Hi,
I think he tends to be mentioned whenever authors give an overview of the history. As an example, the only two QM textbooks I have available right now, Sakurai's (standard) book and Weinberg's "Lectures on quantum mechanics", do mention him and his role when it came to matrix mechanics...
Maybe I am reading this completely wrong then? "A certain automobile manufacturer claims that its deluxe sports car will accelerate from rest to a speed of 44.5 m/s in 8.90 s." To me, that certainly sounds like the car travels at 0 m/s in the beginning, and at 44.5 m/s after 8.90 s.
Your numbers are slightly off (even for what you are doing), but my main objection is this: the question tells you that the car accelerates from rest with the same acceleration all the time - yet you claim that it will travel slower after 10 seconds than it did after 8.9 seconds.
OK. Well, if it makes you feel better, those values (assuming you give α in the usual units) and the linear expansion formula does give me the same answer. Basically, thermal expansion is a small effect, which thus mainly matters for larger things.
Yes, that's right. \Delta x only depends on \rho g h (and vice versa). By the way, this is why we often call this pressure (i.e. absolute pressure minus atmospheric pressure) gauge pressure.
The piston moves due to a change in the pressure, doesn't it?
So, above surface the pressure is already P_0, which corresponds to a certain compression x_0. Under the surface your equation holds and gives a compression x(P)=x_0+\Delta x. What is \Delta x?
It depends in which order you take the points in the potential difference. V(2)-V(1) would be the difference in energy if you go from point two to point one, while you consider the opposite path and thus gets the opposite sign.
I mean, in which direction does the y axis point? Does it point upward? I.e. does higher positions correspond to higher values of y? Because, in that case the formula the gravitational acceleration is opposite to the y axis, and the acceleration a=-g. You would then have
y_1 = ut -...
Yes, you do have momentum conservation.
Sorry, what do you mean? The momentum is certainly imparted to the shot person through a force, but you don't know either the magnitude of the force or the duration of time it acts on the person, do you?
That's basically the reason, yes. In the stable case, the force will push it back towards the equilibrium, and not away from it.
Another way to think about it is that the force on the particle is related to a potential through \mathbf{F} = - \nabla V \left( x \right). If the equilibrium...
Well, the wind is moving the air the sky diver falls through. Does this affect the motion of the sky diver or not? (Hint: think about something light, like a piece of paper or a balloon.)
As for the other question, you are probably expected to use the same wind speed at all heights. That's...
You draw it, and then use trigonometry. (Always draw the problem!) Because you have learnt about decomposition about vectors into components, right? This is more or less the same problem.
There is an easier way to solve it, just factor out one of the t's:
x=ct^2+bt = t(ct+b),
for which x=0 when t=0 or t=-b/c. Both solutions are of course valid (based on the problem text), even if the homework website doesn't think so...
Well, let us take this carefully.
1. The ball loses speed as it moves up. Thus the acceleration is in the opposite direction of the velocity, i.e. downwards. (Think of it as vectors.)
2. The ball reaches a max height, where it has a minimum speed of zero. The acceleration continues to point in...
OK, suppose you throw a ball up in the air. Is the sign of the acceleration negative while it is moving upwards and positive when it is moving downwards then?