New answers tagged

0

The simplest approach to a problem like this would assume that the collision is elastic, and that you have some knowledge of the elastic constant. But a collision between car and human is not that. Instead, let us assume that the "elbow sized object" hits the human in the mid section, and that it doesn't simply go right through him. Then the next thing that ...


1

That's fairly small for an object. It wouldn't have any significant gravitational effect on the moon or the earth. Tidal effects go as the cube of the distance. So the sun has about half the tidal effects of the moon. If this object were in low earth orbit (400km altitude), then the relative tidal effects on the surface when it is overhead would be about ...


1

Now perfect balance between the centrifugal force of orbital rotation and sun's gravity is impossible so the earth's orbit should either be slowly decaying inwards or expanding outwards due to difference in magnitude of those opposing forces. This assumption is incorrect. We could make the same argument about a weight suspended from a spring. ...


2

When you don't have to consider the inertia of the screw (which is always), these problems can easily be analyzed by unwinding the screw to a plane (as you've done in your figure). So think of the screw and the surface of contact as two inclined planes that slide against each other (one on top and one beneath it). Now, pushing on the screw is like pushing ...


1

Gareth, this is a good first question. You've asked it well, and I encourage you to ask more questions in the future. Basically, you can just reverse any usual explanation of a screw. Normally, the wedge transfers torque into thrust; well in this case the wedge transfers thrust into torque — it's exactly the same, but reversed. This is a special example ...


0

If the bottom had tiny holes like the ones on the air hockey games and the floor was smooth, it may slide quite easily just off the floor on the cushion of air.


2

Let x be the instantaneous mass fraction of original water remaining in the lake at time t, w be the mass recharge/discharge rate, and M be the total mass of water in the lake (assumed constant). Also assume that the water in the lake is well-mixed. Then, for a mass balance on the original water, we have: ...


1

Note that the instructions do not claim that the volume of water is changing measurably over the course of the procedure. The volume of water is not changing noticeably. What this procedure allows one to observe is the fact that a salt+water solution has less volume than the total volume of its ingredients. Rather than a demonstration of water changing ...


4

Laser is stimulated emission of highly energetic photons. Fundamental use of laser is heating, propulsion is very distant aim which lasers can achieve. Few kW rating lasers can actually lift the mass (very small values though) because incident energy beam has momentum associated with it. Your assumption is not correct as you are comparing heat energy with ...


44

Your approach is incorrect. You cannot do this calculation by considering that the energy absorbed by the object is converted into a change in gravitational potential energy. For one thing the object would just get hot and radiate away most of the energy and for another this is a dynamical problem, you have to be able to accelerate the object upwards. What ...


8

Lookup Optical Tweezers. The limitations of the technique are due to the damage thresholds; see laser ablation. This idea has been used extensively in science fiction, especially when implemented as solar sails. It's even practical for some applications, as noted in the article. But laser propulsion from the ground suffers losses due to the atmosphere. ...


8

Your consideration implies you have a device that can convert laser light with 100% efficiency and convert it to mechanical power. This is theoretically possible, but there is not much of physics. You still need some ladder to climb along, and building a space elevator requires much tougher materials than we currently know. In practice, you would have to ...


2

Estimation: I want the two densities of vater and vapour to become approximately equal. the density of water is nearly constant the vapour pressure (you can derive this from the above mentioned Clausius-Clapeiron-equation) is approximately exponential in $1/T$. This means, that if you increase pressure by a factor, the inverse of the evaporation ...



Top 50 recent answers are included