium produced. Solve.

    3) Somewhat less than that, because fusion won't meet all the world's energy demand, because of some limitation of resources, if only limitation of money to construct the facilities.

    4) A bit more than that, because fusion will be used for more than just electrical production, for example for large ship propulsion.

    5) A lot more than that, because with a nearly free source of energy, mankind will come up with many more uses for energy, and multiply its energy use many times.

    1 AnswerPhysics8 years ago
  • When fundamental forces unify at high energy, what is this energy "of"?

    For example, they say at 100 GeV the electromagnetic force and weak force unify into an "electroweak force". My question is, what energy has to exceed 100 GeV for the electroweak force to come into play? 100 GeV is actually a tiny amount of energy, much less than 1 joule. There's more energy in an AA battery. But I guess what it means is energy in a single particle, or average energy of particles according to kT.

    What if the temperature is close to the kT cutoff for the electroweak force, so that some particles have energy below 100 GeV and others above 100 GeV? Then are half the particles affected separately by weak and electromagnetic forces, and half by the combined electroweak force? Or is there some fuzzy region where the effect is a mix of both descriptions? Or is it the energy of the bosons that determines which set of rules apply, so that we will get some photons and some B0 bosons, but all the electrons will interact with both? Is it impossible to have a gamma ray photon with more than 100 GeV energy? What about the fact that energy is dependent on reference frame? Some gamma ray might have less than 100 GeV energy in our reference frame, but in a reference frame moving relativistically toward the source, it will be blueshifted and might have more than 100 GeV energy. In that reference frame will the identity of the particle be different, because now electroweak rules apply?

    What about other high-energy cutoffs, like asymptotic freedom in the strong force?

    1 AnswerPhysics8 years ago
  • A bit confused about binding energy and fusion?

    So I've heard either nickel-62 or iron-56 is the heaviest nucleus you can produce exothermically by fusion, because these have the highest binding energy per nucleon or lowest mass per nucleon respectively. But if I look up the masses of these isotopes, and then consider trying to fuse them with helium, it looks like it should be exothermic. E.g.

    56Fe + 4He -> 60Ni

    55.9349375 u + 4.002602 u = 59.9375395 u > 59.9307864 u

    62Ni + 4He -> 66Zn

    61.9283451 u + 4.002602 u = 65.9309471 u > 64.9292410 u

    I considered that I should be using masses of ionized nuclei, but even if I use the mass of the alpha particle instead of neutral helium, I still get more mass on the left hand side. If anything that underestimates the extra left-hand-side mass. For the bigger atoms, subtracting out the electrons should remove more from the right hand side, which has all the same electrons as the left hand side plus two more.

    It occurs to me that there should be no rule about fusion getting past the binding energy per nucleon peak. The only rule should be that you can't fuse two elements that are *both* above the binding energy per nucleon peak. Say you try to fuse two elements with BEPN a and b, and m and n nucleons respectively. Then to be exothermic the resulting nucleus has to have BEPN c

    c > (am + bn) / (m + n)

    If both a > c and b > c then (am + bn) / (m + n) > (cm + cn) / (m + n) = c, so it's impossible. But if either a < c or b < c then it's possible. This is the case where am is for 62Ni, bn is for 4He, and c is for 66Zn. b < c < a.

    But I'm explicitly told here - http://en.wikipedia.org/wiki/Alpha_process - that a similar reaction 56Ni + 4He -> 60Zn is endothermic. I checked that one out too and yeah the reactant masses add up to more than the product mass.

    What's the deal?

    2 AnswersPhysics8 years ago
  • Why do my ceramic bowls get hot in the microwave?

    I have these ceramic bowls with bright color glazes: yellow, orange, lime green, blue. If I try to microwave something in them, the bowl gets scorching hot and the food stays cool. What might be in my bowls that's absorbing microwaves?

    3 AnswersPhysics8 years ago
  • Does dark matter interact with other dark matter, besides by gravity?

    Supposedly dark matter is matter that interacts with normal matter by gravity, and little or not at all by any other force. But how does it interact with other dark matter? Also only by gravity, or could there be other forces that affect it?

    If only gravity affects it, would it tend to collapse to form "dark black holes", and if so, would those be indistinguishable from normal black holes? I would assume they would be pretty similar, since in any case we can only detect black holes through their gravitation. Only perhaps dark matter black holes could occur in other mass ranges besides stellar mass and supermassive.

