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28

The rate at which your home loses heat is proportional to the difference of temperature between the inside and the outside. This is Newton's law of cooling. Hence, a higher temperature home will lose heat faster. This means that if your thermostat is set lower, the furnace will need to turn on less often.


23

Firstly, the specific heat (more precisely the specific enthalpy) of humid air varies a bit with temperature, so going from 72 to 74 degrees F will take a bit more energy than going from 66 to 68 F. However this effect is not very large compared to the effect of heat lost through the house's walls, floor, and roof to the colder outside environment. To a ...


18

Because in the frame of reference that is co-rotating, the object doesn't move, and therefore it has no kinetic energy in that frame, which is the frame in which most problems involving objects on earth are looked at. Note that kinetic energy is evidently not a frame-invariant quantity, but it is not required to be.


14

Both are correct, within the domains for which they are correct. More seriously, the general relation $$ E^2 = m^2c^4 + p^2c^2$$ holds for all objects, whether they have mass or not, whether they are moving or not. The special case $E = mc^2$ is for $p = 0$, i.e. objects which do not move, as you said. The special case $E = pc$ is for objects which have ...


8

I agree with answer of ACuriousMind, but I think it might also help to think about it like this.... $E^2=m_0^2c^4+p^2c^2 =m^2c^4$ where $m_0$ is the rest mass and $m$ is the relativistic mass (or inertial mass), defined as $m = \gamma m_0 = m_0 / \sqrt{1 - v^2/c^2}$. The relatavistic mass increases as the momentum of the mass increases. At rest the two ...


8

Let me clarify some confusions in the notation that other answers have alluded to but not clearly mentioned. Historically, physicists liked to talk about two different definitions of mass The first is the rest mass of a particle $m_0$. This is the mass of the particle when it is at rest. For example, the rest mass of the electron is $(m_0)_{electron} = ...


6

The reason the relationship $$ W=\int\mathbf s\cdot d\mathbf F $$ doesn't work is because Work is defined as the result of a force $\mathbf F$ on a point that moves along a distance. The point follows a curve $\mathbf s$ with a velocity $\mathbf v$. The small amount of work, $\delta W$, that occurs of the instant of time $dt$ is $$ \delta W=\mathbf F(\mathbf ...


6

The answer depends on what the symbols mean. The question does not make it clear how the symbols are defined. The most confusing quantity is $\omega_2$. How is this defined? Is it the angular velocity of the disc relative to the fixed lab axes or relative to the axle about which it is rotating (where this axle itself will be rotating at $\omega_1$)? Also ...


6

First, the concept of matter and energy are entagled since Einstein. Most people have an intuition on the word energy as some untouchable fuild with destructive power,mainly due to hollywood. But energy and mass have a much more specific, mathematical meaning in physics. So instead, let me rephrase the question as "does it makes sense to talk about time in ...


5

The exact quantities of kinetic energy (like momentum) depend on your choice of a reference frame. Don't get too worried though; regardless of your choice of a reference frame, you will find that energy (and likewise momentum) is conserved within the reference frame. Therefore, two observers may not agree on the kinetic energy or momentum of an object, but ...


5

Possibly your confusion arises from not considering that KE is a scalar, whilst momentum is a vector. Yes, there is of course a connection between KE and momentum: $K = p^2/2m$ (for non-relativistic bodies). Thus, for two equal-mass particles heading directly towards each other at equal speeds $v$, their velocities are $\pm {\bf v}$ and their momenta $\pm ...


5

Forget about the furnace kicking in at 2 degrees below the preset or whatever. This is immaterial. The heat is not needed to produce this temperature rise, but rather to compensate heat loss with the surroundings, and thus maintain the temperature. You can clearly see that higher temperature presets need a higher amount of heat to be maintained, because the ...


5

For fluorescence to occur, you need to excite at an energy higher than the emission. According to the spec sheet this plastic scintillator emits at a peak of 420 nm - it may be that 400 nm is just not a high enough energy to excite it. I am used to thinking of scintillators as being used with radiation sources - xray, gamma sources. You need a pretty "hot" ...


4

This is a nice example of one of the foundational issues in relativity: how do we know that energy-momentum transforms like a four-vector, or, essentially, how do we know that $E=mc^2$? A historical overview is given in [Ohanian 2008] and [Ohanian 2009]. As Ohanian points out, there are logical problems if one tries to do what Einstein did in 1905 and prove ...


4

Because, according to your definitions, if I strain a rubber bar with constant force until it rips apart, I haven't done one joule of work to it.


