Recent content by Duarh

Hmm, I think I see where I got confused. . .the forces exerted by the rocket on the fuel and vica versa are just internal forces to the rocket-fuel system - the center of mass of the whole system will still fall or decelerate in the gravitational field the same as if there was no thrust present...

Absolutely, forces remain the same. That doesn't mean that the rate of energy loss remains the same. A force of 10 N will add more Joules of energy per second to an object moving at 100 m/s than to one at 10/s, for instance. (101^2-100^2>11^2-10^2). The rate of energy loss is the same _with...

This is a reference frame problem. Energy is not the same in all reference frames. Both rockets will indeed burn fuel at the same rate. The fuel (burning gas) exiting the two rockets will have the same speed _relative to the particular rocket_ - but, in the Earth's reference frame, the speed...

From the given information, I would assume that the rod doesn't change at all during this process so it can't have any change in entropy. If there is some constant heat flow across the rod, its temperature is ill-defined but you can't really say it's changing in any way, so all the entropy would...

I think it's that in the middle of the curve you've got the tighest bonding (lowest potential energy), so that's where all nuclei would "like" to be - to the right of that point, decreasing mass number will yield a lower potential energy, while to the left of that point increasing mass number...

Yup - and the weak force is many orders of magnitude weaker than the strong, so weak interactions are waaaay less likely (for rough strength breakup see p. 55 of Griffith's book)

The principal decays of the \Lambda^{0} are p\pi^{-} and n\pi^{0}. All of these decay products contain only u & d quarks. The principal decay of the \Delta^{0} is N\pi. This is also just u & d quarks. The \Lambda contains a strange quark, while the \Delta is only u's & d's. So there's a...

Hmm. . .I'm here at Reed (Portland OR), and we've a strong physics program, but we're also a liberal arts school, which means that they force us to spend all too much time on hateful required courses outside of the sciences. . . Great profs here, though - David Griffiths, for instance, who's...

set up the profit function (income minus costs) and then take the derivative of the whole thing and set it equal to zero. looking at marginal cost separately is not that useful (though, of course, by taking the derivative of the whole thing and setting it equal to zero you'll be looking for that...

Just grab some books from a library and read up. . .definitely won't kill you; and the material is quite learnable. Try something like Thomas' Calculus - it is a huge tome that basically gives you an overview of all of introductory calculus, but you don't have to read all of it at once; if you...

well, since the collision is not elastic (the bullet sticks), PE+KE energy is not conserved (dissipated into sound, block heating up/breaking and so forth). so you should calculate the velocity of the system after impact using conservation of momentum and then use this new velocity to determine...

well, all you need to know that the change in potential energy is equal to the change in kinetic energy. that is, mgy - where y is the _change_ in height - is equal to mv^2/2. From that you can extract v in terms of the distance fallen.

allright, diagram looks fine. now if you draw the component of the boy's weight in the direction of the normal force, it'll be exactly on the line connecting the boy and the center of the semicircle. make another small triangle with the full weight as the hypothenuse (sp?) and the components as...

if you've a scanner, show me the diagram you've got. basically, you'll need the sine of that angle, which you can express in terms of R and y (the height). so no new variables.

well, if you draw a straight line down to the ground from the boy's position, it's bound to be perpendicular to the ground, isn't it? that's what I mean by straight - right in the direction of the boy's weight. and, incidentally, that side of the triangle will also be equal to the height of the...