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Based on work that I've done in the past, automobiles moving at "normal" highway speeds of 50-60 mph experience total friction forces of approximately 50% wind drag and 50% rolling resistance. Two identical and separate cars driving at 50 mph could reasonably be expected to get the same gas mileage. If one of those cars was tied to the rear bumper of the ...


2

It is not that easy. Now one engine is delivering about twice as much torque and was probably not designed for that. An engine specifically designed to deliver more torque would like be more efficient than the two combined. Lets look torque alone. Ignore the tow rope and assume the second is far enough back to not get any draft. In this case wind ...


0

Entanglement is the term used for quantum mechanical correlation, and as always "correlation does not mean causation". In quantum mechanics most yes/no correlations come from conservation of quantum numbers. Conservation laws are strict as for example angular momentum conservation. If two electrons are set up to have spin up and spin down, the total spin ...


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Regarding the first question, I believe that you missed a couple of sentences in the Efficiency section of the Wikipedia article: It also requires energy to overcome the change in entropy of the reaction. Therefore, the process cannot proceed below 286 kJ per mol if no external heat/energy is added. If you use 286 kJ/mol, you "don't get something for ...


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Observe that $v\dot v=\frac12\frac{\text d}{\text dt}v^2 = \frac12\frac{\text d}{\text dt}\mathbf v\cdot\mathbf v = \mathbf v\cdot\dot{\mathbf v}$ and therefore the integral of the OP becomes (assuming a constant mass $m$, e.g. a point particle) $$\Delta\left(\frac12mv^2\right)=\int_a^b mv\ \text dv = \int_a^b \frac{\text d}{\text dt}(m\mathbf v)\cdot\mathbf ...


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You are talking about two derivatives, one derivative to time and the other to velocity. If you would derive the kinetic energy two times to the velocity, you would end up with m, not F.


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In fact, the water would act as a neutron moderator, speeding up the reaction. However, reactor pressure vessels are quite sturdy, and it would be very unlikely for the salt water to enter the pressure vessel.


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tok3rat0r probably has the right of it. You have not told us exactly how your timing data was acquired. If it was done by some sort of stopwatch (mechanical or electrical) you should assume an uncertainty of at least 0.1 seconds, and perhaps more for a situation where you have to push a button twice in 1/2 second. If you assume a worst-case error of 0.1 ...


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There doesn't seem to be anything wrong with your calculations given the data and assumptions you've made. So lets consider how long you would expect the block to take in a frictionless situation. By conservation of energy, $mgh=\frac{1}{2}mv_f^2$. Putting in your values I get $v_f=2.63 ms^{-1}$. Then using $d=\frac{1}{2}(v_i+v_f)t$ I get an expected ...


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As user37496 said, the proper way is to calculate the error, using the errors of your measurements. But there is also an alternative, that is you measure often and calculate the mean, and determine the standard deviation as an error. Three times is not really enough to give a reasonable result but it shall do. The mean time is: $t = \frac{1}{3}(0.41 + 0.44 ...


2

Individual photons will not lose any energy as long as they do not interact with any other particle. If you are referring to the intensity of the EM emission, that depends of the angle incidence from their source. So basically, if you imagine a laser that could emit just a single photon in the vacuum of space, that photon would maintain its frequency, and ...


1

The work done by the painter is $1.93$ kJ, which represents the force he applied to the rope times the length of the pull. The work applied to the barrel is $1.47$ kJ. The rest went into friction. I believe the question is ambiguous between the two values, but that is an English question, not a physics one.


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Because the barrel is moving at constant speed. The tension in the rope is equal to the weight of the barrel. So the answer should be your 2nd value.


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Work is transfer of energy from one system to another OR transformation of energy from one form to another. Either way, work does not create energy. When I lift an object, I am transferring energy from my body/muscles to the object-earth system. The energy goes into potential energy of the object-earth system because the separation between the object and ...


0

Work is defined as $\displaystyle\int F(d) \cdot d$ where $F(d)$ is the net force acting on the object. That's it, that's the definition. You might ask; why do scientists talk about energy and work so much? The answer is that experimental science shows that the energy in a system is constant, and thus energy is a useful concept which can be talked about in ...


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Net work is basically defined as $\displaystyle\int F(s) \cdot s \; \mathbf{d}s$, where $F(s)$ is the net force – the sum of all forces acting on the body. For example, when a car holds a constant velocity on a highway the net work done is zero (since the net force is zero) but that doesn't mean work isn't done. Both the engine and drag due to air ...


0

The total work done by all forces acting on an object throughout the motion interval of interest is what the work-energy principle involves. It never says "no forces are doing work." And it doesn't talk about changes in potential energy. The changes in potential energy are involved in the work done by the conservative force attached to the particular ...