    Or would not having other interactions besides gravity prevent accumulations of dark matter from being able to shed excess kinetic energy and angular momentum, so that they couldn't collapse, but instead would orbit the center of mass. Isn't it possible to still shed energy via tidal interactions?

    Supposing it does interact, is it possible that it has an exactly parallel set of interactions to normal matter, so that there is for example a dark electric charge, where dark positive charges attract dark negative charges, but have no effect on normal matter positive and negative charges. And dark charges imply a dark photon to carry the force. Dark strong forces would allow dark fusion in dark stars. Doesn't that blow your mind? There could be a whole parallel galaxy to ours occupying the same space.

    Is there anything that isn't possible about that hypothesis? How could we test it? With sensitive gravity probes could we start to map the dark matter in our nearby interstellar space to a high degree of accuracy?

    If galaxies are filled with dark stars radiating dark light, could dark light be dark energy? Or is it still not enough to account for all the dark energy?

    2 AnswersPhysics8 years ago
  • Why is the cosmic microwave background radiation coming toward us instead of going away?

    I mean, I suppose some of it is going away, but how is there any coming toward us.

    I understand that the matter that became our galaxy was contained inside the matter that emitted the CMBR, therefore the CMBR would reach "us" at some time, but why didn't it all pass by us billions of years ago? Correct me if I'm wrong, but wasn't the universe much smaller at the time the CMBR was emitted, say less than 1 billion light years across. So the most distant emitter of CMBR would be 1 billion light years from "us" or less.

    I get special relativity to some extent, so I know I need to think of an inertial reference frame. Let's pick the frame in which the matter that eventually became our galaxy was at rest at the time the CMBR was emitted, and call its origin the position of that matter. The most distant CMBR emitter was less than 1 billion light years away, and more than 1 billion years has passed in that reference frame, so the CMBR should have passed the origin. Our galaxy might have accelerated, but my understanding is we should have "decelerated" (i.e. accelerated toward the center of the universe, therefore toward the most distant origin of CMBR). Currently the expansion of the universe is accelerating, but I read that it was not always so. So the light should have passed our galaxy too.

    I'm guessing the answer has something to do with general relativity, which I don't understand much. Does the mass of the entire universe affect distance or time in some way to resolve this discrepancy? Is it possible to explain how general relativity affects the way the CMBR reaches us, without teaching me to understand the whole thing? Or if not just tell me it has to do with general relativity and leave it at that.

    2 AnswersPhysics9 years ago
  • Can Alpha Centauri Bb have an atmosphere?

    It's theorized to have a temperature of 1200 degrees C. That, and being so close to its star, I think would strip away any atmosphere consisting of familiar gases: hydrogen, nitrogen, oxygen, methane. However, can it hold on to heavier substances, which would have gaseous forms at that temperature? What would be some examples?

    It's suspected to have a magma ocean. My basic physical chemistry knowledge says that anything that's liquid has a vapor pressure, so would there be some small amount of evaporated silicon dioxide in the atmosphere, if it has one? And therefore would there be lava rain?

    1 AnswerAstronomy & Space9 years ago
  • Silicon-based life on Alpha Centauri Bb?

    No answer in Biology, trying here.

    A terrestrial sized planet has been discovered orbiting Alpha Centauri B.

    http://en.wikipedia.org/wiki/Alpha_Centa%E2%80%A6

    It's not in the habitable zone. Its temperature is estimated as 1200 degrees C, hot enough to melt silicates. As such it might contain magma oceans.

    Silicon has some chemical similarities to carbon. Could life exist on Alpha Centauri Bb based on molecules based on chains of silicon instead of carbon? Can the magma act as a solvent for those chemicals, as water does for carbon-based chemicals?

    4 AnswersAstronomy & Space9 years ago
  • Aircraft "thrust" vs. power?

    A simple question of Newtonian mechanics has long eluded me. Automobile engines have limited power. Gears help, but as you go faster and faster, you can produce less forward force. The strength of jet engines is usually quoted as "thrust", which is a force. As the jet goes faster and faster, it will produce more and more power by the formula P = Fv. Does that mean a jet engine has unlimited power? Surely there must be a limit, based at least on the maximum rate of fuel delivery, and perhaps before that based on a maximum rpms like a car. But maybe the plane reaches its top speed before its maximum power?

