Janus

A lamp typically uses 100w, so at 120v, it draws 833 ma A toaster can use 900w and will draw 7.5 A. The thinner the cord, the greater the resistance per foot. using the formula P=I²R. you can see how much wattage is used by the cord itself. Thus, the same cord as used by a lamp would use 81 times as many watts if used on a toaster. So not only are you wasting power via the cord, but this wattage is lost by heating the cord, and the thinner the cord, the hotter it must become to shed the heat. A thin cord trying too carry to much current runs the risk of overheating and causing a fire. This is also why you need to be sure that when you use an extension cord that it is rated high enough to run whatever you plug into it. If the cord becomes overly warm to the touch while in use, it is too thin. On an additional note, this is why you never replace a circuit breaker or fuse with a higher rated one than is already there. And you never, never, never replace a household fuse with a penny. Those fuses/circuit breakers are there to protect your household wiring from overheating and causing a fire hazard.

It does, but it is insulated. Basically, you have a nichrome wire coil that is encased in a ceramic casing which is in turn protected by an outer metal shell,

Between 100-300ma. 100ma being the lower limit of what could kill you and anything over 300ma definitely being lethal.

Then I suggest that you either get your watch or memory checked, because one of them is wrong.(assuming that you are seeing the sunrises/sunsets from the same latitude each year.)

Yep, just about 1700 km below the surface. Meaning the Earth varies its distance from the Sun by 3400 km due to its orbit around this barycenter, compared to the 5,000,000 km it varies due to the eccentricity of its orbit around the Sun.

No, the shock wave that forms the "boom" is continually produced as long as the craft is supersonic. You hear the boom when the edge of the shock wave passes you.

Actually, unless the balls are on a collision course, they will never collide or slow to a stop because 100 m/s is way more than escape velocity for these masses at these diatances (In fact, the balls would have to have started at a center to center distance of less than 1.3e-14 m for 100m/s to be less than escape velocity.)

You have a train with a relative velocity of 70 kph to the ground. Is the train moving or the ground? Despite your argument, you can't say. The train engine doesn't have to move the the country side, all it has to do it overcome the friction between itself and the rails, and that is the same whether or not you consider it the train or the country side as moving. You can even consider the situation where the train and country side start off at rest WRT each other. The train starts up its engine and the two begin to develop a relative motion. Again, you do not have to resort to the train engine accelerating the whole country side to consider the Train as stationary. The countryside and train instead were both traveling at 70 kph before the engine started and the engine then slowed the train down, while the country side continued to move. Again, the energy exerted by the train is the same no matter what final speed you decide the train has.

The clock is at a lower gravitational potential. Gravitational potential is found by: [math]G_{p}= -\frac{GMm}{d}[/math] Gravitational force is found by: [math]G_f= \frac{GMm}{d^2}[/math] For Earth, The acceleration for a 1 kg object would be: [math]G_f= \frac{G 5.97 x10^24} {(6.378x10^6)^2} = 9.793 m/sec[/math] for Uranus: [math]G_f= \frac{G 8.68 x10^25} {(2.556x10^7)^2} = 8.87 m/sec[/math] Now compare gravitational potential for the same mass: Earth: [math]G_{p}= -\frac{G 5.97x10^24}{6.378x10^6} = -6.245e7[/math] Uranus: [math]G_{p}= -\frac{G 8.68x10^25}{2.556x10^7} = -2.266e8[/math] (A larger negative potential is a lower potential, for example, the potential on the Moon's surface would be -2.822e6)

This implies that the total time dilation is not dependent on its final velocity difference, but on the amount of acceleration/deceleration/gravity it has experienced since the first measurement. You also seem to be suffering from a common misconception about time dilation due to gravity; that is it due to the gravitational force experienced by the object. This is not true. It is due to a difference in gravitational potential. For example, you gave the situation of a clock running faster on Mars than on Earth, which is fine, as long as you don't think it is just due to the weaker surface gravity. For instance, a clock will run slower on the surface of Uranus than on Earth, but the surface gravity on Uranus is slightly less than on Earth.

