# Tag Info

12

"Total energy of the Earth" is somewhat of an odd concept, but there's no reason we can't really entertain it. It brings up some genuinely difficult questions. The right way to approach this is to define the system correctly and then identify forms of energy content and flows. Things to "count" in the Earth's energy: Heat content Nuclear energy ...

7

No. Newtonian gravity is a conservative vector field (or conservative force) which mean that energy that you extract from the field has to be put in first. This is technically stated as the work done around any closed looped is equal to zero. For example, you raise your pet cat up 1 meter (you do work against gravity) you let go and gravity does the same ...

6

The heat generated from the Earth's core is about 4x10^13 W while the Sun provides about 1.7x10^17W so although the Earth's core is slowly cooling this has very little effect on the Earth's temperature. The Earth is in equilibrium between the energy received from the sun and the energy it emits into space. If the amount received changes, then temperature of ...

6

As weird as it sounds, the answer is "yes." Take, for instance, a satellite in gravitational orbit around some heavy body. It's energy is given by $$H=\frac{p^2}{2m}-\frac{GMm}{r}$$ Clearly, there are solutions to this equation which have $0$ energy (look at a slowly moving particle that's really far away), but those solutions necessarily involve a ...

5

No, in a vacuum light sources will appear dimmer as you move further from them because of the inverse square law. In a medium, a light source suffers from both the inverse square law and absorption/scattering. Below is a diagram illustrating the inverse square law: As you move further away from a light source, your pupil (assume it remains the same size, ...

5

It's because the amount of area "covered" increases as the square of the distance. Imagine a sphere, centered on the source, at a radius equal to your eyeball's location. If the source generates X watts (or whatever unit you like) total, the brightness, i.e. the percent of light which hits your eyeball, is X divided by the ratio of your eyball's area to ...

5

To augment what tpg2114 wrote, work is usually defined as: $$W = fd$$ Where $W$ is the total work done, $f$ is the force, and $d$ is the distance (actually it's the $\Delta d$ caused by the force). As long as $d = 0$ no work is being done. For example, if you're standing on the surface of the Earth, energy is not being consumed to keep you from falling ...

4

Consider the time derivative of the Hamiltonian $$\frac{dH(q,p,t)}{dt}=\frac{\partial H}{\partial q}\dot{q}+\frac{\partial H}{\partial p}\dot{p}+\frac{\partial H}{\partial t}=-\dot{p}\dot{q}+\dot{q}\dot{p}+\frac{\partial H}{\partial t}$$ From this you see that the Hamiltonian is conserved if it does not depend on time,$t$, explicitly. $H$ may or may not be ...

3

Most gamma rays from $pp$ collisions come from neutral pions ($p+p\to p+p+\pi^0$), you'd first have to do some relativistic momentum & energy conservation to determine the energy of the neutral pion. It's easiest if you consider the two subsequent reactions: $$p+p\to p+\Delta^+ \\ \Delta^+\to p+\pi^{0}$$ (it's up to you to figure out the kinematics). ...

3

As Kyle implies in the comments, mechanical energy is generally defined only up to a constant. Therefore, if you choose your constant as a large, negative number, you could have a total energy that is negative even with a very fast moving particle. Likewise, if you choose your potential energy to equal zero at, say, the top of a cliff, then anything you ...

3

X thread to the inch (or centimeter) means that if you measure an inch of the screw, you will find X grooves. The figure of 126 is an approximation of 3,14 (pi)* 2 * r (in this case 20 inches). The circumference of the circle inscribed by the end of the handle when you turn it one revolution.

3

As long as the $\angle \theta_n$ for each is $0^\circ < \angle \theta_n < 180^\circ$ then their $\angle \theta_n$ and trajectories don't really matter. They will start at at the same $h_0, v =v_0$, travel to some $h_\mathrm{max}, v_\mathrm{max}=0$ and then fall back down to $h$ traveling the same speed $v = v_0$. At this point they continue to fall ...

2

I'd say depends on the definition of 'total energy' - see Alan's answer above. The total thermal energy is not in equilibrium, it is increasing: global warming, that is an imbalance of around 0.5 W/m^2, corresponding to a total imbalance of 2.5x10^14 watts (if I did the multiplication correctly...)

2

I would think that a hardened hole saw approach would be most efficient. Bore a short distance into the rock with a self-augering hole saw the diameter of the desired tunnel and then impacting the center of the core to break out the center, remove the broken material and then take the next bore. Where is the sense of pulverizing the center material to dust ...

2

Look up something called the Carnot efficiency. That is the theoretical limit of how effecient any heat engine can be at converting heat power to some other form. This maximum possible efficiency is    Carnot efficiency = Tdiff / Thot = (Thot - Tcold) / Thot By simple 8th grade algebra, you can see that you get a higher value by decreasing ...

