cyeokpeng

I actually dig out my previous Physics notes and textbook to refresh my concepts. No much of a problem since I also learn a lot in the process. We help each other, that's the spirit.

NB: We consider an ideal blackbody so that the range of frequencies emitted is only due to its own temperature and NOT through reflection off the body's surface. An ideal blackbody has this property that it will absorb all the frequency of EM radiation falling on it and none gets reflected. The frequency (or wavelength for that matter. wavelength=c/frequency) of emitted radiation due to an ideal blackbody according to quantum theory consists of a range/multitude of frequencies in the whole EM spectrum. However, the range of frequencies plotted against the intensity of the emitted radiation has this bell shape curve (NB: Correct me if I m wrong this bell-shape curve follows the Rayleigh distribution) which has the intensity of radiation peaked at a certain frequency or wavelength following the Wien's Displacement law: lamda(max)*T = 0.2898*10^-2 m.K where lamda(max) = wavelength of the emmited radiation having the highest intensity T = temperature in Kelvins Note that the above only calculates the frequency of radiation emmited with the highest intensity. There are other range of frequencies emitted too, but the intensity tapers off quickly as you move away from the lamda(max) value following the bell-shape. This is the reason why in total darkness, you will not see your buddy emitting visible light from his body, shining like a star. Most of the intensity of EM radiation emmited by a person's body due to its own temperature 300K is at the frequency range of infrared region, and negligible intensity of EM radiation emmited is in the visible light region. (You can calculate it yourself using the Wien's dispacement law shown above) This is also the reason why many high-end operations to discover the enemies position uses infrared imaging detectors. The actual relationship of the intensity-wavelength plot (Your bell-shape curve) is given by the Planck's blackbody radiation law: I(lamda,T) = (2*pi*h*c^2)/(lamda^5*(exp((hc/lamda)*kT-1))) where I(lambda,T) = intensity of emmited radiation at a particular wavelength of radiation and temperature of the blackbody. lamda = wavelength of radiation in m T = temperature of blackbody in K h = Planck's constant 6.626*10^-34 J.s c = speed of light/EM waves 3*10^8 m/s k = Boltzmann constant 1.38*10^-23 J/K No, it does not depend on the characteristic of the material. The Planck's law of blackbody radiation is independent of the property of the blackbody material unless its atomic structure of material is degenerate, as in the case of neutron stars. This is because normal object containing matter in our universe are not degenerate in their atomic structure, and therefore can use the same Boltzmann distribution statistics to model its atomic structure, together with the application of Quantum theory since it involves small quantum particles like the atoms, and get the famous Planck's law of blackbody radiation. If the microstructure of the blackbody is degenerate (like in neutron stars where they are neutron-degenerate), then Planck's law does not hold and the blackbody radiation intensity-wavelength curve is obtained by other theories. If you have understood the above, this should be easy to answer. The tungsten in light bulbs is normal matter and not degenerate matter, so the Planck's law is applicable here. To get an idea in which region of frequencies most of the intensity of emmited EM radiation is in, we can apply the Wien's displacement law first to get the wavelength of max. intensity, which is found to be in the high-frequency end of infrared region if you use the max. steady temperature of the filament to be roughly 1000 K. Since the peak is nearer to the visible spectrum, you can deduce that there is observable intensity of emmited EM radation that is in the visible light range. This explains why it lights up as tungsten filament heats up. But careful here, the intensity of radiation peaks at infrared, which means that a large portion of power is waste heat (infrared) and this explains why light bulb using tungsten is not efficient to be used as lighting. More efficient ones uses metallic gas lamps known as flourescent lamps.

Please define your forces consistently. I'm confused with which one is your externally applied force AND which one is the force of friction. Do your f refer to the friction force acting on the object or the externally applied force???

