I disagree. In Harry Potter for example, when referring to a Patronus as the name of the magical creature (not a specific Patronus like Harry's), they capitalized it, i.e., "Lupin knows all about Patronuses". I don't think there's any proper rule for this, just do whatever seems best stylistic.
So, I have to do a presentation in Physics, arguing that Tidal Power is better than any other power source. Does anyone know of any good sources that I could use? Thanks in advance.
Use Wikipedia, then, instead of citing Wikipedia, cite the citations that Wikipedia cites.
24) A matter wave is a concept that properly models the dual nature of matter. To properly account for quantum behavior, particles have to be described as a propagation of a wave. The quantity that "waves" is a purely mathematical entity called the wavefunction. It has no physical meaning, though the absolute value of the wavefunction squared gives the probability density of finding the particle.
25) Question is self-contradictory because there're no electron orbits when the electron is treated as a wave.
Ok, the answer your teacher wants is probably this: the electron forms sort of a standing wave around the nucleus; this wave satisfies the uncertainty principle; if the electron were to fall into the nucleus, you would know exactly its position, which is forbidden. So the electron standing wave just wobbles around the nucleus. Its allowed energies are discrete because the frequencies of a standing wave are discrete.
The centripetal force is the resultant inward push that is exerted on any rotating body. It's what actually causes the rotation. It's a real force because it's associated with physical interactions: a rotating yo-yo has the tension of its string as its centripetal force; the Earth's gravitational attraction towards the Sun is its centripetal force. Because it's produced by actual physical interaction with "stuff", all observers perceive it.
The centrifugal force is an outward push exerted on all bodies when the observer is rotating. It's a fictitious force because it's not due a physical interaction; it's not a mechanical, electromagnetic or gravitational force. Because it's produced by the observer's rotation, only this specific observer perceives it.
Basically you have to ask two questions:
1) Is this object rotating? If yes, then I know that all the real physical interactions on it will result in an inward push (centripetal force).
2) Am I rotating? If yes, then I know that for every single object, I will have to add, on top of all real physical interactions, an outward push (centrifugal force).
They are not a pair of forces due to Newton's Third Law.
First, centripetal force is actually a bad term; centripetal resultant is clearer.
For an object to perform a circular motion (with respect to you), it can be mathematically shown that it needs to have an acceleration that is always directed towards the centre of rotation. If you're in an inertial reference frame (you aren't being accelerated), you can use Newton's Second Law to deduce that if the acceleration is towards the centre, then so is the resultant force. We say this resultant is centripetal. There's no "centripetal force" in the sense that a force doesn't appear just because the object is rotating. There's also no centrifugal force in this case nor is there anything balancing the centripetal resultant.
But what if you are in a non-inertial reference frame? In particular, imagine that you're rotating along the object you're observing. Now, this object is at rest with respect to you. But wait! For the inertial observer, there was a centripetal resultant, so you, too, should observe a resultant force on the object, even though it's at rest! Newton's First Law fails in non-inertial frames or you can simply state that the law still holds by producing a "fictitious" centrifugal force that, when added with the already existing centripetal resultant, results in a net force of zero. In this case, a force does appear just because you are rotating.
Note that, no matter how you phrase things, Newton's Third Law breaks down in a non-inertial frame because the centrifugal force appears by itself without a reaction pair. Its pair can't be the centripetal resultant because it exists even when the centrifugal force doesn't. But, hey, Newton's Third Law is wrong anyway even in inertial frames so who cares?
To sum up:
- The physical configuration of the problem may produce a centripetal resultant and this resultant doesn't change as you change reference frames. For instance, if you're rotating a ball attached to a string, the tension in the string will end being a centripetal resultant; this tension is invariant for all observers.
- If your reference frame is inertial, then the centripetal resultant, together with Newton's Second Law, is all you need to explain how the object moves with respect to you.
- If your reference frame is rotating along the rotating object, you need the centripetal resultant, Newton's Second Law plus the centrifugal force to explain why the object remains at rest with respect to you. Alternatively, you accept that Newton's Laws are invalid in this context.
Obviously, this only apply to non-relativistic motion.
Why does the number of protons and neutrons determine whether if an atom is stable or radioactive?
