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## Hot answers tagged energy

52

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 ...

27

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 ...

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 ...

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 ...

6

Let us calculate the energies. For the bomb, Google gives $3.8\times 10^{16}\,\text{J}$. For the stone, we use the Einstein equation $$KE=mc^2(\gamma-1)\qquad\gamma\equiv\sqrt{1-\beta^2}^{-1};\quad \beta=v/c$$ Plugging in the numbers, this gives a stone energy of $3.8\times 10^{18}\,\text{J}$. The stone is actually more powerful than the bomb. This is ...

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 ...

5

The E=mc^2 formula only applies to an object at rest, and light is never at rest. You want to use the more general formula: $E^2={m_0}^2c^4+p^2c^2$ Then you can set the mass to zero. $E=pc$ What this says is that light has momentum, which is related to its energy.

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, ...

4

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 ...

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 ...

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 ...

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 $$\frac{d\mathcal{H}}{dt} = \dot{x}\ddot{x} + ... 2 As you've worked out, since v=0 then the kinetic energy must also be zero. Potential energy is a little more dubious. At school you are usually taught that the gravitational potential energy is E=mgh but that's not quite right; this equation is the work that you must do to lift an object of mass m a height h or, equivalently, the work done by ... 2 We know that solar cells generate electricity by utilizing the energy of the photon, This is an every day language, electricity. It means things electrical in general every day language. but how does it generate electricity forever? What is generated when the photons hit any material, is heat, and the sun's energy is at maximum 1300Watts per ... 2 The fact that the apparatus is using outer space as a heat sink or just using the atmosphere should not have any significant impact on radiative cooling properties. Is this reasoning correct? No, it is not. The cooling properties of the apparatus depend on the heat sink to which the heat is aimed, as the final result will be an equilibrium of ... 2 Is it possible to generate energy using gravity? Yes and no. It depends what you mean by generate. What gravity does, is convert energy from one form to another. When you lift a bucket of water you do work on it. You add energy to it - we call it potential energy. Then when you upend the bucket, the water pours out, and gravity converts potential energy ... 2 The author confused things in section 3.1. In macroscopic EM theory of radiation (radiometry), technically irradiance at a given point of a plane is defined as time average of the normal component of the Poynting vector (normal to the plane):$$ I = \overline{\mathbf S \cdot \mathbf n} $$As you have shown, this is function of |\mathbf E|^2 only in some ... 2 No, Chemical Processes and Mechanical Processes can also release more energy than is input. See Exothermic Process and Catalysis. Think of an explosion of, say, dynamite: for the low energy input of lighting a wick which can be done with a lighter, you can output a large explosion. Also, Palladium is not Fissile, though it is a Fission product (aka ... 1 Is it possible to make an electromagnetic shield to protect something (or ourselves) against everything which is attacking? It is possible to shield very well against some types of electromagnetic radiation. For example, putting a cell phone inside a "Faraday bag" will block enough radio waves from reaching the cell phone that it can not receive ... 1 In nuclear fusion power plants, electromagnetic fields might be used to contain plasma. But I am not sure what would happen if you threw a solid at the electromagnetic field. 1 No, the energy of a black hole is not infinite. It depends on its mass, angular momentum and charge. Infinite density at a point does not translate to infinite energy in the E=mc^2 sense. It is in fact possible to extract energy from black holes by exploiting certain properties of accretion disks or ergospheres, but this is a finite process. 1 The problem with this is that$$E = mc^2$$. Density, however, is given by$$density = \dfrac{mass}{volume} Thus, if volume = 0, then density is infinite. Black holes have a finite mass. It is there density which is finite because all the mass is at a single point (singularity, volume = 0).

1

First, the curve you are looking at is binding energy per nucleon, so it might not be the best thing to look at. The other thing is that binding energy is negative. Going to a more negative value means that energy is released.

1

Two equations that may be helpful to you, depending on exactly what you're trying to do with your spring: Hooke's Law, which is written in equation form as $F=-k\Delta x$ states that the restoring force that returns a spring to equilibrium is proportional to its distance stretched, where $k$ is the spring constant. The spring constant is a measure of how ...

1

I'd like to add two points to describe the history after the events and people described in Svetlana'a answer: After James Joule discovered the relationship between mechanical work and the generation of heat in about 1850, in the hands of the nineteenth century thermodynamicists (Clausius, Gibbs and especially Thomson (Lord Kelvin)) the notion of a budget ...

1

There is no net energy being transferred. The energy each wave has is cancelled by the other wave! You might be confusing this with the "two slit" experiment. In this case you do have both, destructive and constructive interference. This is the result of two simultaneous waves with a given spatial separation causing the waves to "cancel" each other at ...

1

You're looking at solar cells for terrestrial operation. The main efficiency number is not Power_electric/Power_solar, but Power_electric/investment. Capturing the last few bits of blue light just isn't worth it. In space applications, the investment is dominated by the launch costs. Using a more exotic material to capture 1% more energy might shave a ...

1

When you pluck the string, you impart energy into it that's slowly radiated as sound. There are ways to radiate the energy faster, in which case the string loses energy faster. You're increasing the power and decreasing the time, so energy stays constant.

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