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Landauer's principle (original paper pdf | doi) expresses a non-zero lower bound on the amount of heat that must be generated by computers. However, this entropy-necessitated heat is dwarfed by the heat generated through ordinary electrical resistance of the circuitry (the same reason light bulbs give off heat).

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Computers manipulate internal stored values "0" and "1" represented as different voltages. Every change 0-to-1 and 1-to-0 involves an electric current I passing through a circuit resistance R, which gives rise to ohmic or "Joule" heating.

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You almost answered it on your own! Essentially it's the ratio of the viscous force to the gravitational force. As $\beta \rightarrow 0$, the gravitational force dominates and the damping due to air friction is very small. Likewise, as $\beta \rightarrow \infty$, the air friction dominates the solution. This isn't really all that illustrative physically, ...

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There are two types of frictional force, the static friction and kinetic friction. Kinetic friction is the force experienced when you drag an object on the floor. Static friction is what enables you to hold objects without it slipping away from your fingers. Similarly, as you drive, assuming that the wheels don't spin, your wheels are pushing backwards ...

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In general, smoothing the surface, changing the interaction from sliding to rolling, adding (air) space between the surfaces, and adding lubrication (oil, graphite, teflon, ball bearings, air cushion...) are the most common techniques.

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Regardless of whether you are driving at 65mph or 1mph, as long as your steering angle is the same, you will travel the same path. (Of course in real life understeer or oversteer has to be considered) However, due to the much greater centripetal acceleration experienced, where it is proportional to the square of the velocity, $$a_c = \frac{v^2}{r}$$ You ...

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Yes, adding mass to a toy car should at least in principle make it accelerate down a ramp faster. The total force on the car is in the "forward" direction, with magnitude $$F=m g \sin\theta\ -\ m g C_{rr}\cos\theta\ - \tfrac12 \rho v^2 C_D A\ ,$$ where $m$ is the car's mass, $g$ is the acceleration due to gravity at Earth's surface, $\theta$ is the angle ...

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Various physical properties of materials including shear strength are connected to the strength of attractive forces between molecules -inter-molecular forces. The type of inter-molecular force determines the force and energy by which the molecules stick to one another. Inter-molecular force types include ionic bonds, hydrogen bonds, dipole forces and ...

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every system that oscillated and that should not osscillated would be an application for damped oscillation. There are many examples in control engineering, but to give you a more daily concurrent example: car shock absorbers. If you drive along a road and hit a chuckhole you don't want your car to jump up and down for half a minute. Since car shock ...

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A kinetic friction force that acts on a body always has an opposite direction to that in which the body is moving. So, if you have a system Surface+Body A+Body B, where B is on top of A and A is moving to the right, both the surface and B will exert a friction force on A, pointing to the left. Moreover, according to Newton's third law, A will exert a ...

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Think of it like this and you will never be confused: Friction always tries to keep the two objects together. This regards both static and kinetic friction ($f_s$ and $f_k$) . If something is sliding to the right over asphalt (kinetic friction), friction will try to stop this relative motion, and hold the object and the asphalt together. So $f_k$ will ...

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Yes, the force $A$ exerts on $B$ has to be equal to the one $B$ exerts on $A$. And when $F$ vanishes the forces between $A$ and $B$ vanishes too. More specifically, if in the time $F$ is acting no movement occurs (i.e. $F$ does not overcome the total friction of $A$ and $B$) the only thing causing $A$ and $B$ to interact is $F$, so this action reaction ...

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I suppose you could try to think of the problem in another way. What would happen to a cylinder (or sphere) if you put it on a frictionless inclined plane? Would it still roll or just slide? The imbalance in forces acting on the cylinder at different points, with respect to its center of mass, are what lead to the rotation. Gravity acts on the center of ...

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I would try to explain very briefly. When you apply a force onto a body, it causes the body to move in the line of application of force. This introduces translational motion into the body. When you apply a torque onto a body, it causes the body to rotate about a point. This introduces rotational motion into the body. Torque in rotation is analogous to ...

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Force represents an energy transfered (and applied in general in a straight line). On the other hand Torque represents a force acting on a point on a straight line but the effects are applied elsewhere (in rotating the body). As such there is (at least) this subtle difference. The force is applied to a point of the body which then (by internal constraints ...

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The heat generated in a computer has nothing to do with the reversibility condition in Landauer's principle. Computations can be carried out reversibly, if required. What can not be made reversible is the RESET of the computer. The first time we turn the machine on, the memory is in a random state, and it takes energy and entropy to turn that random state ...

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Obviously the friction on the top box will be along the direction of force(because of which the top box moves) and friction on the bottom box will be opposite to the force.

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You're totally correct with the m*a vector! Since the wedge is accelerated and we're assuming that the mass is sort of stuck to it (moves on its surface but stays with the wedge), in a frame of reference attached to the wedge it's going to experience a fictitious inertial force (imagine sitting next to that mass; the bigger the acceleration, the more you'd ...

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Can we assume that this a non-deformable problem? i.e. Are we talking about something like a metal rod inside a metal tube? If the materials are deformable - e.g. a wine cork in a bottle, then I think there are all sorts of factors to do with the elastic properties of the material that come into play. So leaving that aside; applying a torque to the stuck ...

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Note that it is not motion, but relative motion that the friction opposes. An example: Consider a box lying in the compartment of a train that is accelerating. If the box is stationary relative to the train,it is in fact accelerating with the train. Clearly,the only force in the horizontal direction is the friction. If it were not so, the floor of the train ...

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When you pull on the cylinder you apply a force $f_\mathrm{pull}$ between the rod and the shaft. When you twist you apply an additional force $f_\mathrm{twist}$ which acts perpendicularly to $f_\mathrm{pull}$. The total force acting between the rod and shaft is then $\sqrt{f_\mathrm{pull}^2+f_\mathrm{twist}^2}$ which is greater than $f_\mathrm{pull}$ alone. ...

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Friction is an adhesive process. When you touch two surfaces together the atoms at the two surfaces come into contact and form interatomic bonds. In the extreme case of very clean and smooth surfaces you can get cold welding. So reducing static friction is basically a matter of stopping the surfaces from adhering to each other. For example applying a thin ...

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Friction coefficient for ice and ice is minimum. The very high pressure on the ice due to the very narrow blades of the skates causes the ice immediately below the blades to melt. The thin layer of water (melted ice) between the blades and the ice surface reduces the friction between the blades and the ice surface and so the ice skater can glide easily on ...

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