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From a very fundamental point of view one can see this with respect to Noether's theorem. Every symmetry is related to a conservation law, e.g. the symmetry of space with respect to rotational symmetry (space does not change if we rotate our coordinate system) results in conservation of angular momentum. In this framework energy is related to time symmetry. ...


1

First, it's not true that energies are generally in the form $\frac{a_1a_2^2}{2}$. Take the gravitational potential energy $U = \frac{GMm}{r}$ as an example. However, it is generally true that kinetic energy takes that quadratic form. Why? Kinetic energy is the energy traded when some agent applies a force on some system that causes it (the system) to ...


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It all depends on the amount of radiation to which one is exposed due to the day-to-day things used. Radium releases alpha particles which have very low penetrating power. Hence we are not in danger of those radiations.


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Is there any fundamental reasons seemingly many physical phenomenon has the energy in that form? Yes. Think about a 10kg cannonball in space. If it’s travelling at 10m/s relative to you, you say it's “got” kinetic energy KE=½mv², and you say it's "got" momentum p=mv. Now brace yourself, then apply some constant braking force with your spacesuit jets, and ...


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These terms of energy are defined so because the expression keeps on popping up in various scenario. Its similar to the reason why Moment of Inertia is defined.


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When the sloth starts falling, it's potential energy starts getting converted to kinetic energy, and just before it hits the ground, all of its potential energy has been converted to kinetic energy. Now, the problem you're facing is because of the fact that you're considering just the sloth, which is wrong, because the sloth is interacting with the ...


7

One has to make clear that the watches we are using now are no longer using radium , because of radiation danger awareness. Radium dials are watch, clock and other instrument dials painted with radioluminescent paint containing radium-226. The 1900s (decade) were the peak of radium dial production, as radiation poisoning was then unknown; subsequently, ...


1

By far the most common isotope of Radium is 226Ra, which decays by emitting an alpha particle. Alpha particles have almost no penetrating ability, and in general externally- occurring alpha particles are absorbed by the outer layers of skin which are naturally sloughed off, so no permanent damage occurs. If you swallow it, that's a whole other (very sad) ...


-2

Humans can tolerate a certain amount of radiation. The watch contributes less radiation to our bodies than the soil. Radium emits x-rays, so yes, they can excite an electron.


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There is an external agent that removes mechanical energy from the sloth, namely the normal force exerted by the ground. You are right that no net work is done, but remember it is the work-kinetic-energy theorem: The net work equals the change in kinetic energy of the sloth.


3

Conservation of energy, as you note, holds for "the system." For instance, if you push on a ball, that ball gains energy, but the energy of the ball is not conserved--only the energy of you and the ball. In this case, the system needs to include more than just "the sloth" because the sloth is not an isolated system--there are external forces at work. Here, ...


3

I think the answer has to be yes, it would be much easier to detect the cosmic neutrino background if you could arrange for your laboratory to be travelling at close to the speed of light in the local, co-moving reference frame defined by the cosmic microwave background. The current energies of neutrinos in the cooled cosmic neutrino background are expected ...


2

It would certainly require a material that allows electron release from energies lower than those of the visible spectrum. The energy of a wave is given by E=hf where h is the planck constant (6.63 x 10^-34) and f is the frequency. The wavelengths of IR light range from 0.001 m to 750 x 10^-9 m. (Hyperphysics.com, infrared) Using this knowledge you can get ...


0

I am sorry to say this but surface area is not proportional to the force of friction. Friction is given by the coefficient of friction (static or kinetic) multiplied by an objects normal force. What may ne useful is the theory that the coefficient of friction is equal to the tangent of the angle at which an object starts to move on a plane. Also something ...


0

The rotational motion theory, the kinetic motion theory, the friction theory, and added to further too in special conditions


3

One need not follow these steps. Indeed let $\gamma : I\subset \mathbb{R}\to \mathbb{R}^3$ be the trajectory of a particle. It's position at time $t$ is $\gamma(t)$, it's velocity is $\gamma'(t)$ and it's acceleration is $\gamma''(t)$. It's easy to see that $$(\gamma'\cdot \gamma')'(t) = 2\gamma'(t)\cdot \gamma''(t),$$ so the work done by the resultant ...


1

Formally, we have the standard dot product identity $$\frac{\mathrm{d}}{\mathrm{d}t}\mathbf{A}\cdot\mathbf{B}=\frac{\mathrm{d}\mathbf{A}}{\mathrm{d}t}\cdot \mathbf{B}+\mathbf{A}\cdot \frac{\mathrm{d}\mathbf{B}}{\mathrm{d}t}.$$ Inserting $\mathbf{A}=\mathbf{B}=\mathbf{v}$ gives $$\frac{\mathrm{d}\mathbf{v}^2}{\mathrm{d}t}=2\mathbf{v}\cdot ...