Are you familiar with the electron shell model for atoms? Remember that due to the way it works, if an atom has a number of electron that fills exactly all its shells, the atom is chemically inert (you could say it's chemically "stable" through a gross abuse of language). When that happens we get a noble gas, which have 2, 8, 18, etc electrons total.
Well, atomic nuclei are quantum systems as well, so they, too, have a shell model of sorts. The closer the total numbers of nucleons (a proton or a neutron) of a nucleus is to the number needed to nicely feels all its shells, the more stable it is. These numbers are called magic numbers.
This is only a basic answer though. There're other factors that affect the stability of a nucleus, such as the fact that a neutron is unstable by itself and the fact that the nuclear attraction between nucleons is short ranged whereas the electromagnetic repulsion between protons is not (It's...less straightforward to explain how these things affect stability).
Number two is wrong. Jostling an electron can create an EM wave, but it's not involved in propagating the EM wave. From Wikipedia, an EM wave is "a self-propagating transverse oscillating wave of electric and magnetic fields". Basically, a moving charge will create a magnetic field, and that magnetic field will keep the charge moving, which will create a magnetic field, which will keep the charge moving... and on and on and on.
Hopefully that makes sense.
No quite. An EM is maintained solely by itself; in fact in can do this even in free space when there's no charge present.
What happens is this: two of Maxwell's equations (basis for classical electromagnetism) dictate that changes in the electrical field produce changes in the magnetic field and vice-versa.
So, in an EM wave, the oscillation of the magnetic field causes the electric field to oscillate which causes the magnetic field to oscillate which causes the electric field to oscillate...etc. The wave sustains itself, even in regions without charges because the electric and magnetic fields are everywhere.
Now, this only covers the propagation of EM waves. How are they created in the first place? Well, I'm afraid this question's hopelessly open-ended. There's probably dozen of different ways: accelerating/oscillating charges, black body radiation, electron transition between atomic levels, particle-antiparticle annihilation...the list goes on...
dcartist, you're absolutely correct.
According to the internet, a lightbulb filament is generally at a temperature of about 3000 K. Using Planck's law, it turns out that the filament releases most of its electromagnetic radiation as infrared, not visible light (though it releases sufficient visible light to shine with an orange/yellowish hue). Glass being able to absorb infrared radiation, the lightbulb will most certainly heat up, considerably so!
Arnnaria, whoever assigned this problem needs to get a reality check. This entire problem is hokum.
Arnaria, your overall thought process is correct, but you made some mistakes along the way.
The light bulb is clear so it gives off most of it's radiant energy through radiation
This is circular. The only reason why the bulb is clear in the first place is that it doesn't absorb visible light. It would be better to say that the fact that the glass bulb is clear is evidence that glass doesn't absorb radiation along the frequencies of visible light (which is the main frequency emitted by a lightbulb).
sends waves of heat
No such thing. It sends electromagnetic waves; a heat wave is something completely different and has nothing to with this.
If I left the bulb on long enough it would become slightly hotter than room temperature, but it is sending off it’s radiant energy through waves and the filament is not heating up the actual bulb
I would rephrase this as "As the glass bulb barely absorbs the light emitted by the filament, the bulb's temperature won't increase very much; perhaps only slightly above room temperature."
If a lightbulb was murky or colored in some way less radiant energy would be able to escape and the bulb would be hotter than a transparent bulb.
...or if you connect the lightbulb to a weaker power source so that it generates infrared radiation instead of light. Glass absorbs infrared most excellently.
Heat transfer is the energy that flows from a substance of higher temperature to a substance of lower temperature through conduction, convection, and radiation.
No, because it's possible to transfer heat from a cold object to a hot one.
Temperature is the measure of average speed of molecules per area per size given other considerations.
The temperature of an object is the average kinetic energy of molecules of the object. It doesn't depend on the size or area of the object.
Another thing, the problem the way you phrased seem sorta weird. The bulb will feel hot regardless of whether you touch it or just approach it with your hands; either way your hand is absorbing lots of radiant energy. I think the problem is better posed as "if you touch a bulb while it's on, it feels hot, but if you turn it off then touch it immediately, it feels cold, so the bulb's temperature never increased while it was on".
Depends on your time scale. For the purposes of all life on Earth, atoms can't be destroyed. Unstable ones will decay into lighter and more stable atoms and that's it.