    What about rockets in space? Do they also produce constant thrust? I'm not even sure how to make sense of power here. By the formula, a rocket going 8000kph would use more power than one going 100kph, for the same thrust. Therefore I'd think it would use more fuel. But according to relativity, we can regard either as at rest, therefore using no power. Does it have to do with the exhaust gas? The rocket "at rest" would emit exhaust gas at nonzero speed, with thrust equal to that applied to the rocket. Therefore some power is spent there.

    4 AnswersPhysics9 years ago
  • Prove |-V> = -|V> (vector spaces)?

    I'm having a little trouble with this proof. Here are the axioms we may use:

    1. Vector sum |V> + |W> is uniquely defined.

    2. Scalar multiplication is uniquely defined.

    * Closure: the result of sum and multiplication is also an element of the vector space.

    * Scalar multiplication is distributive in the vectors: a(|V> + |W>) = a|V> + a|W>

    * Scalar multiplication is distributive in the scalars: (a + b)|V> = a|V> + b|V>

    * Scalar multiplication is associative: a(b|V>) = (ab)|V>

    * Addition is commutative: |V> + |W> = |W> + |V>

    * Addition is associative

    * There exists a null vector |0> such that |V> + |0> = |V>

    * For every vector |V> there exists an inverse under addition, |-V>, such that |V> + |-V> = |0>

    We also previously proved

    |0> is unique

    0|V> = |0>

    so you can use those in your proof.

    First, I interpret -|V> in what we are supposed to prove as -1|V>, since otherwise no operator has been defined that it could be interpreted as, other than the inverse operation itself which would make the proof trivial.

    The book suggests to start with

    |V> + -|V> = 0|V> = |0>

    I can kind of see how they got that.

    |V> + -|V>

    = 1|V> + -1|V> ???

    = (1 + -1)|V> associative

    = 0|V>

    But I'm not quite sure how to justify

    |V> = 1|V>

    in the line marked with ???

    In fact I can't even figure out how to rule out the possibility of strange vector spaces that have some subset that aren't even in the domain of scalar multiplication, i.e. there exist vectors |V> such that |V> =/= a|W> for any a or |W>.

    I imagine you would go about trying to prove |V> = 1|V> based of the multiplicative identity properties of 1, i.e. 1a = a for scalars, and perhaps the associative axiom. In particular the fact that 1*1=1 might be useful. However I can't figure out how to get it.

    2 AnswersMathematics9 years ago
  • Strong force potential dependence on color charge?

    I have found this formula for the potential of the strong force between quarks:

    V(r) = −(4/3)α(r)ℏc / r + kr

    However, nothing is mentioned about color. The above is always attractive. It seems to me, for sets of three color charges to neutralize, the effect between some pair must be repulsive.

    Let's assume we're in the range where the +kr term dominates. So V(r) ~= kr. The force on a test quark, which let's say has green color, from a hadron of balanced colors, should be zero. So my guess is that k is different for different pairs of colors, and negative for some pair.

    Vtot(r) = k[rg]r + k[gg]r + k[bg]r

    = (k[rg] + k[gg] + k[bg])r

    k[rg] + k[gg] + k[bg] = 0 so that force is zero.

    Possible solutions:

    k[rg] > 0

    k[gg] = -2k[rg] = -2k[bg]

    Unlike colors attract, and like colors repel twice as strongly.

    k[rg] < 0

    k[gg] = -2k[rg] = -2k[bg]

    Unlike colors repel, and like colors attract twice as strongly. But that's inconsistent with a hadron holding together, since each quark is affected only by unlike color quarks.

    k[rg] =/= k[bg]

    For example

    k[rg] > 0

    k[gg] = -k[rg]

    k[bg] = 0

    That is red and green quarks attract, like quarks repel, and blue quarks have no effect on green quarks. It seems wrong and also it turns out it's impossible to get a neutral effect on all colors of test quarks unless the red-red and/or blue-blue coupling constants are different from the green-green, which seems even more wrong. The same is true even if k[bg] is not zero but is any value different from k[rg].

    I know that color is not a permanent property of quarks and that they change color as they interact, but I'm not really sure what to make of that. I also don't understand non-abelian gauge groups or whatever, so if that is the only way to explain the problem, never mind.

