# Tag Info

1

Of course in theory nothing stops you from doing the same thing on a bigger scale. Consider however that the effect is rather feeble and you need really strong field for heavier objects, which is expensive and quite cumbersome. The relevant equation is given here. If you are mainly looking for a fascinating project, I would consider magnetic levitation, ...

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Although there are excellent answers, I think a more "simplistic" answer is required to correct your thinking. If you start with a piece of lead (1 kg) on the floor, grab and lift it 1 meter, it will gain (1 x 9.8 x 1 =) 9.8 J of energy. If you now open your hand (release it), it will fall by "it self" and hit the ground and "loose" the energy it had ...

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Newton's 1st law states, as you rightly say, that a force-free body maintains its state of motion. (This also holds in relativity.) Newton's 2nd law states, that if there is a net force on the body, the body will accelerate in the direction of that force, $\vec{F}=m\vec{a}$. (This will be altered a bit in relativity, but for simplicity, let's stick with ...

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Science never answers "why" questions, so in a strict sense there is no such explanation, but one can try to triangulate where we stand, at the moment. In classical physics space, time and the existence of massive bodies are inexplicable pre-physical facts. Inertia then becomes an observed property of massive bodies that allows to differentiate them by ...

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Answer:1 Mass is just a form of potential energy of a gravity wave of relativistic bent time space and light is a gravitational wave of bent space, when it travels at the velocity of C, where time relativistically slows down. At that velocity (with time dilation) it makes a gravity wave into the photon. because it is traveling relativistic-ally it is slowing ...

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The book is wrong and you are correct. $$\begin{pmatrix} -\frac{L}{2} \cos 60° \\ \frac{L}{2} \sin 60° \\ 0 \end{pmatrix} \times \begin{pmatrix} 0 \\ -W \\ 0 \end{pmatrix} + \begin{pmatrix} -L \cos 60° \\ L \sin 60° \\ 0 \end{pmatrix} \times \begin{pmatrix} F_3 \\ 0 \\ 0 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}$$ $$\rightarrow ... 0 If you do a free body diagram you will see that a moment needs to be added in order to balance the forces and moments. The forces will be equal and opposite and the moment will depend on the distance between the forces as well as their magnitude.$$ M = F d

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I want to elaborate on MAFIA's correct and important "potential energy is a property of the whole system". The potential energy is not a property of just one of the involved objects, like the lifted ball. That in our common experience all the potential energy somehow seems to be "attached" to the ball is just a consequence of the very different masses.1 ...

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They take a symmetry axis about B So the angle DBE is split into two 30° angles.

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I haven't really worked it out, but my first impression is that in your torque equation the Fw term should be cos(90-theta) not cos(theta). See if that helps.

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In the case of the ball on the table, it is in a state of stable equilibrium, where the table is pushing up against the ball to counteract the gravitational force pulling it down. If the table and ball were to be moved to a planet with much stronger gravity, the gravitational force on the ball could be strong enough to break the table and the ball would move ...

3

This answer is only about Where does the stored energy stay in the object, and why does it only convert into vertical motion and not horizontal motion? because I think your other questions have been well-addressed, but this one has only been answered in highly technical terms that may not have clarified anything for you. Think about what happens if ...

2

The first thing is to note that the gravitational potential energy is associated with both the object and the Earth. You may think that only the object has the potential energy because when you drop the object you see it accelerate downwards and gain kinetic energy. At the same time the Earth is accelerating upwards at a rate of $\frac {\text {mass of ... 20 It is wrong to think potential energy is stored in the object. The earth pulls the object down, but the object pulls the earth up. They share the potential energy. The object fails to fall down because the tabletop pushes it up. The earth fails to fall up because the bottom of the table legs push the earth down. The table pushes up and down because it is ... 3 [...] when it already has energy, then why doesn't it fall off from the table top onto the ground by itself? Because it is being held back. It wants to fall straight downwards, but the bookshelf applies a normal force to hold up the book, which is stronger than the downwards force (gravity). Just as the rubber band holds back the spring from elongating, ... 0 In a nutshell: it takes an infinite time to reach perfect equilibrium. For realistic measurements though, the time until it is 'near enough' depends on material constants and what you consider 'near enough'. For measuring human body temperature, an exactness of .05 K is probably good enough, and the thermometers are made for good heat transfer, so it would ... 1 In the case of the mercury thermometer we can model its temperature rise based on the following assumptions and parameters.$T(t)$is thermometer temperature,$T_m$is mouth temperature.$C$is total heat capacity of the part of the thermometer placed in mouth.$h$is the Heat Transfer Coefficient between mouth and thermometer,$A$is the total surface area ... 0 If the see-saw is hanging from the pivot point, at equilibrium we would have the following situation: The beam's overall length is$2L$and is attached to the pivot point$P$at the centre of the beam. I'll also assume the centre of gravity of the system is in$O$($mg$is the total weight of the system: beam plus$PO$) and that the distance between$O$... 0 It looks like the pivot is slightly displaced from the beam's center of mass. If the center of mass of the beam is displaced slightly below the pivot by some distance$D$, then just pick the pivot as the center of mass and calculate the torque from that small displacement as the angle of the beam changes. The force$g\, M$of the center of mass is ... 0 The object is not in equilibrium. However, you can still calculate$F$. You know that the string isn't lengthening or shortening, so the net force in the direction of the string is zero. So you can decompose the three forces in the problem into components along the string and components perpendicular to the string, and set the net force along the string to ... 2 I am not entirely happy with the given answer so I will provide mine. First of all there is not just a single entropy to talk about, even at equilibrium. So it is misleading to talk about a single entropy. To give an example of what I mean, let us consider an ideal gas in a box with fixed energy$E$, number of particles$N$and volume$V\$. What equilibrium ...

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