# Tag Info

## New answers tagged estimation

85

A "perfectly efficient" computer can mean many things, but, for the purposes of this answer, let's take it to mean a reversible computer (explained further as we go). The theoretical lower limit to energy needs in computing is the Landauer Limit, which states that the forgetting of one bit of information requires the input of work amounting $k\,T\,\log 2$ ...

0

It really depends on the strength of the warheads. Bombs are measured by their TNT equivalent. A kiloton atom bomb is equivalent to that of a 1000 tons of TNT. However, that is only for the blast energy generated. Blast energy only accounts for 85% of the total destruction. The remaining 15% is the long term effect, radiation, pollution, and potentially ...

2

This is a surprisingly difficult question to answer properly, but if you read the report from the Ricardo group (thanks @pentane for the reference!) it includes the following table: They do a lot of analysis to show that if you reduce the weight of a car, you can get away with a smaller (more efficient) engine; but I am assuming that you just want the ...

1

our aging is directly proportional to metabolism of body and division,growth and death of cells of body and if gravity has some effect on the rate of above aspects astronauts will be definitely younger

2

The first answer has all the results, but I will try to show some calculations, cause I have been writing them since there was no answer. It is known from General Theory of Relativity (GTR) that the closer you are to a massive object - the slower the time goes. On the other hand Special Theory of Relativity (STR) gives us the next statement: the faster you ...

4

This is anwered in Gravity on the International Space Station - General Relativity perspective, where we learn that time dilation in the ISS with respect to Earth equator is 1.00000000028655. So after 17 years for us, the astronauts would come back younger by about 0.15 seconds than if they have stayed on the ground. Note that a full GR treatment is ...

10

Note that this is an incomplete answer. Imagine an object of mass $m$ at a distance $r$ from the centre of a black hole of mass $M$. The gravitational potential energy is $$U(r)=-G\frac{Mm}{r}$$ This has its highest value when $r=\infty$ and its lowest value when $r$ is at the event horizon of the black hole, i.e. the Schwarzschild radius  R = ...

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.

Top 50 recent answers are included