QUESTION: Is it possible to add the friction force on the object by hand? Friction force is characteristics of the system in equilibrium i.e the static force of friction is responsible in keeping the object from moving. And static force of friction will always be acting opposite in direction to the direction of tending motion. The other type of friction, which is the friction force acting on the object when the object is on the move, is the kinetic force of friction, which is also acting opposite in direction to the motion direction. So question is, is it even possible to add the static force of friction by hand, that is in the case of your question? You may argue that by somehow increasing the roughness of the surface evenly throughout the surface, i.e the increase of roughness follows a uniform distribution. BUT NOTE THAT THE STATIC FORCE OF FRICTION IS ALWAYS ACTING IN THE OPPOSITE DIRECTION TO THE INTENDED MOTION. So in your question, let's say we can somehow increase the static force of friction continuously. If no external force is applied and the surface is horizontal (no gravity effects come in), the direction of static friction can be to the left or to the right for the object-surface system in equilibrium. If an external force of 1N is applied to the right, the static force of friction will be ACTING TO THE LEFT.(It cannot be to the right due to the nature of friction) So it is impossible to have the applied force of 1 N to be in the same direction of the f force. Unless I misunderstood your question, this should be the case.

The answer is that such a system will be impossible. The speed of the rocket with respect to the Earth will be zero if you do the relativistic time-dilation calculations, so it is impossible for the rocket to be moving in the opposite direction with respect to the train.

The answer is that such a system will be impossible. The speed of the rocket with respect to the Earth will be zero if you do the relativistic time-dilation calculations, so it is impossible for the rocket to be moving in the opposite direction with respect to the train.

Suppose we have a Aluminium on p-Si wafer contact. Calculations have shown that it will form a Schottky junction since the p-doping will move the fermi level towards the valence band, and hence the fermi level of the p-Si before contact will be lower than the fermi level of Al before contact (condition for a Schottky contact for metal-semiconductor). In a non-ideal contact, a thin insulating layer of Si oxide is between such contacts. If the applied voltage with respect to the Si-wafer is 10V, estimate the amount of field-assisted emission current. Also estimate the amount of the tunnleling current. Given that the thickness of the SiO2 layer is (i) 2nm. (ii) 10nm. My answer: Since holes are the majority carriers mainly responsible for the forward and reverse conduction currents for the system, we shall neglect the electrons as the carrier in the calculations. The schottky barrier formed is such that if a positive voltage is applied on the n-Si, it will be forward bias because the Schottky barrier is lowered by the external applied field, resulting in more holes moving from p-Si to Al, as compared to the holes from Al to p-Si. V = applied voltage = 10V Wonder if because of this, we can apply the Schotkky diode equation: I = I0*exp[-eV/kT] neglecting the -1 term where I0 = reverse saturation current = A*Richardson constant*T^2*exp[-effective barrier/kT] How do we calculate the effective barrier in the equation? I am fumbled on how to calculate the tunnelling current too.

Suppose we have a Aluminium on p-Si wafer contact. Calculations have shown that it will form a Schottky junction since the p-doping will move the fermi level towards the valence band, and hence the fermi level of the p-Si before contact will be lower than the fermi level of Al before contact (condition for a Schottky contact for metal-semiconductor). In a non-ideal contact, a thin insulating layer of Si oxide is between such contacts. If the applied voltage with respect to the Si-wafer is 10V, estimate the amount of field-assisted emission current. Also estimate the amount of the tunnleling current. Given that the thickness of the SiO2 layer is (i) 2nm. (ii) 10nm. My answer: Since holes are the majority carriers mainly responsible for the forward and reverse conduction currents for the system, we shall neglect the electrons as the carrier in the calculations. The schottky barrier formed is such that if a positive voltage is applied on the n-Si, it will be forward bias because the Schottky barrier is lowered by the external applied field, resulting in more holes moving from p-Si to Al, as compared to the holes from Al to p-Si. V = applied voltage = 10V Wonder if because of this, we can apply the Schotkky diode equation: I = I0*exp[-eV/kT] neglecting the -1 term where I0 = reverse saturation current = A*Richardson constant*T^2*exp[-effective barrier/kT] How do we calculate the effective barrier in the equation? I am fumbled on how to calculate the tunnelling current too.

