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Can a "Floppy Hammer" apply more force/energy than a regular hammer?

Looking at the video linked in the OP I don't believe the intent is to hit harder, but rather to transmit less shock to the user. If you're going to be hitting over and over and over the depicted ...
Loren Pechtel's user avatar
9 votes
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Can a "Floppy Hammer" apply more force/energy than a regular hammer?

Most of the kinetic energy of a sledge hammer is in the head. It will have more kinetic energy if the head is more massive and/or moves faster. This handle looks longer than the usual sledge hammer. ...
mmesser314's user avatar
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0 votes

How much forest must be cut down to produce $N$ petajoules of energy?

The specific energy of wood is roughly $20$ megajoules per kilogram - the exact figure depends on the type of wood, how dry it is etc. So one petajoule is equivalent to about $50,000$ tonnes of wood. ...
gandalf61's user avatar
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0 votes
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Final velocity of block pulled along a surface using work and energy

You need to know the net force acting on the block, i.e. the difference between the upward force of $100N$ and the opposing frictional force. So lets find the magnitude of the normal force. The ...
Albertus Magnus's user avatar
1 vote

Radiative Energy in a Gravity Well

This isn’t a distinction between mass and radiation. Indeed, if a small portion of the mass were converted to radiation but instead of escaping were used to propel matter, then the matter would escape....
Dale's user avatar
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0 votes
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Fusion in the sun for 4 hydrogen to Helium-4. How is the energy produced

(0) This process is called "the proton-proton chain", and should be referred to as such in question titles and what-not. Two protons do not fuse into ${}^2_2{\rm He}$, that has a half-life ...
JEB's user avatar
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2 votes

Electrical energy is $I^2Rt$, and heat dissipated is also $I^2Rt$?

A key phrase is, "a conductor of resistance 𝑅" All of the electrical power that is dissipated in a resistor is converted to heat. The same is not true of, for example, an electric motor. ...
Solomon Slow's user avatar
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3 votes

Electrical energy is $I^2Rt$, and heat dissipated is also $I^2Rt$?

The first law of thermodynamics tells us that $$\Delta U=Q-W$$, where U is the internal energy of the system, Q is the heat added to the system, and W is the amount of work done by the system on its ...
Chet Miller's user avatar
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2 votes
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Electrical energy is $I^2Rt$, and heat dissipated is also $I^2Rt$?

"If a source provides some energy, and all of it is dissipated as heat, is it of any use?" Do you not consider electric heaters, (eg fan heaters, convectors) useful? They consist, ...
Philip Wood's user avatar
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1 vote

Deduction of Kinetic energy operator in quantum mechanics

Actually mathematical part of kinetic energy derivation is Ok : $$ \begin{align} \hat T &= \frac {\hat p^2}{2m}\\ &=\frac {1}{2m}\left(\hat p \cdot \hat p\right)\\ &=\frac {1}{2m}\left(-i\...
Agnius Vasiliauskas's user avatar
1 vote

Percentage change in K.E for a given change in momentum

Just to follow up on the answer from @Thomas Fritsch that your dE/E = 2dp/p is only valid for infinitesimals, let's integrate. $\int dE/E$ = 2 $\int dp/p$ where we take as limits E$_1$ to $E_2$ and $...
Not_Einstein's user avatar
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0 votes

Why does internal energy depend only on a set of two state variables?

In addition to the answer given by Chemomechanics, I just wanted to add some detail. In general work can be done on a system by jdX where j is a "generalized force" and X is a "...
McKinley's user avatar
2 votes

Percentage change in K.E for a given change in momentum

When you want to calculate the percent change of a function consider that: $$\text{percent change}={|f(x_f)-f(x_i)|\over f(x_i)}\times 100\%.$$ So, considering that $E(p)=p^2/2m$: $${{p^2(1+0.5)^2\...
Albertus Magnus's user avatar
1 vote

Percentage change in K.E for a given change in momentum

Just to show that 125% is the correct answer: $E_0$ = m$v_0^2$/2 Assuming constant mass, $v_1 = 1.5 v_0$ $E_1$ = m$v_1^2$/2 = 2.25m$v_0^2$/2 ($E_1$ - $E_0$)/$E_0$ = 1.25
Not_Einstein's user avatar
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3 votes
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Percentage change in K.E for a given change in momentum

The equation $$\frac{dE}{E}=\frac{2dP}{P} \tag{1}$$ is only valid for infinitesimal small differentials $dE$ and $dP$. Thus using differences (instead of differentials) in $$\frac{\Delta E}{E}\approx \...
Thomas Fritsch's user avatar
2 votes
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Deduction of Kinetic energy operator in quantum mechanics

In operatorial language for any operator $A$, $A^2$ means $A \circ A$, that is $A$ composed with itself. In the particular case of $A = \hat{p} = -i\hbar \frac{d}{dx}$, indeed $$\hat{p}^2 = -\hbar^2 \...
lucabtz's user avatar
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4 votes

Deduction of Kinetic energy operator in quantum mechanics

Once a differential operator represents the momentum $p$, $p^2 f$ should be intended as the operator $p$ acting on $pf$. Therefore, the second expression is the correct one. Edit after some exchange ...
GiorgioP-DoomsdayClockIsAt-90's user avatar
0 votes

How much more fuel to double the speed?

