The iron will modify the magnetic field as it moves upward. This will induce a voltage in the winding, which will oppose the current. So the iron will extract energy from the magnetising current. This will end up as kinetic energy of the iron, but then turn to heat (and sound) as the iron hits the magnet.
I am not sure how this works for a permanent magnet instead of an electromagnet. It sounds like the sort of problem Feynman will have covered in his lectures on physics. However, I do believe in conservation of energy so the energy has to come from somewhere. In both cases, once the iron stops moving no further energy is needed to hold it in place.
You need to realise that your questions relate to your/our understanding of physics, not to flaws in the physcis itself.
QUOTE
But when it is in the active phase of magnetization, electromagnet can lift a piece of iron. This is a further work, without spending more energy elsewhere.
Please let me know if i am wrong.
hello.
the piece of iron is lifted against gravitation force (x distance = work ,in direction of the gravitation lines). if the iron falls back to earth you will get back that energy/work............
Q Nowdays it is known that astronauts while are in weightlessness for a long time, then they lose muscle mass and tone overall. Isn't this a proof that on the earth there is more trouble or effort compared to an environment like empty space?
Does from this follows that one guy spends more energy to stay on earth proportionally to its gravity?
...........a big fat man needs more energy to move his body (against gravitation force) than a small one........not only your hands and feet move,even the blood circulation (heart work) needs energy to move blood from toe to tip...........
greetings
But when it is in the active phase of magnetization, electromagnet can lift a piece of iron. This is a further work, without spending more energy elsewhere.
Please let me know if i am wrong.
hello.
the piece of iron is lifted against gravitation force (x distance = work ,in direction of the gravitation lines). if the iron falls back to earth you will get back that energy/work............
Q Nowdays it is known that astronauts while are in weightlessness for a long time, then they lose muscle mass and tone overall. Isn't this a proof that on the earth there is more trouble or effort compared to an environment like empty space?
Does from this follows that one guy spends more energy to stay on earth proportionally to its gravity?
...........a big fat man needs more energy to move his body (against gravitation force) than a small one........not only your hands and feet move,even the blood circulation (heart work) needs energy to move blood from toe to tip...........
greetings
If you consider the initial assembly of the lamppost, there actually was energy stored in the arms & post when the lights were hung. These structures are not infinitely stiff (they have mechanical compliance) so that as the lights were hung, their weight caused mechanical strain to be induced in the arms, which are cantilever beams. The weight of the lamps acting in this manner resulted in energy being stored as the arm is displaced by this force. These forces are then transferred to the post by the reaction forces and moments at the base of the arm, which act to both compress and twist the post, resulting in small displacements of the post which also stored energy. TANSTAAFL!
But very likely you're right, just like all physics books.
It would be very convenient from the physical laws established. I will venture to say one thing out of the chorus and maybe making fool
Mine is an attempt to show just the opposite of what you say: I'm not that be confusing, but the basic laws of physics to be unclear.
May be willing to insist on these poor bases, recently were born exotic theories, perhaps just to right a wrong starting assumptions.
A thought experiment:
We live in an expanding universe: For simplicity we assume that there are only 2 galaxies and that these are moving away. A source on the galaxy 1 emits a signal at a frequency X so that its energy is equal to hv (Planck's constant x frequency). This signal is received on a galaxy 2 with frequency X '<X due to the stretching frequencies for the Doppler effect. Thus the energy received on the second galaxy is less than that transmitted by a galaxy 1.
Where did the missing energy? Maybe it went to enhance the dark energy !!!!!!!
The two energies would be equal if we kept in mind the different durations of the signals transmitted and received. But in the physics book there is not reference also to lifetime but only Freq * K Planck.
mos, I'm hoping you took my post the way it was meant- a list of terms whose definitions you should familiarize yourself. It's easy to confuse this stuff and that leads to unnecessary puzzlement.
As to the galaxy problem, what happens to its velocity when a galaxy emits a photon? And when the other galaxy absorbs it?
edit: here's a nice explanation, complete with equations:
http://www.newtonphysics.on.ca/einstein/chapter8.html
Don't take the rest of that site too seriously.
As to the galaxy problem, what happens to its velocity when a galaxy emits a photon? And when the other galaxy absorbs it?
edit: here's a nice explanation, complete with equations:
http://www.newtonphysics.on.ca/einstein/chapter8.html
Don't take the rest of that site too seriously.
One more similar problem: suppose you both have different metal plates in hands (of course, you wear rubber gloves rated for 2KV!). They are charged in respect to each other to say 100 Volt. You keep it on the distance say 10 centimeters between them. Now you go to the distance 20 centimeters. What is the voltage? You keep increasing distance between plates until voltage between them is enough to find path through rubber gloves and kill you.
Now, does an energy you spent on increase of the distance enough to kill you?
Now, does an energy you spent on increase of the distance enough to kill you?