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Work is a definition of the expenditure of energy over time. That's it. It's not contained in the definition of any other physical quantity. One could make lots of examples that generate work with chemical reactions (internal combustion engines, explosives), mechanical dynamics (piston firing), electromechanical transformation (electric motor), and so ...


1

When you lift an object (with no increase in kinetic energy), you're doing positive work to lift it up, while gravity is doing negative work (trying to pull it down). In total, the net work is zero. Another way to think about it: total work is net force times displacement. If the object isn't accelerating, you know from $F=ma$ that net force is zero.


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You're completely correct, letting an object fall under the influence of gravity will not change its kinetic+potential energy, it just transforms potential energy into kinetic energy, while leaving the total constant. However, OTHER forces could cause the kinetic energy to increase without changing the potential energy. Imagine a flat ice rink, and a puck ...


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I like to think if it as connected to potential in the way described below, but transmitting or otherwise erasing it might be called kinetic energy. The one thing for sure is that there is a minimal entropy generated when a bit is erased. Landauer's principle says at least kT ln(2) energy must be lost as heat when a bit is erased. At least k ln(2) entropy ...


5

The issue is that there are multiple crater scaling laws, each with different assumptions, as Horedt & Neukem (1984) show (title of paper is Comparison of six crater-scaling laws). Your first equation is a naive approach that the volume of the hole is linearly related to the kinetic energy, hence the $d= k\cdot E^{1/3}$ relationship. This particular ...


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The electric field shows the gradient (slope and direction of change) of the potential. If the electric field is high magnitude, the potential is changing quickly with a change in position. If its magnitude is small, the potential is changing slowly with a change in position. If the electric field is zero, the potential is either at a maximum or a minimum. ...


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Potential is not same as electric field,electric field is zero doesn't mean potential is zero too. your calculation is right,total potential is double the potential of each charge. Edit:For the 2nd part of your question ,there is nothing wrong in potential surfaces criss crossing like that(but electric field lines shouldnt criss cross like that).


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The effect you describe is unphysical. The energy for a randomized, undriven system should never rise when the system moves into equilibrium. If your energy rises (and especially, if it diverges!) there is a problem in your code. This is most likely due to numerical instability in your method for integrating the equations of motion. This is easy to check by ...


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$E=m(g-a)h$ is the energy spent in a Climb, where $a$ is your own acceleration. Consider yourself a forklift/lift trying to lift yourself at every step. You will notice it's harder to climb a steeper staircase than one that is bent closer to the ground (creates a smaller angle with horizontal axis). With a smaller angle the height is greatly lessened at ...


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I think you're drawing a parallel where none exists. Mutual inductance is where a changing current in one inductor influences the current in an adjacent inductor; this happens because it's easy for the (also changing) magnetic field of one inductor goes through another inductor (e.g. a transformer). Mutual capacitance, instead, is just a measure of the ...


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It seems to be your assignment, problem set or so: So instead I will not be spoon feeding you much, here is the method: Use this formula: Work = FxD = KEf - KEi where: KE - Kinetic energy F - Friction D - Distance Remember: KE = mxv/2 Note: Convert units to SI


3

So the earth won't gain any significant energy There's your error. The Earth will gain a significant amount of energy, enough to compensate for the loss of energy of the ball. But because the Earth is so massive, its speed won't significantly change.


1

A simple solution can be like this- At first you find the rate of change of Kinetic Energy. The process is as follows. $$d/dt(1/2 mv^2)$$ =$$F.v$$ Now imagine this case of a constant force acts on the body which equals to -mg(minus indicates downward direction). So, the equation is -mgv.Now the velocity is the rate of change of vertical position of the ...


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Argon is not used because it is too easily activated by neutrons, forming Ar-41, which has half-life of 1.8 hours. ...


2

Why it is colder in mountains, at high altitudes? One answer is that mountains on Earth aren't all that tall. An impossibly tall mountain would see temperatures vary with altitude as depicted below. Tall as it is, even Mount Everest doesn't extend into the stratosphere. This is a question about the lowest layer of the atmosphere, the troposphere. ...


0

The right way to derive the total energy is to put the particles one by one. In a space centered at $e_1$ and having nothing else, the energy to bring $p_1$ from far away to a distance $r_0$ is $E_{e_1,p_1}$, this you know how to calculate. The energy to put $e_2$ at its position will have two terms coming from $e_1$ and $p_1$. By superposition, this ...


0

Where does the energy go when light is redshifted? It doesn't go anywhere. Let's say we're motionless with respect to some source which is emitting a stream of photons. We agree that the photons have some energy E=hf. Now let's say I push you such that you're moving away from the source. You now claim that the photons are redshifted, but those photons ...


1

Nobody ever said that different observers have to agree on the energy of a photon (or anything else). The invariant quantity is energy minus momentum (i.e. rest mass), which is equal to zero whether the photon is red or green. (Edited to add: I see now that userLTK already said as much in a comment.)