4

For starters, these are not the same thing. The integration by parts rule makes this fairly obvious: $$\int_i^f y\,\mathrm{d}x = y_f x_f - y_i x_i - \int_i^f x\,\mathrm{d}y$$ But then you might be wondering what makes $\int \vec{F}\cdot\mathrm{d}\vec{s}$ the "right" definition for work while $\int \vec{s}\cdot\mathrm{d}\vec{F}$ is the "wrong" one. In a ...


4

The equation $$E^2=(mc^2)^2+(pc)^2$$ represents the correct energy-momentum relationship. It gives the total energy $E$ for an object of invariant mass (rest mass) $m$ that is observed to move with momentum $p$. This equation is applicable regardless whether the object is observed to be in motion ($p \ne 0$), or is observed to be at rest ($p = 0$). In the ...


4

I'm 27 and since I was about 15 I had the same doubt you do. Only a couple of years ago I realized why momentum is always conserved in a collision, whereas the same is not enforced for energy. (They must have told me this at some point -- or points --, but I guess sometimes I just don't pay much attention) First of all, let's make it empirically clear that ...


4

Is there no relation between the momentum and kinetic energy? Are they not linked with each other? No, they're not. Conservation of momentum only says that $$m_1 v_{1_-} + m_2 v_{2_-} = m_1 v_{1_+} + m_2 v_{2_+}$$ I used subscripts 1 and 2 denote the particles involved in the collision and the subscripts - and + denote the pre- and post-collision ...


4

Here's a paper with a proof that the ground state must be l=0 for spherically symmetric potentials for a single particle, assuming there's a bound state. Abstract: The variational principle is used to show that the ground-state wave function of a one-body Schrödinger equation with a real potential is real, does not change sign, and is nondegenerate. As a ...


3

If the velocity decreases by a little bit, it goes into an elliptical orbit with the apogee the same as the original orbit. The orbit will be stable if no other changes happen. Only when the velocity decreases to the point where the elliptical orbit intersects the Earth's atmosphere will the object crash into the Earth.


3

If the string is massless and taut, then the wave velocity is infinite - that is, a component of the force at one mass will immediately be felt at the other mass. But to answer your first question, not all the force of the impact will be transmitted along the string, as made clear by this diagram: As for your second question: momentum for the system is ...


3

1) From the equation Work Done = Force x Distance, if a force is exerted, but no distance moved, then no work is done. Hence, the statement is false? 2) However, when a person exerts a force, is he not expending chemical potential energy? In other words, there is an energy change from chemical to thermal. Hence, the statement is true? I ...


3

You can get original Fiestaware pieces online or at some antique stores. The uranium glaze is a decent low-level x-ray and gamma emitter that is safe to handle. My orange salt-shaker reads a couple hundred Bequerel on a good Geiger tube held a few centimeters away. You can also order nano-curies of a number of isotopes without any paperwork (in the US, no ...


3

I'd like to add to dmckee's answer PLEASE heed his last paragraph. There are heaps of experiments on the internet that blithely instruct you to take a smoke detector apart. Another good "safe" source is uranium marbles: they are traded by marble collectors and are easier to come by than Fiestaware or Annagr√ľn / Annagelb pieces (and the last two are too ...


3

Let's make the following assumptions: One needs to add $1$ log / $30$ min. to keep the fire burning, One log weight approximately $1$ kg. Under these conditions, with your 50% efficiency estimation, the power consumed by the fire is: $15$ x $10^6$ joules x $0.5$ / $1800$ seconds $=~ 4.2$ x $10^3$ W (which is consistent with the estimated heat output ...


2

The change in internal energy is not a relevant quantity for spontaneous evolution of a system. Consider an isolated system made of two blocks of the same material at two different temperatures such that $T_1>T_2$. Heat will flow from $1$ to $2$ but the total change in internal energy is $\Delta U_{1+2}=0$. This information is therefore not useful. On ...


2

$B$ is constant once it has been established. But it takes energy to establish the field, and while it's being established the field is not constant. The integral that you present is the energy required to establish the field, and gives the correct energy. $\frac{1}{2\mu}B^2$ is the energy required to establish the constant field $B$.


2

The first statement is an idealisation when no environmental effects are taken into account (nor heat). The second is more realistic and takes into account environmental effects in a thermodynamic setting. So for an idealisation it would be true while in a thermodynamic setting it would be false. Now, one basic issue here is the definition of Work, ...


2

1.) As you assumed, since the object does not move any distance, then by the definition of work, no work is done and thus the statement is true - no matter how much force is exerted, if there is no displacement, no work is performed. 2.) If a person exerts a large force on an object and the object does not move, there is no external work done. The muscles ...



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