But things change if you're willing to sit down and wait a few aeons. Some grand unified theories predict that protons are unstable and also decay so that after a very, very, very long time all atoms will be destroyed.
Another possibility is that if gravity ends beating the dark energy repulsion, then all matter will eventually be collected into black holes. End of atoms as we know them.
But if the dark energy repulsion beats the gravitational attraction, a possible scenario is the "Big Rip". The repulsion is so strong that galaxies, planets and eventually atoms are torn apart and destroyed.
For some reason I don't even remember where to start for resonance problems I get that I'm going to need to find the angular velocity of the wheel and convert that into the car's velocity but aside from that I really don't know what to do.
This is not damped oscillation, so resonance frequency = natural frequency.
Natural frequency is just the one you would calculate for a mass-spring system with the given mass and spring constant. Call the natural angular frequency w0.
The wheels rotate and so have an angular frequency w = v/R, v is the speed and R the radius. For resonance, w = w0 because resonance frequency = natural frequency.
@Arnnaria
Base your essay on the Wikipedia articles that Viricide linked, but instead of citing Wikipedia, cite the sources that Wikipedia cites. Voila, stupid restriction imposed by teacher circumvented.
Use Wikipedia, then, instead of citing Wikipedia, cite the citations that Wikipedia cites.
25) Question is self-contradictory because there're no electron orbits when the electron is treated as a wave.
Ok, the answer your teacher wants is probably this: the electron forms sort of a standing wave around the nucleus; this wave satisfies the uncertainty principle; if the electron were to fall into the nucleus, you would know exactly its position, which is forbidden. So the electron standing wave just wobbles around the nucleus. Its allowed energies are discrete because the frequencies of a standing wave are discrete.
The centrifugal force is an outward push exerted on all bodies when the observer is rotating. It's a fictitious force because it's not due a physical interaction; it's not a mechanical, electromagnetic or gravitational force. Because it's produced by the observer's rotation, only this specific observer perceives it.
Basically you have to ask two questions:
1) Is this object rotating? If yes, then I know that all the real physical interactions on it will result in an inward push (centripetal force).
2) Am I rotating? If yes, then I know that for every single object, I will have to add, on top of all real physical interactions, an outward push (centrifugal force).
First, centripetal force is actually a bad term; centripetal resultant is clearer.
For an object to perform a circular motion (with respect to you), it can be mathematically shown that it needs to have an acceleration that is always directed towards the centre of rotation. If you're in an inertial reference frame (you aren't being accelerated), you can use Newton's Second Law to deduce that if the acceleration is towards the centre, then so is the resultant force. We say this resultant is centripetal. There's no "centripetal force" in the sense that a force doesn't appear just because the object is rotating. There's also no centrifugal force in this case nor is there anything balancing the centripetal resultant.
But what if you are in a non-inertial reference frame? In particular, imagine that you're rotating along the object you're observing. Now, this object is at rest with respect to you. But wait! For the inertial observer, there was a centripetal resultant, so you, too, should observe a resultant force on the object, even though it's at rest! Newton's First Law fails in non-inertial frames or you can simply state that the law still holds by producing a "fictitious" centrifugal force that, when added with the already existing centripetal resultant, results in a net force of zero. In this case, a force does appear just because you are rotating.
Note that, no matter how you phrase things, Newton's Third Law breaks down in a non-inertial frame because the centrifugal force appears by itself without a reaction pair. Its pair can't be the centripetal resultant because it exists even when the centrifugal force doesn't. But, hey, Newton's Third Law is wrong anyway even in inertial frames so who cares?
To sum up:
- The physical configuration of the problem may produce a centripetal resultant and this resultant doesn't change as you change reference frames. For instance, if you're rotating a ball attached to a string, the tension in the string will end being a centripetal resultant; this tension is invariant for all observers.
- If your reference frame is inertial, then the centripetal resultant, together with Newton's Second Law, is all you need to explain how the object moves with respect to you.
- If your reference frame is rotating along the rotating object, you need the centripetal resultant, Newton's Second Law plus the centrifugal force to explain why the object remains at rest with respect to you. Alternatively, you accept that Newton's Laws are invalid in this context.
Obviously, this only apply to non-relativistic motion.