Our space is either finite or infinite, depending on the nature of the curvature of the space-time fabric taken on the average of the whole universe. Remember that if the average density of matter in the universe (normal matter plus dark matter) is greater than the critical density, then we have a closed universe. This means that eventually, the rate of expansion of the universe will eventually slow down as time passes and start to contract, resulting in the Big Crunch in the end. A closed universe has a Riemannian geometry (positive curvature in the space-time fabric), and one characteristic of such geometry is than it has finite volume, meaning the universe is finite. (Imagine a sphere like the earth has positive curvature, and it has a maximum finite volume of 4/3*pi*r^3) If the total average density of matter in the universe is very much smaller than the critical density, then we have an open universe. This means that eventually, the rate of expansion of the universe will eventually speed up as time passes and continue to expand forever.An open universe has a Lobachevsky geometry (negative curvature in the space-time fabric), and one characteristic of such geometry is that it has infinite volume, meaning the universe is infinite. (Imagine a saddle which has a negative curvature, and its volume increases at a increasing rate as you go farther and farther from the centre without any upper bound. This implies that the rate of expansion of such an open universe will increase with distance (analogous to the Hubble Law) and will continue to expand forever.) However, up till even now astronomers have no idea that our universe is a closed or open one. The average total density of the universe is found to be very close to the critical density, and this is suggestive of the nature of our universe. However, bear in mind that there is highly-likely that we have not taken into account much of the dark matter in the calculation of the total average density of matter. (since dark matter cannot be seen using normal EM radiation/light, so not easy to discover them) But still, some scientists believe that even accounting all the dark matter of the universe, our universe is still a closed universe. This is debatable. Of course, there is the other school of thought that after accounting all the dark matter into the calculation of the total average density of matter, the universe is an open universe. This is much more likely as experimental evidence have shown that even up till now (age of the universe is now xxx billion light years---forgot exactly how much), the universe is still expanding according to the Hubble's law. Any comments are welcome.

Our space is either finite or infinite, depending on the nature of the curvature of the space-time fabric taken on the average of the whole universe. Remember that if the average density of matter in the universe (normal matter plus dark matter) is greater than the critical density, then we have a closed universe. This means that eventually, the rate of expansion of the universe will eventually slow down as time passes and start to contract, resulting in the Big Crunch in the end. A closed universe has a Riemannian geometry (positive curvature in the space-time fabric), and one characteristic of such geometry is than it has finite volume, meaning the universe is finite. (Imagine a sphere like the earth has positive curvature, and it has a maximum finite volume of 4/3*pi*r^3) If the total average density of matter in the universe is very much smaller than the critical density, then we have an open universe. This means that eventually, the rate of expansion of the universe will eventually speed up as time passes and continue to expand forever.An open universe has a Lobachevsky geometry (negative curvature in the space-time fabric), and one characteristic of such geometry is that it has infinite volume, meaning the universe is infinite. (Imagine a saddle which has a negative curvature, and its volume increases at a increasing rate as you go farther and farther from the centre without any upper bound. This implies that the rate of expansion of such an open universe will increase with distance (analogous to the Hubble Law) and will continue to expand forever.) However, up till even now astronomers have no idea that our universe is a closed or open one. The average total density of the universe is found to be very close to the critical density, and this is suggestive of the nature of our universe. However, bear in mind that there is highly-likely that we have not taken into account much of the dark matter in the calculation of the total average density of matter. (since dark matter cannot be seen using normal EM radiation/light, so not easy to discover them) But still, some scientists believe that even accounting all the dark matter of the universe, our universe is still a closed universe. This is debatable. Of course, there is the other school of thought that after accounting all the dark matter into the calculation of the total average density of matter, the universe is an open universe. This is much more likely as experimental evidence have shown that even up till now (age of the universe is now xxx billion light years---forgot exactly how much), the universe is still expanding according to the Hubble's law. Any comments are welcome.

There is no length contraction in this case. This is because length contraction only occurs if the observed length of the system you are measuring is moving as a whole respective to the stationary observer. In your scenario, the measured length of the two points are not moving together with respect to the stationary observer, and hence no length contraction results as oberved by the stionary observer. Quote me if I m wrong.

There is no length contraction in this case. This is because length contraction only occurs if the observed length of the system you are measuring is moving as a whole respective to the stationary observer. In your scenario, the measured length of the two points are not moving together with respect to the stationary observer, and hence no length contraction results as oberved by the stionary observer. Quote me if I m wrong.

