elfmotat

I'm not really sure what you mean by "strongest." The "g" value (which on Earth is ~9.8 m/s2) approaches infinity near the event horizon of any black hole.

The centripetal force is the force that keeps a rotating object in orbit. For example, if you were to tie a ball to the end of a rope and whirl it around, the tension in the rope would be the centripetal force. The centrifugal force is what is known as a "fictitious force." A fictitious force is one that arises because the observer is in a non-inertial reference frame. Fictitious forces occur because the reference frame itself is accelerating. If you were to be put inside a giant elevator that moves up with constant acceleration, you would feel a "force" pulling you to the ground. What's "really" happening is that the floor of the elevator is accelerating up towards your feet, though from inside the elevator it seems like some invisible mysterious force is pulling you down. Similarly, if you were to be placed inside a rotating spaceship then you would observe a mysterious force pulling you to the walls. What's "really" happening is that the walls are pushing on your feet, keeping you in orbit. I.e. they're applying a centripetal force. But from inside the ship, all you can tell is that you're being pressed to the walls with a force of magnitude [math]m \omega^2 r[/math].

No, I don't think that makes any sense.

The "right hand rule" is merely a simple way to determine the direction of a vector that is the product of one vector crossed with another. The cross product of two vectors is: [math]\mathbf{A}=(a,b,c)[/math] [math]\mathbf{B}=(d,e,f)[/math] [math]\mathbf{A} \times \mathbf{B} = (bf-ce,cd-af,ae-bd)[/math] So if, for example, [math]\mathbf{A}[/math] is the unit vector in the x-direction [math]\mathbf{A}=(1,0,0)[/math] and [math]\mathbf{B}[/math] is the unit vector in the y-directiontion [math]\mathbf{B}=(0,1,0)[/math], then the cross product is: [math]\mathbf{A} \times \mathbf{B} = (0,0,1)[/math] This is the unit vector that points in the z-direction. When translated into your fingers, your index finger represents [math]\mathbf{A}[/math], your middle finger represents [math]\mathbf{B}[/math], and your thumb represents [math]\mathbf{A} \times \mathbf{B}[/math].

The time for two point masses to come together under gravity from rest is: [math]T=\frac{\pi}{2}\frac{r_0^{3/2}}{\sqrt{2G(m_1+m_2)}}[/math] For two 1-kg masses starting out 5 meters apart, time time for them to come together is about 12 days, so you'd definitely notice if you were patient enough.

Your friend is right. There is a gravitational force between any two massive objects. So yes, if you put any two objects relatively close to each other in deep space then they will eventually come together under gravity. The equation that tells us this is called "Newton's Universal Law of Gravitation:" [math]F=\frac{Gm_1 m_2}{r^2}[/math] where [math]F[/math] is the force on one object due to the other, [math]G[/math] is the gravitational constant (it's just a number that's always the same), [math]m_1[/math] is the mass of object 1, [math]m_2[/math] is the mass of object 2, and [math]r[/math] is the distance between the two objects. What this says is that larger masses create larger gravitational forces (which should make sense), and that the farther away two objects are, the smaller the force between them (which should also make sense).

Why would an additional dimension make electrons distinguishable? I can't think of any good reasons.

First, get a good textbook on SR. Taylor/Wheeler's Spacetime Physics would be my recommendation. You don't need to know too much math to learn SR. If you want to get into GR, you'll need to be familiar with at least vector calculus and linear algebra. Schutz's GR textbook is good for a beginner, and it's available online at <link removed by mod pending copyright check>.

This is a basic Ohm's Law problem. Show us what you've done so far and we'll give you some hints if you need them.

It depends mainly on the material of the object being dropped. From 45 meters, the object will have a final velocity of about 30 m/s before it hits the ground. The ground will apply a net impulse of 1500 Ns, but the actual force as a function of time F(t) will depend on what the object is made of. A big rock, for example, will have a much faster impulse than a 50 kg bundle of feathers.

This whole thread is very strange. Of course you can "model" mass in any number of dimensions. It's just a scalar.

What do you mean by "explained?"