    Is there any sense to my idea that k[gg] = -2k[rg] = -2k[bg]?

    1 AnswerPhysics9 years ago
  • Classical electromagnetism; acceleration; momentum; energy?

    In reading on quantum mechanics from a qualitative point of view, certain issues relate to the fact that in classical electromagnetism, accelerating charges produce radiation, and a charge absorbing radiation accelerates it. For example the Bohr model of the atom is supposed to fix the idea of a circularly orbiting electron, which would lose energy by EM radiation; and the momentum of the photon is anticipated by classical theory. I would like to have a better understanding of those ideas. I have a fair understanding of the simple sine wave solution to Maxwell's equations in free space, but I don't know how to proceed when the charge and current terms are involved. Can anybody help me with any of these concepts?

    1) Why does an accelerating charge radiate EM waves? What are the characteristics of those waves? Say for example an electron decelerates in a straight line from 1m/s to 0m/s over 1s. Which direction will it radiate waves, what will be their polarization, etc.

    2) Why does a moving charge NOT radiate EM waves when it is moving with constant velocity? What is the shape / formula for the E and B fields near its path?

    3) Does any acceleration of a charge produce EM waves? What about a charge accelerating because of gravity? What about a charge accelerating because of the electric field of another charge? What about accelerating because of an EM wave in the first place? What about because of the strong force? What about when two atoms collide in a gas, and their electrons and protons experience accelerations? Any different if it's an ion colliding with a neutral particle?

    4) What is the classical derivation of the momentum of an EM wave? Does conservation of momentum imply that an EM wave will accelerate a positive or negative charge the same way, and how is that possible, when most effects are opposite on positive and negative charges.

    10 points if you even answer part of one of those questions, if no one can do better.

    2 AnswersPhysics9 years ago
  • At low energies, could there be more fundamental forces?

    At high energies, some of the fundamental forces merge into one. For example the electromagnetic force and the weak force merge into the electroweak.

    Is it possible that their are yet unknown fundamental forces, which merge into the known ones above the lowest energies we've ever been able to achieve? Would there be any hints of that above that energy, for example some anomalies in the known forces, or do they act totally consistently right until they split?

    2 AnswersPhysics9 years ago
  • Is THIS what the Higgs boson is?

    There's tons of questions on here asking what the Higgs boson is. I haven't seen a very adequate answer. I think I have a vague grasp of it, but I need confirmation.

    The important thing is first the Higgs field. The Higgs field interacts with certain particles. In a similar way to how being in an electric field gives an electron potential energy, being in the Higgs field gives any particle that interacts with it potential energy.

    Energy IS mass. By having Higgs potential energy, the particle has some energy that is not associated with its momentum, p.

    E > pc

    E^2 > p^2 c^2

    E^2 = p^2 c^2 + X

    The full relativistic energy-momentum equation is

    E^2 = p^2 c^2 + m0^2 c^4

    So

    X = m0^2 c^4

    X > 0

    m0 =/= 0

    The particle has invariant mass a.k.a. rest mass.

    Also,

    v = pc^2 / E

    = (pc / E)c

    pc < E so

    v < c

    The particle travels at less than light speed.

    So that's generally how the Higgs field results in some particles having mass. Quantum theory says that every field has to have a particle. For example, the electric field is associated with the particle, the photon. The particle for the Higgs field is the Higgs boson.

    In an atom, there is an electric field holding the electrons close to the protons on the nucleus. There are photons associated with that electric field, but they're virtual photons. Analogously, particles that have mass are associated with a Higgs field and Higgs bosons, but they're virtual Higgs bosons.

    Given the right energy, an atom can emit a real photon. Likewise, a particle can emit a real Higgs boson, but it requires much more energy than a photon.

    Is that the gist of it?

    2 AnswersPhysics9 years ago
  • What does "help" with your homework mean to you?

    I see a lot of questions from people asking for help with their homework, but then the content of the question is just the homework problem. To me, the word "help" means you do part and I do part. So, where's your part?

    Just FYI, if you make an effort to answer the question yourself, and show what you've tried, I'm willing to help. If you are just posting every question from your assignment and expecting someone else to do it for you, I'm not. How about any other answerers here? Do you also skip questions if the asker hasn't shown that they've tried anything?

    2 AnswersPhysics9 years ago