Senior Members
  • Content count

  • Joined

  • Last visited

  • Days Won


Janus last won the day on April 3

Janus had the most liked content!

Community Reputation

673 Glorious Leader

About Janus

  • Rank
  • Birthday August 28

Contact Methods

  • Website URL
  1. For one, you can accelerate while maintaining a constant speed, just travel in a circle. Secondly, kinetic energy is a relative and not absolute measurement. When you say you are gaining kinetic energy, what frame are you gaining it relative to? Different frames will say that your kinetic energy at any given moment have different values. When Einstein says the gravity and acceleration are equivalent (not "indistinguishable") He means as measured from the rest frame of the system concerned. The elevator is not gaining kinetic energy relative to itself nor are you gaining kinetic energy relative to the elevator. It doesn't matter if you are or are not according to some other frame.
  2. You seem to be assuming a "mechanistic" cause for Relativistic effects. This is not how Relativity deals with them. As I stated above, Relativity is about the geometry of Space-time. As an analogy, imagine that you have a 4 ft long board and a 3 ft wide doorway. The board won't fit through the door cross-wise, but if you rotate it some, you change its effective "width"with respect to the door's width and it will it fit through. You haven't altered the "structure" of the board in any way, you just changed its orientation with respect to the door. In Relativity, relative motion changes the relative orientation in space-time between frames. Objects moving relative to you undergo length contraction because you are measuring its length from a different space-time "angle" than it measures its own length. And since this angle difference is in space-time, it also causes you to measure clocks traveling on that object to run slowly and a disagreement over simultaneity between the two frames.
  3. What to you mean by "traveling into the future"? Whether or not an orbiting clock runs faster or slower than one on the surface of the Earth depends on its orbital altitude. Below a certain height it will tick slower and above it it will tick faster.
  4. With sound, The reason its speed is independent of the velocity of the source is that sound requires a medium to propagate and its speed is fixed with respect to that medium. So when we say that the speed of sound does not depend on the velocity of the source, what is meant that the speed of sound relative to the medium does not depend on the velocity of the source with respect to the medium. To illustrate, consider the following scenario: You have a completely enclosed railway car traveling down the tracks. As its back end passes an observer, the observer hits it with a hammer. The sound travels through both the air in the car and the air outside. Both sounds travel at a fixed speed relative to the air carrying them. Therefore, the sound traveling in the car will reach the front end before the sound traveling outside the car does. An observer in either the car or outside will measure the speeds of the two sounds to be different relative to himself. If you replace the sound with light and perform the same experiment with light, Both observers would measure both beams as traveling at the same speed and both would measure them as moving at c with respect to themselves. That being said, I think you are putting to much emphasis on light itself. The fact that light travels at c and has an invariant speed is a symptom of Relativity not its cause. It is the result of there being a speed 'c', which is the invariant speed of the universe. Light simply happens to travel at this speed. And the existence of c come from the rules of geometry the relationship between time and space follow. In the end, Relativity isn't about light at all, but the fundamental nature of time and space. Light just allowed us to uncover this nature.
  5. No. The given radius gives an orbital speed of 7673.556779 m/sec. At that speed and radius, the centripetal force would be 7673.556779^2 x 50/7=6,771e6 = 434.8211... N The force of gravity would be 3.987e14 x 50/6,771e6^2 = 434.8211.. N They are equal to the same degree of accuracy we have for the given parameters The 3.9987e14 is the "gravitational parameter" for the Earth which is equal to GM,where M is the mass of the Earth. This gives a more accurate answer as we know it to a better accuracy than we know either the gravitational constant or the mass of the Earth separately.
  6. A person moving at 5 mph relative to the space station, would not travel toward Uranus or anywhere else in a straight line they would just enter a different type of orbit around the Earth than the space station. If he were free to do so, (He wasn't inside the space station when he started.) this new orbit would take him just a bit further from the Earth than the space station before returning to the same distance from the Earth one orbit later. This orbit will also take a bit longer than the space station orbit. In order to permanently leave the vicinity of the Earth, he would have to increase his speed by another ~41 percent that of the space station itself (for the ISS this is 7.67 km/sec, so he would have to increase his speed by nearly 3.2 km/ sec. Even then, he wouldn't travel in a straight line, but on a parabolic trajectory, And once he got far enough away from the Earth that its gravitational effect was negligible, it would still be constrained to an orbit around the Sun, This drift with respect to the ISS would be hard to notice for an object inside the station. Imagine you are at one end of a 10 meter long section of the station and you toss an object from one end to the other at 5mph. The "outward" drift of the this object in the time it took to travel the length of the 10 meters works out to be just 2 inches. This is a smaller distance than likely to be caused just by the inaccuracy of the thrower. For a person traveling from one end to the other at typical speeds, his tendency to drift in towards the Earth or away from it is so small, you would need very sensitive measurements to notice it. It is a bit silly to argue that Newton's equation is wrong when there are countless examples that show its accuracy. Every satellite and space probe arrived where it is by virtue of precise calculations that are derived from the equation, If it was in error, these calculation would give the wrong answers and those craft would have ended up way off course instead of exactly where we planned to put them.