I didn't make any assumptions. From the rest frame of the ground, the man is running horizontally relative to the rain. In the man's rest frame the rain is therefore falling with a horizontal component relative to the man.

In the man's rest frame the rain has a horizontal component to its velocity, so the man should tilt his umbrella at an angle.

Both your image an your words look like meaningless nonsense to me.

You can parametrize the curve as simply [math]\mathbf{r}(t)=(t,t^2,0)[/math]. The tangent vector to the curve is then [math]\mathbf{r}'(t)=(1,2t,0)[/math], and the acceleration vector is [math]\mathbf{r}''(t)=(0,2,0)[/math]. The tangent component of the acceleration to the curve is the dot product of the acceleration vector with the unit tangent, the normal component is therefore: [math]a_\perp = \frac{|\mathbf{r}'\times \mathbf{r}''|}{|\mathbf{r}'|}[/math] So we have: [math]a_\perp (t) = \frac{|(1,2t,0)\times (0,2,0)|}{|(1,2t,0)|}=\frac{|(0,0,2)|}{\sqrt{1+4t^2}}=\frac{1}{\sqrt{t^2+1/4}}[/math] Plotting a graph: This looks as we'd expect. The maximum acceleration occurs at t=0, when the point is at the minimum of the y=x2 graph, and it tapers off in either direction.

"Expanding faster than light" is ill-defined. Nothing is actually moving during expansion, distances are just getting larger due to the time-dependent metric. There's no "velocity of expansion," so "expanding faster than light" makes no sense.

The basic laws of currents tell us that the power dissipated by a resistor is [math]P=I^2 R[/math]. Resistivity [math]\rho[/math], which is proportional to resistance, is related to conductivity [math]\sigma[/math] by the equation [math]\rho \sigma =1[/math]. So yes, that equation looks right to me.

It may be a bit beyond what the OP has in mind, but classical unification of gravity and EM is very easy with various Kaluza-Klein theories.

Any point on the band, moving or stationary, will have an attributed coordinate acceleration: [math]\frac{d^2x}{dt^2}=v_r v_a(L_0+v_r t)^{-1}[/math] In the case where the point is stationary w.r.t the band (i.e. if we drew a little dot on the band with sharpie), the coordinate acceleration is therefore zero. This means that all points on the band are moving with constant velocity, so each point has its own attributed inertial reference frame. A bug not moving w.r.t the band will not be able to measure anything with his accelerometer. However, as soon as the bug begins to move he will experience a measurable acceleration. Each point on the band will move with constant velocity, however the bug will be accelerating w.r.t every point on the band. The "constant velocity" of the bug refers to the velocity the bug would have w.r.t the band if it were not being stretched. It's easy to see this by considering the reference frame of, for example, the middle of the band. The new coordinates are: [math]x'=x-\frac{1}{2}v_r t[/math] [math]t'=t[/math] Therefore [math]\frac{d^2x'}{dt^2}=\frac{d^2x}{dt^2}[/math], so the bug is accelerating w.r.t the middle of the band in exactly the same way that the bug is accelerating w.r.t the left end of the band.

Field excitations are called "particles." So, we get particles from fields by definition. Some fields are coupled to each other in various ways. For example the photon field and the electron field are coupled.

I'm not sure what you mean by "in 4 dimensional space-time." Are you talking about four-vectors? If so, then in a sense yes, they all have a magnitude zero four-momentum: [latex]p^\mu=(E,\mathbf{p})[/latex] with [math]p^\mu p_\mu = 0[/math]. They also have an associated wave-vector which contains the information about the frequency and wavelength of the light: [math]k^\mu = (\omega, \mathbf{k})[/math]. Since [math]p^\mu = \hbar k^\mu[/math], the magnitude of the wave-vector is also zero: [math]k^\mu k_\mu =0[/math]. EDIT: I'm not sure what's going on with the LaTeX, but that </span> thing shouldn't be there.

"File not found."

This is a rather strange thread. "Aether," as far as physics is concerned, was a hypothetical medium through which light was suspected to propagate. It has, since the early 20th century, been disproven. The rest of your thread has nothing to do with physics, and everything to do with mystical nonsense.

What does that have to do with anything?