How is the reation force at the microscopic level related to the electromagnetic force?

How is the reation force at the microscopic level related to the electromagnetic force?

What about light that is swept past like the lighthouse effect, observed at a very very far distance away? Ideally, if we can observe the sweeping of light in an arc at increasing distance from the light source, the observed speed of sweeping action will increase until the observed speed is faster than the speed of light. From what I know, light rays does not possess finite rigidity and hence it will not bend inwards due to limiting rigidity of the light path. Real objects will appear bent if the same experiment is done using a long object instead of light rays.

What about light that is swept past like the lighthouse effect, observed at a very very far distance away? Ideally, if we can observe the sweeping of light in an arc at increasing distance from the light source, the observed speed of sweeping action will increase until the observed speed is faster than the speed of light. From what I know, light rays does not possess finite rigidity and hence it will not bend inwards due to limiting rigidity of the light path. Real objects will appear bent if the same experiment is done using a long object instead of light rays.

Another possibility of a system faster than the speed of light. We know from our observable radius of the universe to be defined by the Hubble's radius. Hubble radius = speed of light c/Hubble's constant. That means that if there are galaxies further away than the Hubble's radius (should be since we know from the Big Bang theory that our universe is still expanding), the galaxies should be travelling away from us faster than the speed of light. How do we explain this phenomenon? Does this violate Einstein's postulate?

Another possibility of a system faster than the speed of light. We know from our observable radius of the universe to be defined by the Hubble's radius. Hubble radius = speed of light c/Hubble's constant. That means that if there are galaxies further away than the Hubble's radius (should be since we know from the Big Bang theory that our universe is still expanding), the galaxies should be travelling away from us faster than the speed of light. How do we explain this phenomenon? Does this violate Einstein's postulate?

From the physics of optics, as light is reflected to the angle near the normal incidence, it will suffer a 180 degrees phase change if the refractive index of the reflecting material is larger than that of the surrounding material. So, at the point of contact with the planoconvex lens and the plane surface, due to 180 degrees phase change, the two reflected waves should "destroy" each other. This should result in the centre point being dark followed by white bands, then dark lines. Any explanation for that?

From the physics of optics, as light is reflected to the angle near the normal incidence, it will suffer a 180 degrees phase change if the refractive index of the reflecting material is larger than that of the surrounding material. So, at the point of contact with the planoconvex lens and the plane surface, due to 180 degrees phase change, the two reflected waves should "destroy" each other. This should result in the centre point being dark followed by white bands, then dark lines. Any explanation for that?

In a particular well-known experiment known as Newton's ring interference pattern, we get a dark fringe at the point of contact of the glass lens and the plane glass surface, followed by rings of bright fringes and dark fringes. This seems to show that at that point of contact, the monochromatic light reflected from the glass surface is 180 degrees out of phase with the light reflected from the lens surface. How can that be possible?

Another possibility of a system faster than the speed of light. We know from our observable radius of the universe to be defined by the Hubble's radius. Hubble radius = speed of light c/Hubble's constant. That means that if there are galaxies further away than the Hubble's radius (should be since we know from the Big Bang theory that our universe is still expanding), the galaxies should be travelling away from us faster than the speed of light. How do we explain this phenomenon? Does this violate Einstein's postulate?

What Swatson means is this: If the slope is at an inclination of theta to the horizontal, the vertical acceleration consists of two components. One component is the acceleration due to gravity g which is acting downwards. The other component is your normal reaction force acting on the trolley by the slope. This normal force is always acting at right angles on the trolley upwards, and it will always be the same magnitude as the force that is acting downwards perpedicular to the slope so that you will not see the trolley tunnelling into the slope. This normal force will then have the magnitude of g*cos(theta), using the above reasoning. The normal force component in the vertical direction will then be g*cos^2(theta), resolving the component of the normal force in the vertical direction. So the vertical acceleration will have g in the downward direction, and g*cos^2(theta) in the upward vertical direction. So the effective reduced vertical accleration will be g - g*cos^2(theta) m/s^2 in the downward direction.