Well, the Magellanic Clouds are considered irregular dwarf galaxies in their own right, so this is not strictly true.

And it takes the whole 6 x 10^24 kg of the Earth's mass to do so. Whereas, as pointed out in the last post, it only takes a small magnet yo lift up a paper clip against the pull of the entire Earth. This is what is meant by gravity being a weak force. It takes a huge mass to create a gravitational force equal to the electromagnetic force created by a very small charge or magnetic pole. Where gravity holds the advantage is in the fact that it only attractive, whereas the electromagnetic forces can be either attractive or repulsive. On a large scale, the negative and positive charges and the North and South poles tend to cancel each other out, leaving gravity to hold sway. One can say that gravity is both rearmost and foremost. Rearmost because of its relative weakness, but foremost because it has the greatest combined effect over large scales.

Generally such drawings are of ideal circuits. IOW, the wires are considered as have zero resistance. The lamps are considered as loads with some finite resistance. when current divides between two branches, the amount of current that travels through each branch depends on the relative resistances of the branches. If they are equal, equal current travels through each branch.. If the total current is 10A, 5 A travels through each branch. If the resistance in one branch is half that of the other, twice as much current travels though it. When one branch has zero resistance and the other has any resistance, then all of the available current passes through this branch. In real life the wire has some small resistance, but it will still be very small compared to the resistance of the lamp. Thus the vast majority of the current will pass through the branch on with the switch, leaving only a very small current to pass through the lamp. This current will be so small that it won't be enough to cause the light to glow.

Put quite simply, because it would take an infinite amount of energy just to accelerate up to the speed of light. IOW, no matter how close to the speed of light you are or how much energy you add, it will only get you closer to the speed of light and never equal to or faster than.

Not any change in speed, any change in acceleration (see above) In an elliptical orbit it is the decreasing distance that results in the increase in pull. Since the mass of the planet does not change neither does its inertia. Inertia only changes if you add more mass to the planet. just increasing the pull does not add more mass. Yes.

I have feeling here that part of the problem revolves around the use of the word "acceleration" . Acceleration is a change in velocity, where velocity is measured by both speed and direction( IOW a car driving 30 kph heading East has a different velocity than a car driving at 30 kph heading West. They have the same speed, but have different headings)A planet in a circular orbit is constantly accelerating even though it never changes its speed, by virtue of the fact that it is constantly changing its heading. So what Swansont is saying is that increasing the mass of the object does not change its acceleration (either its speed or heading), and it continues to travel in the same orbit it did before. Just as above, both objects have equal acceleration (this time as a change of speed), fall at the same rate and hit the ground at the same time.

The difference in the second situation is that the second case the inertial mass increases by the same factor as the force. Try looking at it this way: Without the Sun's gravity, the planet would fly off in a straight line. It is the centripetal force supplied by the Sun's gravity that causes the planet to follow the curve. So what happens if you double the mass of the planet? For one, you double the gravitational force. At the same time you double the inertia of the planet, and double the amount of force it takes to hold the planet to the same curve. Mathematically it looks like this. The centripetal force needed to hold an object in a circular path is [math]F_c = \frac {mv^2}{r} [/math] The gravitational force is found by [math]F_g =\frac{GMm}{d^2} [/math] If m is the mass of the planet M is the mass of the Sun d is the distance between Sun and planet and is equal to r (the radius of the orbit. Then for a circular orbit: [math]F_c=F_g[/math] [math]\frac {mv^2}{r}=\frac{GMm}{d^2}[/math] Substitute r for d: [math]\frac {mv^2}{r}=\frac{GMm}{r^2}[/math] Reducing [math] v^2=\frac{GM}{d}[/math] Notice that the mass of the planet (m) cancels out of the equation, so that the orbital velocity for a given orbital distance "r" remains the same no matter what the mass. (within limits)