  7. No joke. Look up "Newton's Cannon". For an object in orbit, gravity is the centripetal force that prevents it from flying off in a straight line.
  8. I'm not sure if the OP is still around to see this, or if he'll even bother to read and try to understand it, but: Have you ever ridden in a fast moving car that reaches a dip in the road, and you get the "leaving your stomach behind" feeling? This is just a lesser magnitude example as to what is happening for the astronauts. At that moment, you are just slightly "lighter" with respect to the car than you are normally, and this is what your stomach is reacting to. If you where to be traveling so fast that the car's wheels left the road for some time, you would feel "weightless" and objects would "float" around in the car. This is because your "weight" is a function of gravity pulling down on you and the car, but the ground pushes back on the car and through it to you. When the car leaves the ground, both you and the car are still subject to the same force of gravity as before, but are both free to respond to it equally and are both in free fall. There is no force pushing up on you to resist gravity and you are "weightless". The "Vomit Comet" which is a plane designed to simulate this effect for longer than you could achieve in the car, does it by flying in an arc that follows such a free-fall path. Again, while following this arc, you, and object in the plane, float around in a weightless condition, even though the force of gravity on you has not let up one bit. The ISS is in a state of continuous free-fall. While near the surface of the Earth and at normal speeds, such a free fall path will always end up intersecting with the ground eventually, at the altitude and speed of the ISS, as it curves towards the ground, due to the spherical shape of the Earth, the ground curves away also. It basically end up in a state of constantly falling towards the Earth but it keeps missing it. Since both the ISS and astronauts are following the same free fall path, the astronauts end up in the same weightless condition as in the car in the air or the Vomit Comet during the arc, even though the force of gravity acting on them is not much weaker than it would be at the surface.
  9. Just to elaborate on the answer already given. Gases are just collections of fast moving particles bouncing off of each other. The pressure of a gas is a result of the collision of those particles. If you were to have a container, divided in two with a vacuum on one side and a gas under pressure on the other, The pressure on the dividing wall by the gas is the result of the constant collisions of those fast moving particles with the wall. If you cut a hole in the wall, the vacuum does not "suck" the gas from its side of the container, but rather, the particles that would have collided with the wall where the hole is have nothing to stop them from passing through to the other side so they will cross over into the vacuum side. Eventually (how long depends on the size of the hole) there will be just as many particles on one side as the other. Of course now the same number of particles will be in a larger volume with a larger inner surface area. While this will not change the force of collision for any given particle with the wall, it does decrease the frequency of collisions with the wall. This, in turn, decreases the net pressure felt by the wall. Now the Earth's atmosphere doesn't have a wall to hold it atmosphere, but here is how gravity does the trick. Take a ball and throw it upwards. It leaves your hand at a certain speed, slows as it climbs, and then eventually comes to a stop and falls back down. Air molecules are no different, They are bouncing off of each other at high speeds, but in order to escape into space they have to climb just like the ball, and just like the ball they lose speed as they do. Eventually they lose all their speed and can climb no further. Now if They had started with enough speed (known as escape velocity), they could have kept going, as gravity gets weaker the higher you go, and it would never be able to rob it of its last bit of velocity. However, the typical speed for air molecules are much lower than this required speed. (except in the case of light elements like helium and hydrogen, the Earth has a more difficult time holding on to these.) The speeds of the air molecules are a function of their temperature. With hotter gasses the molecules have faster speeds. If you were to heat our atmosphere considerably, you could cause it to lose its atmosphere. (Going back to hydrogen and helium, the reason the outer gas giants atmospheres contain them is bot because the planets are large and have strong gravity and they are further from the Sun and thus cooler. The Sun, on the other hand, which is mostly made of hydrogen, is very hot, but is also very massive, on the order of 1000 times the mass of Jupiter, and it only needs its gravity to hold together. In fact, if it weren't for the heat generated by fusion at its core, its gravity would compress it even smaller than it is.) Attempting to invoke buoyancy as an explanation for the effects we attribute to gravity is a bit of a non-starter as buoyancy relies on gravity to work. It works on the comparison of different weights and volumes of the object and fluid involved, and you can't have different weights without gravity. It also fails to explain why our space probes follow trajectories influenced by the gravity of the various bodies in the solar system while traveling through a vacuum, which has no buoyant effect. Nor does it explain why object still fall in a vacuum chamber, such as shown here:
  10. Mentalism uses gimmicks and tricks also, just because you don't know that that they are there doesn't mean they aren't. As an (out of practice)amateur magician myself, I assure you that it has nothing to do with the magician being "smarter". He is just taking advantage of certain aspects of human psychology and have learned the "tricks of the trade". I have one trick that uses no special gimmick to it at all other than the fact that at a certain point, the person I'm performing the trick for takes my word for what happened. It would take just a moment for them to ask for me to verify it, and force me to show them that it didn't, But in all the countless times I've performed this trick, no one ever has. There they are, knowing that I am trying to trick them, but they never consider that I might lie to them. Magicians also benefit from the retelling of their tricks, the trick always seems to grow and become more impressive in the retelling as the little details that are the key to how the trick was done get left out.
  11. Here's your problem. If you are using lbm, then the unit for force is the poundal, and there are 0.13255 Newtons to a poundal. Thus if you change the 4.44822 to 0.13255, yo get an answer of 26689 N If you wanted to use lbf, then the mass unit is the slug. lbm and poundal belong to the absolute FPS system of units, while the lbf and Slug belong to the gravitational FPS system of units.
  12. That's because, with the pivot point of the beam being inline with the points from which the pans hang,and unequal weights in the pans there is no equilibrium unless the beam is allowed to reach a vertical position. No, as the arm approaches vertical with the heavier weight being the lower, F1 and f2 extend until they equal Fg1 and Fg2 respectively and Fl1 and Fl2 go to zero. If the scale as I drew it were released from the shown position, the beam will rotate clockwise until it reaches the vertical position. Momentum will carry it past vertical, Fl1 and Fl2 will grow again, but now the net torque will be counter-clockwise. It will slow, stop, reverse and swing back to vertical. It will keep swing back and forth in smaller arcs until friction brings it to rest in the vertical position. This is the equilibrium point for this set up. In order to get the standard beam balance effect where the equilibrium when the weights are equal is with the beam horizontal (or at some less than vertical angle if the weights are unequal), the pivot point of the beam and "hang points" of the pans cannot be on the same line.
  13. You have a problem with your image On the left you properly show F1 as being the vector sun of delta F and FL, but on the right, you show FL as being the vector sum of F2 and Delta F, which is incorrect, F2 is the vector sum of the respective forces of delta F and FL on that side. The vectors, (here drawn as if the right weight was twice that of the left) would look like this: Here Fg2>Fg1 and respectively F2>F1 and Fl2>Fl1 by the same factor. F1 and F2 both act along the length of the beam and through the pivot and apply no torque to the beam. Fl1 and Fl2 apply torque to the beam with the net torque being in the direction of F2. In this situation the right weight will simply settle at the lowest point allowed by the scale. (If the cross beam had no restriction on its rotation limit, the final resting state would be with the beam perfectly vertical) This is different than the case where the pivot point is above the points where the hanging pans exert their forces on the beam. Now F1 and F2 do not act through the pivot point, but through a point where, combined, they exert an torque on the beam. This torque will be opposed to the net torque exerted by F1 and F2, and as the beam rotates clockwise, the torque grows while the Net F1 and F2 torque diminishes. At some angle these torques equalize, and if it is within the limits of the scale's motion, this is where the scale will settle.(Even if the beam has full rotational freedom, this point will be somewhere before vertical.)
  14. Of course this only applies if the hanging points of the Pans are in line with the pivot point of the arm. In the case where the hanging points are lower than the pivot, you get the following The red lines mark the center lines of the weights. If you try to tilt the scale, the right weight weight goes down and to the left and the left weight goes up and to the right. However, the Left weight/pan does not move as far right as the Right weight/pan moves to the left. There is a net movement of the CoG's of the weights to the left. If the weights were equal and the scale balanced before tilting, this will supply the torque that acts to return the scale to the balance point.
  15. The trick is to make sure that the scale is balanced without weights. Once it is properly calibrated, then it will only perfectly balance if the weight on the pans are equal. Something like this: The hanging arrow is affixed to the pivot point. There is a weight on a threaded rod which can be adjusted from one side to the other. First you adjust the weight so that the arrow points straight down with empty pans, you have now "zeroed" scale and it is ready for use.