A mass of 1Kg has a gravitational force of 9.8 Newtons acting on it. If falling from rest it is travelling at 9.8 m/s at the end of the first second, its average speed is (9.8+0)/2 metres/second and it has travelled 4.9 metres. Work is being done at a rate of 9.8*4.9 = 48.02 Newton metres/sec or 48 Watts.
Check:-
A mass of 1 pound has a force of 32 poundals acting on it. If falling from rest it travels 0+32/2 = 16 feet in 1 second, work is being done at the rate of 16/550 = 0.029 HP. If there are 2.2 lbs per Kg, then for 1 Kg work would be being done at 2.2*0.029 = 0.064 HP. At 746 Watts/HP this represents 47.7 Watts, near enough considering the rough figures used.
So with 100% efficiency, a 1 Kg mass requires 48 (say 50) Watts or six-and-a-half hundredths (less than a tenth) of a horsepower to achieve a stationary hover.
Most people, however, recognise that there is a fundamental difference between a helicopter that is hovering and one which is sitting on the ground, or, for that matter, hanging from a lamppost.
w
Check:-
A mass of 1 pound has a force of 32 poundals acting on it. If falling from rest it travels 0+32/2 = 16 feet in 1 second, work is being done at the rate of 16/550 = 0.029 HP. If there are 2.2 lbs per Kg, then for 1 Kg work would be being done at 2.2*0.029 = 0.064 HP. At 746 Watts/HP this represents 47.7 Watts, near enough considering the rough figures used.
So with 100% efficiency, a 1 Kg mass requires 48 (say 50) Watts or six-and-a-half hundredths (less than a tenth) of a horsepower to achieve a stationary hover.
Most people, however, recognise that there is a fundamental difference between a helicopter that is hovering and one which is sitting on the ground, or, for that matter, hanging from a lamppost.
w
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I am not angry with you: when the criticism is calm and fair it is always appreciated.
The physics does not seem complete and therefore is allowed other that necessary to feed the healthy doubt, with the hope of overcoming its current limitations.
So what I want to understand you in this thread is that I sometimes say things that may not be always true because they may be "just my speculation".
But these last the first step to be taken seriously, then they must pass the test-wise compared with Skeptics (B. Russell). Endeavoring yourself on this role I can appreciate you and all who have done or will do in the future as I already wrote above.
Thanks again for the link: I'm reading
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Not enough charge to kill. Let's say that you're holding a pair of 1 square foot plates. Start with them 1 inch apart and charged to 100V. That's 32pF of capacitance, thus charge is .0032 microcoulombs. Move them apart a foot, the potential increases to 1200V, but charge remains the same.
No, you are confusing power and energy, and your calculation is wrong. At the end of 1 second acceleration at 9.8 m/s^2 the mass is moving at 9.8m/s, but it has moved 4.9m. It will then have kinetic energy of 0.5*9.8^2=48 Joules. This comes from the potential energy the mass had before it fell. The rate of doing work increases as the mass speeds up, because the force stays the same (approximately) but the speed increases. This says precisely nothing about the power needed to maintain a particular height, which depends entirely on what technology is used and how quickly it turns fuel into heat. For example, a hovering helicopter produces mainly heat and sound from its fuel. When sitting on the ground it does not need to use any fuel.
I'm sorry, I did not well understand.
Can you explain better?
My English is very poor.
Wavebourn's variable capacitor increases its store of energy as you move the plates apart.
charge=CV, energy=0.5*CV^2.
So if you halve the capacitance, you double the stored energy. If you calculate the Coulomb force between the plates, then you should find that the mechanical energy used in separating the plates is exactly equal to the gain in electrostatic energy.
charge=CV, energy=0.5*CV^2.
So if you halve the capacitance, you double the stored energy. If you calculate the Coulomb force between the plates, then you should find that the mechanical energy used in separating the plates is exactly equal to the gain in electrostatic energy.
I'm sorry, I did not well understand.
Can you explain better?
My English is very poor.
Mine is poor as well, but it is too late: DF96 already explained everything you need to know.
Mos, I don't follow your objection. You can treat either of the frames equivalently as the reference, and the emitted particle will either red shift or blue shift.
Maybe you'll like this treatment better?
Relativity Calculator - Law of Energy Conservation and the Doppler Effect
Maybe you'll like this treatment better?
Relativity Calculator - Law of Energy Conservation and the Doppler Effect
Wow, thank you for the idea!
1. Charge plates from an external capacitor through diode.
2. Flip diode's polarity
3. Increase distance between plates.
4. Flip the diode back.
5. Decrease the distance
6. Go to step 2 while it finally kills you.
I would call it, "A Capacitive Pump", an alternative source of an energy!
Odd that you should say that, considering your result is identical to mine.
1 Watt = 1 Joule/Sec. 48 Watts = 48 Joules/Second. 48 Joules at the end of 1 second = 48 Watts.
Jeez...
w
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