2

You are right that in this theoretical problem the breaking force to achieve constant velocity is the same for all velocities. You either stipulate that the bike has the desired speed as a initial condition, or that it coasts with no breaking applied until the desired velocity is reached. In the real world, there will always be some friction or drag forces ...

2

It is because you wouldn't hide in the corners like your kitty does! A electric radiator is designed to be directional and therefore it doesn't heat the unnecessary part of your room. It makes you feel warming in front of it, but some part of the room don't get heated like those corner and the ceiling. In comparison, a vacuum cleaner heating the gas ...

2

Sure, a particle can have potential energy in one dimension. Just look at Hooke's Law or the gravitational force. Both of those are conservative forces in one dimension ($x$ and $r$, respectively) that have a corresponding potential energy. In higher dimensions, nothing has to change, but it is possible to have potential energies which depend on the value ...

2

As I cannot post any comments I have to post this as an answer although the essential points were already given: In classical mechanics energy itself was no meaning. Only energy differences have a physical interpretation. Thus in the classical case energy is only defined up to an arbitrary constant. So any fixed state's energy can be set to zero (but of ...

1

It's up to you whether or not to include elastic potential energy. You'll get the same answer as long as you're careful about what you define to be inside and outside of your system. The basic idea to use here is the work-energy theorem when only mechanical energy is of concern: $$W_\text{net,external}=\Delta K_\text{total}+\Delta U_\text{mechanical}.$$ ...

1

On your first question: absolutely, energy gravitates (or induces curvature in spacetime) the same way that mass gravitates. If you read general relativity, you will learn that it is in fact the Stress-Energy Tensor that is the source of gravitational interaction (or equivalently spacetime curvature). Energy can be localized very easily; a parallel-plate ...

1

Draw a force diagram and determine the braking force required to counteract the (portion of) gravitational force pulling the bike down the slope. If the net force is zero, delta-v will be zero. I think the problem you were given assumes the bike starts out with velocity v . If it doesn't then you'll need to derive a braking force whose magnitude varies ...

1

Here in case of electron, it has already emitted absorbed energy as quanta. So, is it that electron losses some energy other than the energy absorbed from the source, to come down to ground state. I thought, if it was a possibility, then electron would constantly need to lose energy, whenever excited, at last, it would collapse into the nucleus. But, this ...

1

Based on some of your comments, I think what might be tripping you up is the first statement you started with: From the Bohr's atomic model, it is clear that electron can have only certain definite energy levels. and ...If suppose, we assume electron losses total energy, electron can't stay in any particular shell, as it would not have that ...

1

Reading your comment in reply to mine, I understood what you wanted to ask. This is where so many people are confused. Since we start middle school chemistry, we are taught about electron orbits which are like concentric circles. Everyone innately assumes that the 1st orbit is inside the 2nd orbit, which is inside the 3rd and so on. It isn't like that! ...

1

If $\Delta \lambda$ is much smaller compared to either $\lambda_1$ or $\lambda_2$(it doesn't really matter, it should be much smaller than both), then we can make the following approximation: $$|\Delta E|= \left|\Delta \left( \frac {hc}{\lambda}\right)\right|\approx\left|\frac {hc\Delta\lambda}{\lambda^2}\right|$$

1

You're almost right. But... 1138 kilowatts power output will give you 1138 kilowatt-hours in, well... one hour, not 1 second. Just leave out the $\times 3600$ It's better to avoid weird units(like kilowatt-hours as much as possible, so another longer way is this: $1138 \text{ kilowatts}$ mean $1,138,000\text{ joules/sec}$ So in one year you'll get ...

1

They're not the same thing. They have very different implications. You can imagine Force and thus Momentum as the "push" that will happen to the target, while Kinetic energy is the damage it causes. E(k) is equal the Work the object will perform, let it be penetration, fracture, etc. As soon as the object hits the target, the E(k) applies (i.e. the ...

1

It's impossible like anna v. said, but let's entertain the possibility for a second. Assume you have a star-trek like device, which is capable of transforming every atom in your body into energy. However, before this process is even started, the device would first have to register and store every bit of information about every atom, their exact configuration ...

1

No, you can't do that. You can write $\lvert x\rvert = \sqrt{x^2}$, and you can then go to $\langle\lvert x\rvert\rangle = \langle\sqrt{x^2}\rangle$, but it's not valid to say $\langle\sqrt{x^2}\rangle = \sqrt{\langle x^2\rangle}$. You can pick almost any wavefunction, $\psi(x) = \pi^{-1/4}e^{-x^2/2}$ for instance, and show that the two are not equal. In ...

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