On the other hand, once the car (or spaceship) is at 𝑣1 in the initial frame of reference R0, we can consider a new frame of reference R1 where the car/spaceship is at rest. Then we are exactly in ...
BowlOfRed's user avatar
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0 votes

How much more fuel to double the speed?

The argument that the frame of reference matters when determining energy expended to accelerate to a new velocity different by $\Delta v$ is valid. Consider that the total motion of the Earth, ...
eshaya's user avatar
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1 vote

Why does internal energy depend only on a set of two state variables?

The internal energy of a system can be changed by heating it, doing work on it, and adding matter to it. These correspond to a shift in entropy $S$, volume $V$ (if only pressure–volume work is ...
Chemomechanics's user avatar
1 vote
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Applicabilty of the definition of thermodynamic temperature

I think you are mixing Thermodynamics and Statistical Mechanics concepts. From the point of view of Thermodynamics, there is a set of definitions and relations between functions of the state that hold ...
GiorgioP-DoomsdayClockIsAt-90's user avatar
5 votes

Total kinetic energy confusion

These are not silly questions and have puzzled many students in the past and probably will so in the future. The idea of splitting the kinetic energy into two parts is to help with the analysis of ...
Farcher's user avatar
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3 votes

Total kinetic energy confusion

Imagine a wheel with radius R that is rolling along in the vacuum of free space and advancing in a straight line by $2 \pi R$ for every complete revolution, just like a wheel rolling along a road, but ...
KDP's user avatar
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10 votes
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Total kinetic energy confusion

lets say we have this object which we can make roll really fast (above ground) and so it has a lot of rotational kinetic energy but no translational kinetic energy. Great example. We then put it on ...
BowlOfRed's user avatar
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1 vote

Total kinetic energy confusion

Aren't we adding the same energy twice ? No, we are not - rotational kinetic energy is intentionally defined so that this does not happen. Let's consider a small part $P$ of the ball. $P$ has a ...
gandalf61's user avatar
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2 votes

Total kinetic energy confusion

Lets say we have a rolling ball, which is purely rolling, and there is no friction (is that even possible lol) Actually, this is not possible, no :) Rolling requires static friction with the surface. ...
Steeven's user avatar
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0 votes

Work Done on a rotating thin rod by hinge Forces

There is a horizontal force at the hinge and a vertical force. The horizontal force does work to accelerate the rod in the horizontal direction. It also exerts a torque about the center of mass. The ...
Rod Cross's user avatar
2 votes

When is the internal energy of a system not considered potential energy?

Is it correct to always view the internal energy as part of the potential energy? Yes, but only the potential energy associated with internal molecule forces, as opposed to the potential energy of ...
Bob D's user avatar
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1 vote
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When is the internal energy of a system not considered potential energy?

This is a serious edit to rectify my mistake of not explaining things correctly or more precisely I must rewrite this answer again as I misdirected the OP to understand only one side of The story, ...
Dheeraj Gujrathi's user avatar
1 vote

When is the internal energy of a system not considered potential energy?

In general, Energy is not as well-defined as it may seem. In thermodynamics, we commonly attribute a certain type of energy as the "Internal Energy" of a system, which we don't bother to ...
QuantumBrachistochrone's user avatar
-3 votes

When is the internal energy of a system not considered potential energy?

Internal energy is typically referred to the energy that is stored inside an object or contained. Potential energy or what I would consider kinetic energy, is what that object can produce. For example ...
Richard Morgan's user avatar
1 vote

Solving thermodynamics with classical mechanics

Mechanics (it doesn't matter if Newtonian or Hamiltonian) and Thermodynamics are different theories, working at different levels of description of the physical world. Different doesn't mean they are ...
GiorgioP-DoomsdayClockIsAt-90's user avatar
2 votes

Solving thermodynamics with classical mechanics

Lagrangian and Hamiltonian mechanics requires holonomic constraints, and non-conservative effects like friction introduce non-holonomic terms. There have been some clever ways to rephrase friction ...
Cort Ammon's user avatar
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0 votes

Does Accelerating to Double the Speed Require Quadruple the Fuel Consumption, or Double the Fuel Consumption?

Work done (which is a measure of energy input) has the formula $(F \ d)$ where F is force and d is the displacement or distance travelled. Distance travelled in time t with acceleration a is $d = 1/2 ...
KDP's user avatar
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5 votes

How can the Klein-Gordon equation have negative-energy solution if its Hamiltonian is positive-definite?