2

An atmosphere in absolute equilibrium in fact is isothermal (see below for more detailed analysis of your cannonball). However, if the atmosphere is mixed by wind, gas expands and contracts adiabatically. If the mixing is fast enough, it obeys relatively well the adiabatic invariant, which multiplied by suitable form of ideal gas law ($(T/(pV))^\gamma = ...


1

The air becomes colder because of the ideal gas law, $PV=nRT$. where $P$ is pressure, $V$ is volume, $n$ is the number of moles of the gas, $R$ is the ideal gas constant, and $T$ is the temperature of the gas in Kelvin. If we rearrange $PV=nRT$, we can solve for $T$. By looking at $T=\frac{PV}{nR}$ you can see that reducing pressure will reduce the ...


2

Imagine wind blowing along a plane with the air by the ground all a nice and steady temperature. Now this wind encounters a mountain range, so is forced upwards. The pressure is lower at higher altitude since there is less remaining atmosphere above it. The temperature of gas decreases when the pressure is lowered, which is why this same air gets ...


1

as molecules jump higher, they loose energy/speed due to gravity. This results in molecules slower at heights and therefore you have lower temperatures at the heights, boy. Molecules do not jump up, they scatter off each other every which way. The difference in gravitational energy within the nanometers of the molecule's path before a scatter on ...


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I guess we all agree that a cannonball does not get cold when flying upwards, it loses kinetic energy, not heat energy. Why would ascending air not do the same thing? So equivalently to an ascending cannonball we get: When one molecule in a rising column of air bounces upwards, it loses kinetic energy while moving upwards. That loss of kinetic energy is a ...


0

You can integrate the velocity values over $r$, the radius, using whatever integration technique you would like (trapezoidal rule, Simpson's rule, etc) and use the mean-value theorem to get the average velocity for each radial line. You then have to account for all $2\pi$ radial lines. Assuming the flow is incompressible, you can then find kinetic energy by ...


1

Starting from the work around a close curve $$W = \oint \vec F(r)d\vec s = 0$$ now be stokes throem: $$ \oint \vec F(r)d\vec s = \iint_A \vec \nabla \times\vec F(r)d\vec A = 0$$ where $A$ is the area surrounded by the closed curve. Now if you do the math you will see that $ \vec \nabla \times \vec \nabla U( r) = 0 $ for any field $U( r)$ thus we can ...


1

The reason for this is that, in your notation, $W=C_1-C_2$. This means that the work performed in moving about a circuit from point $A$ to point $B$ along curve $\mathcal C_1$ and then back along $\mathcal C_2$ is exactly the difference between the work performed in moving the particle along $\mathcal C_1$ versus moving it along $\mathcal C_2$. This is ...


0

If a mass M is at a point travelling at a given velocity (= speed and direction vector) then it has a given amount of energy (potential and kinetic energy summed) relative to being stationary* at the same point with respect to a given frame of reference. ie solely by knowing position and velocity vector the PE & KE are defined. (* or at some zero ...


0

It looks like you're confusing vector and scalar $R$. The potential between two charges depends on the magnitude of $R=|\vec{R}|$, not on the vector. So in this case $R_{13}=R_{31}=|\vec{R}_{13}|$.


-2

My question: are there other types of energy (corresponding to, say, acceleration, jerk, etc.)? There are other types of energy, but not the way you're suggesting. Non-gravitational F=ma acceleration is the result of a change of energy, and a jerk is just a rapid acceleration. Why are there only two fundamental types of energy? I wouldn't say ...


3

Higher derivatives of position don't have their own corresponding types of energy, because they're not independent quantities. (See this or this or any of several similar questions.) $F = ma$ relates the second derivative ($a$) to the position itself ($F(x)$), so if there were a type of energy that depended on acceleration, you could just re-express it as a ...


2

What is mass of elementary particles? It is the "length" of the four vector (p_x,p_y,-_z,E), for complex systems it is called rest mass. As the length of three dimensional vectors is not additive ( think adding two opposite momenta), rest masses are not additive, vector algebra has to be used. The invariant mass of your two gammas must be larger than the ...


0

Let $E = KE +U$ be the total energy . We know the momentum operator & total-energy operator are $$\hat{p} = \frac{\hbar}{i} \frac{\partial}{\partial x} \\\\\\\\\\\\ \hat{E} = i\hbar\dfrac{\partial}{\partial t}$$ . This prompts us to write $$\hat{E} = \hat{KE} + \hat{U} \implies \hat{E} = \dfrac{\hat{p}^2}{2m} + U \implies \hat{E} = ...


3

"Rest mass" is probably a bit more abstract in modern science than what you seem to be thinking. It is simply this: if you can put yourself into an inertial frame such that a body is at rest relative to you, then that body's rest mass is defined as that body's total energy measured in this particular inertial frame - multiplied by $c^2$, if you want to ...



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