Are you familiar with the electron shell model for atoms? Remember that due to the way it works, if an atom has a number of electron that fills exactly all its shells, the atom is chemically inert (you could say it's chemically "stable" through a gross abuse of language). When that happens we get a noble gas, which have 2, 8, 18, etc electrons total.
Well, atomic nuclei are quantum systems as well, so they, too, have a shell model of sorts. The closer the total numbers of nucleons (a proton or a neutron) of a nucleus is to the number needed to nicely feels all its shells, the more stable it is. These numbers are called magic numbers.
This is only a basic answer though. There're other factors that affect the stability of a nucleus, such as the fact that a neutron is unstable by itself and the fact that the nuclear attraction between nucleons is short ranged whereas the electromagnetic repulsion between protons is not (It's...less straightforward to explain how these things affect stability).
No quite. An EM is maintained solely by itself; in fact in can do this even in free space when there's no charge present.
What happens is this: two of Maxwell's equations (basis for classical electromagnetism) dictate that changes in the electrical field produce changes in the magnetic field and vice-versa.
So, in an EM wave, the oscillation of the magnetic field causes the electric field to oscillate which causes the magnetic field to oscillate which causes the electric field to oscillate...etc. The wave sustains itself, even in regions without charges because the electric and magnetic fields are everywhere.
Now, this only covers the propagation of EM waves. How are they created in the first place? Well, I'm afraid this question's hopelessly open-ended. There's probably dozen of different ways: accelerating/oscillating charges, black body radiation, electron transition between atomic levels, particle-antiparticle annihilation...the list goes on...
According to the internet, a lightbulb filament is generally at a temperature of about 3000 K. Using Planck's law, it turns out that the filament releases most of its electromagnetic radiation as infrared, not visible light (though it releases sufficient visible light to shine with an orange/yellowish hue). Glass being able to absorb infrared radiation, the lightbulb will most certainly heat up, considerably so!
Arnnaria, whoever assigned this problem needs to get a reality check. This entire problem is hokum.
This is circular. The only reason why the bulb is clear in the first place is that it doesn't absorb visible light. It would be better to say that the fact that the glass bulb is clear is evidence that glass doesn't absorb radiation along the frequencies of visible light (which is the main frequency emitted by a lightbulb).
No such thing. It sends electromagnetic waves; a heat wave is something completely different and has nothing to with this.
I would rephrase this as "As the glass bulb barely absorbs the light emitted by the filament, the bulb's temperature won't increase very much; perhaps only slightly above room temperature."
...or if you connect the lightbulb to a weaker power source so that it generates infrared radiation instead of light. Glass absorbs infrared most excellently.
No, because it's possible to transfer heat from a cold object to a hot one.
The temperature of an object is the average kinetic energy of molecules of the object. It doesn't depend on the size or area of the object.
Another thing, the problem the way you phrased seem sorta weird. The bulb will feel hot regardless of whether you touch it or just approach it with your hands; either way your hand is absorbing lots of radiant energy. I think the problem is better posed as "if you touch a bulb while it's on, it feels hot, but if you turn it off then touch it immediately, it feels cold, so the bulb's temperature never increased while it was on".
Depends on your time scale. For the purposes of all life on Earth, atoms can't be destroyed. Unstable ones will decay into lighter and more stable atoms and that's it.
But things change if you're willing to sit down and wait a few aeons. Some grand unified theories predict that protons are unstable and also decay so that after a very, very, very long time all atoms will be destroyed.
Another possibility is that if gravity ends beating the dark energy repulsion, then all matter will eventually be collected into black holes. End of atoms as we know them.
But if the dark energy repulsion beats the gravitational attraction, a possible scenario is the "Big Rip". The repulsion is so strong that galaxies, planets and eventually atoms are torn apart and destroyed.
This is not damped oscillation, so resonance frequency = natural frequency.
Natural frequency is just the one you would calculate for a mass-spring system with the given mass and spring constant. Call the natural angular frequency w0.
The wheels rotate and so have an angular frequency w = v/R, v is the speed and R the radius. For resonance, w = w0 because resonance frequency = natural frequency.
@Arnnaria
Base your essay on the Wikipedia articles that Viricide linked, but instead of citing Wikipedia, cite the sources that Wikipedia cites. Voila, stupid restriction imposed by teacher circumvented.