0

Ok... I see the confusion... See when you bring a ball from infinity to a position at a distance r from the center of the earth with uniform velocity, you need to apply some external force on it ; otherwise naturally the ball would be accelerated (because of gravitational force of earth ) and the decreased potential energy would appear as the kinetic energy ...


1

I think what may be referred to in the question is the determination of gravitational potential energy (GPE). The definition for GPE is the work that was done in bringing the two masses to a distance r apart from an initial separation that was infinite. (Tsokos, 396) What is important here is that this happens at a very slow uniform speed so the Kinetic ...


1

When a ball of mass m is brought with uniform velocity from infinity into the g field of the earth at a distance r from it, the potential energy of the ball earth system decreases from 0 to -GMm/r. What does this lost energy appear as? Kinetic energy, which is typically radiated away into space. Imagine your ball starts off a long long way from Earth, and ...


-1

as we know the energy of an elctron increases as it moves away from the nucleus..that means potential energy is directly depending on distance between electron and nucleus..but when we derive an mathematical expression we get energy inversly proportrional to the radius...to compensate diz we add MINUS sign to show that lesser negative enrgy means more energy ...


0

As it turns out, adding mass (while keeping the dimensions of the car fixed) will make the car go faster. This sounds contrary to what all physics students are taught, but the reason is while friction scales with mass, air resistance doesn't. That's why a thirty-foot rock will fall faster than a thirty-foot parachute in an atmosphere. It's also why the ...


7

Assuming an ordinary hinged door (without any springs), would it take more energy to open it when applying force in the middle of the door (point b), rather than at the end of the door (point a), where the door knob is? When you push a body it will always rotate around the center of mass (white arrows) if you apply a force at the handle you are ...


1

So the first law for an system where we don't have mass flows in or out ( a closed system ), is $\Delta Q + \Delta W = \Delta U$ Where $Q$ is your net heat added, and $W$ is your net work added, and $U$ is your net internal energy change: internal energy being like the sum of all the different kinetic energies of the molecules. This is why when we add ...


1

Velocity is defined as distance over time, so based on that premise, you are moving through time at the rate of 1 hour per hour, or 1 minute per minute, or 1 second per second. You cannot go faster than 1 hour per hour relative to your own "clock". You are simply experiencing the one-directional "arrow of time" (Sean Carroll), whereas in space you have a ...


5

Moving through the other three dimensions necessitates energy. But why doesn't moving through time necessitate energy? Like OrangeDog and peta said, it doesn't take any energy to move through space. The Earth is moving through space, but it isn't consuming any energy. And like what ACuriousMind said, moving through time doesn't make much sense. To be blunt, ...


53

Moving through space at a uniform pace does not require energy, or force (Newton's 1. law), but accelerating through space does (Newton's 2. law). Similarly, moving through time at a uniform pace does not require a force, but if you're accelerating, your time will change wrt. a non-accelerating observer, so in a way you might say that you accelerate through ...


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I think a simple view is this: The solar cell must have a PN junction, which is a junction between p-type (many holes, no electrons) and n-type (many electrons, no holes) materials. Right where they meet there is actually a "depletion width" within which there is hardly any of either. Within this region, as photons come in they generate electron-hole pairs, ...


0

If you open a door by gripping it near the hinge, you apply GREATER energy for a SHORTER time than when you grip it near the outside edge, which requires LESSER energy for a LONGER time. As the friction of the hinge and the weight of the door are equal in both cases, the total energy applied is the same. It's only the time applied and the intensity of ...


8

The physics 101 answer is no: it takes more force, but it is compensated by the smaller displacement so the energy stays the same. If we start with a static door, and we end up with a door rotating at some speed, the energy into the door is the work done by the force and it must be the same independently from the point where the force was applied But let's ...


0

Torque=(R) x (F) Energy req. to rotate=(T).(Theta), Ta=Tb, ONLY Fa is less than Fb. & T at Hinge=0, it'll not rotate there. The force required will increase from A to hinge point. Energy needed hence is const from A to hinge(except hinge point). [Ta means torque applied at A]


0

This is possible but not 100% efficient, so you will always get back less power than you put into it - at best, maybe 80% with a large-scale plant designed for best efficiency. So why would you want to? If you can buy electricity at 3p/kWh overnight, and sell it at 30p/kWh during the early evening peak demand, then even factoring in the losses, that's ...


1

As @Gautham said, W = Torque * angular displacement which is also equal to the energy needed. here the angular displacement is same as we know and torque will be also same(Large distance implies less force needed,less distance implies large force needed but torque will be same) So energy needed will be same in both cases.