I going to go a little deeper in to the following quote from Eric 4 and show why it is a mis-interpretation of time dilation as described in the book. The above contends that what Einstein meant by a clock in a moving frame ticking slower was that the light received from a receding clock would take longer and longer times to reach an observer, causing an observed slowing of the receding clock. Example: Clock B recedes from clock A at 0.1c. If at the point when clock B was zero distance from clock A both clocks read a time of zero, then after 1 sec clock B will be 1/10 of a light sec from A. The light carrying the image of Clock B reading 1 sec, will then take 0.10 sec to reach clock A. Therefore clock A will read 1.1 sec when an observer next to clock A sees clock B read 1 sec. As B continues to recede, The observer will see Clock A read 2.2 sec when clock B reads 2 sec, clock A read 3.3 sec when clock B read 3 sec, etc. Always a 1.1 to 1 ratio of clock A to B. Accordingly we can get other ratios for different speeds for clock B such as below. 0.25c gives a 1.25:1 ratio 0.5c gives a 1.5:1 ratio 0.75c gives a 1.75:1 ratio 0.99c gives a 1.99:1 ratio So the higher the speed of B, the slower the observer sees it run. But is this the effect Einstein is really writing about in the book? Let's see, why don't we? In the book he gives an equation for time dilation that predicts the ratio between the rates of a "stationary" and moving clock. The equation is as follows: [math]t = \frac{t`}{\sqrt{1-\frac{v^2}{c^2}}}[/math] In our example above t` would be the time reading on clock B and t the reading on clock A as determined by an observer with clock A. If we plug in the velocities from our previous example into this equation we get: 0.1c gives a 1.005:1 ratio 0.25c gives a 1.033:1 ratio 0.5c gives a 1.155:1 ratio 0.75c gives a 1.512:1 ratio 0.99c gives a 7.089:1 ratio These values disagree with the previously values above by several percent, with the last value disagreeing by 150%! So it is obvious that Einstein was not considering the increasing time lag for signals traveling between the clocks as the mechanism behind time dilation, because this mechanism is inconsistent with the equation he uses for time dilation. To claim otherwise is a mis-interpretation of his writings.

That's just ridiculous. It's like saying that anything written on evolution that wasn't written by Darwin is just an interpretation. This book was written as an introduction of Relativity to laymen. It does a fair job of that. But it is far from a complete description of the theory. This is just wrong, and a complete mis-interpretation on your part of what the book says. I have read (and re-read) the book. I've owned a copy for 30+ years. And you are again mis-interpretating What the chapter says.

I've haven't gone to school within the last 3 decades, but I still use metric when talking science.

Are you talking about this book? http://bartleby.com/173/ If so, you do realise that this is meant as an introduction to Relativity, not an exhaustive text on the subject. Also, I see nothing in this book that supports your view.

Wavy line. Not only that, but it is a wavy line that always curves toward the Sun.

Positrons are emited naturally during some radioactive decays, this is known as beta+ emission. It happens when a proton in a nucleus converts to a neutron and positron. PET scans work by injecting such a positron emittng isotope into the person, then scanning the body for the gamma radiation emitted when the positrons mutally anihilate with electrons.

That equation only gives a a good answer at relatively low speeds (conpared to light). More acturate would be: [math]E = mc^2 \left ( \frac{1}{\sqrt{1- \frac{v^2}{c^2}}}-1 \right )[/math]

Why do keep saying this when you have been repeatedly told that it is WRONG! Time dilation has nothing to do with increasing distance between you and another object or the time it takes light to travel between you. This is easily seen by just looking at the equation for time dilation: [math]t = \frac{t`}{\sqrt{1- \frac{v^2}{c^2}}}[/math] There is no difference in the answer when v is negative (a velocity towards you) or positive(a velocity away from you). The time dilation remains the same, the clock moving relative to you runs slow. Compare this with the equation for Doppler effect: [math] f = \left( \frac{c}{c + v} \right) f` \,[/math] Where you do get a different answer for postive and negative velocities. The Doppler effect is due to the increasing distance and the time it takes light to travel between the two objects. But it is a completely different mechanisim from that which results in time dilation. In fact, when dealing with higher veloicites, the Doppler effect equation is modified to take the separate effects of Relativity into account and one gets the Relativistic Doppler shift equation: [math] f_o = \sqrt{\frac{1-v/c}{1+v/c}}\,f_e, [/math]