You have a valid point. 'Negative energy solution' is a misnomer. Negative as well as positive frequency solutions have positive energy, as the Klein-Gordon hamiltonian correctly implies. As an ...
my2cts's user avatar
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6 votes

How can the Klein-Gordon equation have negative-energy solution if its Hamiltonian is positive-definite?

Briefly speaking, there are at least two notions of energy $E$ and momentum ${\bf p}$: One notion stems from the Fourier-transform $({\bf x},t)\to ({\bf p},E)$ of spacetime, or equivalently, a plane-...
Qmechanic's user avatar
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3 votes

How can the Klein-Gordon equation have negative-energy solution if its Hamiltonian is positive-definite?

I sometimes think that the whole negative energy thing is a bit overrated. For example one can solve the classical Klein-Gordon equation and find via the appropriate ansatz that the solutions will be ...
Albertus Magnus's user avatar
9 votes
Accepted

How can the Klein-Gordon equation have negative-energy solution if its Hamiltonian is positive-definite?

If you interpret the Klein-Gordon equation as describing a field theory (which is the modern way to look at it), then yes, the energy of the field is positive definite. The problem is if you try to ...
Andrew's user avatar
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1 vote

If an ice cube moved really fast in a vacuum, would it melt?

No, the orderly movement of the body as whole does not contribute to its thermal energy. Only the random movement of its particles does. Quoted from "Thermal Energy | Energy Fundamentals" at ...
Thomas Fritsch's user avatar
0 votes

Is energy density and pressure fundamentally the same thing?

Gwhiz wrote: "I came to this same question when thinking about the ideal gas equation, PV=NkT. If temperature is a kind of average kinetic energy per gas molecule, then the right side of the ...
Mr_Magoo's user avatar
2 votes

How big is light pressure as a fraction of light energy?

"I'm looking for a dimensionless expression/formula for the fraction of energy turned into kinetic energy ..." Here's a simple treatment... Suppose that the photon of frequency $f_1$ (and ...
Philip Wood's user avatar
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2 votes

Why radial quantization gives different spectrum?

I guess I’ve figure out what’s going on, the notion of translation is different (translation on cylinder is rotation and dilation on plane), so we are working with different momentum operators in this ...
Peter Wu's user avatar
  • 249
6 votes

Why radial quantization gives different spectrum?

TL;DR: The spectrum is discrete (continuous) since space is a compact circle $\mathbb{S}^1$ (non-compact line $\mathbb{R}^1$), respectively, cf e.g. this Phys.SE post. In more details: The radially ...
Qmechanic's user avatar
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3 votes

Why is entropy a quantification of the typical fluctuation of internal energy around the expected value?

In the following, I'll set $k_B=1$. Usually, the fluctuations of energy are rather captured by heat capacity. This is captured by the following formula saying that heat capacity is proportional to the ...
LPZ's user avatar
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3 votes
Accepted

Equation for relative Kinetic energy

In special relativity, the total energy of an object is given by; $$E = \sqrt{ (pc)^2 + (m_0 c^2)^2} = \sqrt{(\gamma m_0 v c)^2 +(m_0 c^2)^2}$$ where $p$ is the relativistic momentum $(\gamma m v)$. ...
KDP's user avatar
  • 2,175
2 votes

Equation for relative Kinetic energy

No. Relativistic kinetic energy is difference of relativistic energy and rest energy of object. To check that your proposal does not hold,- simply try to equate these expressions, like : $$ \tag 1 (\...
Agnius Vasiliauskas's user avatar
1 vote
Accepted

Why is the work here positive and are we using 2nd law for this conclusion?

I don't think there is any first-law basis for saying that the work done on the gas is positive since we do not know how the externally applied force varies as the volume changes (without applying the ...
Chet Miller's user avatar
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0 votes

Why is the work here positive and are we using 2nd law for this conclusion?

I don't think there is any need to invoke the Second Law here. When the piston is pushed in, the environment does work on the gas and the internal energy of the gas increases. If the piston is pulled ...
d_b's user avatar
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0 votes

Why is the work here positive and are we using 2nd law for this conclusion?

what I am confused about is why is it saying that there will be a net increase in the internal energy of the gas? It all makes sense but can someone prove the mathematically using 2nd law. I believe ...
Bob D's user avatar
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-1 votes

Is energy density and pressure fundamentally the same thing?

When all else fails, do dimensional analysis. Force: [M1 L1 T-2] / Area: [M0 L2 T0] = Pressure: [M1 L-1 T-2] / Length: [M0 L1 T0] = Pressure Gradient: [M1 L-2 T-2] Energy: [M1 L2 T−2] / Volume: [M0 ...
Mr_Magoo's user avatar

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