0

Yes, this is possible. You can totally build the described contraption and it will probably even run. It will however, not run indefinitely. "feeding the grid" means to take energy out of the system, which means it will eventually stop its operation and that assumes a machinery without any other "losses" (see @Jimmy360's answer on that)


3

The answer is NO. Change of energy is work i.e $W = \Delta E$ and here the work done is $$W = \text{Torque} \cdot \text{angular displacement}$$ which is equal in both the cases. The only change is one needs to apply more force to achieve the same amount of torque at a smaller radius. $$\text{Force at "b"} > \text{Force at "a"}$$ but not the work done or ...


14

You are right. To open the door during the same time interval (for pushing at $a$ and $b$), you should induce the same angular acceleration. Since rotation in both cases is about the same axis, this means you need the same torque, this gives $$ \frac{F_a}{F_b} = \frac{r_b}{r_a} $$ where $r$ is the distance from the point of contact to the axis. However the ...


0

The problem is electrical resistance and conservation of energy and friction. You lose force due to mill friction. You lose electrical energy due to electrical resistance. Lets say you have 25% of the energy left. You would need 100% to lift the water back to the top.


1

You start with $$ E' = \frac{E-up}{\sqrt{1-u^2/c^2}} = \\ \frac{E-up}{\sqrt{(1-u/c)(1+u/c)}}. $$ You know that $p = \frac{E}{c}$, so $$ \frac{E-uE/c}{\sqrt{(1-u/c)(1+u/c)}} = \\ \frac{E(1-u/c)}{\sqrt{(1-u/c)(1+u/c)}} = \\ \frac{E\sqrt{1-u/c}}{\sqrt{1+u/c}} = \frac{E\sqrt{c-u}}{\sqrt{c+u}}, $$ now using $E = hf$, $$ f' = f\sqrt{\frac{c-u}{c+u}}. ...


5

What causes sunburns is UV radiation, which damages our skin cells. Heat on the other hand is the same as the one felt when near an incandescent light bulb (doesn't cause sunburns). Most of the UV radiation coming from the sun is absorbed by the earth's atmosphere. During sunset and sunrise the radiation emitted by the sun passes through more air until ...


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In general, the sun's light (particularly the UV that causes sunburn) has to pass through a lot more atmosphere (or a greater amount of air mass) in the morning and evening to get to a vertical surface than it does when it is at zenith to a horizontal surface. An example is shown in the generalised image below (all graphs are obviously generalised): The ...


0

It is surprising that the thermal resistances are so different. What is being suggested makes sense if the main resistance is conduction through the wall. In that case, each small patch of surface area has a thermal resistance. All these resistances are in parallel, so the total thermal resistance is inversely proportional to the total area. This will be ...


1

It all comes down to the band structure. In a solid the energy levels are not the same as an isolated atom, there are bands instead of levels. So if you are thinking about an electron jumping between discrete energy levels then that is probably not going to happen. Particularly if you are thinking that an electron will de-excite emitting a photon, then that ...


2

You need to use the equations of motion in the expression for $dE/dt$. Lets take a simple example. For the Hamiltonian $$\mathcal{H} = \frac{1}{2}\dot{x}^2 + V(x)$$ (which is equivalent to the Lagrangian $\mathcal{L} = \frac{1}{2}\dot{x}^2 - V(x)$) we have that the time-derivative of the energy is $$\frac{d\mathcal{H}}{dt} = \dot{x}\ddot{x} + ...


2

First of all, you can look at the translation of his paper here. As was already noted Planck firstly discovered the correct blackbody radiation formula by simple interpolation of $R=-\Bigl(\frac{\partial^2 S}{\partial U^2}\Bigr)^{-1}$ where $S$ is entropy and $U$ - mean energy of the oscillator in the bath. He knew that $R=\alpha U$ gives Wien law for ...


1

This is from David Bohm's book, Quantum Theory might be a help to you. Not sure it's the complete derivation, but I think it is. also not sure if it's the orginal method P. used to find it, or Bohm's version (probably the latter) (not sure about copyright either, but the book is 70 odd years old :) any problem with that, i will need to delete it


0

was I really feeling the sound in my heart and all over my body? It is definitely possible to feel sound. This occurs when the pressure is high enough and the frequency is low enough for the sense of touch. The heart can definitely produce a sensation of pain, perhaps also that of external pressure albeit with a rather low sensitivity.


1

For mass $2m$, let $x$ be its displacement to the right from its initial location. For mass $m$, displacement to the right from initial location is $y$. This means the extension of the spring, $e$, equals $y-x$. So look at the forces acting on each mass, and apply Newton's 2nd Law to get the follow equations: $$3F - ke = m\ddot y$$ $$ke - F = 2m\ddot x$$ ...


1

At the instant of maximum extension of the spring both the blocks are moving with same accelerations. Hence the acceleration of (two blocks + spring) system can be found simply by $a=\frac{F^{ext}_{net}}{m+2m} = \frac{2F}{3m}$. Now since the $2m$ mass block is accelerated by only spring force, the spring force is given by, $F_s = F + 2m \times ...



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