Recent content by Starwatcher16

Well, I am not quite as old as you, but I also a freshmen taking my first physics class. What has helped me a lot is banging my head against more advanced problems then we are currently talking about in class.

I've heard it said that it is impossible to determine an objects absolute velocity, all one can do is find it relative to some other object... but, if the CMB is everywhere, why could you not just measure the doppler shift in all directions, and adjust your velocity so the CMB has no doppler...

Oh, btw I would probably start out going for an EE, if that makes a difference. I don't really know much theory yet, just random stuff I have picked up reading Nuts & Volts. My brother says go for it, but I would like to research this a little first, as I got a few lucky breaks, and was able to...

Hi, I'm 19. I graduated when I was 17, I worked as an apprentice for 18 months, before getting pulled into my companies estimating division, where I have been for the last three months. Anyways, I am bored as hell, and was thinking about going back to school for engineering, but I have some...

Not really sure if this is the place for this, but can overhead tranmission lines be located via sattelite imagery?

x_f=x_i+V_iT+\frac{at^2}{2}, where a=\frac{-kx}{2m}, x_f=-A,x_i=+A -2A=\frac{\frac{-kx}{2m}*t^2}{2}=-2A*\frac{2}{\frac{-kx}{2m}} =T^2=\frac{8m}{k}, T=(\frac{8m}{k})^(1/2) What did I do wrong with the algebra? As for the physics, your saying if I have a non constant a, I can't...

Hmm..still not really making sense to me. I must be doing something basic wrong. Okay, let's just ignore the first method. If I had a block attached to an unstretched string, and then I stretched it to x=A, how would I go about solving for T. When T is the time it takes the spring to get the...

...oopsie. I see now what I did. My x=0 point is not the point where the spring is all the way compressed, my x=0 is where the spring is only halfway compressed,

No, F_ave is over 1/4 a cycle, from x=A to x=0. F_ave=kx/2 is kx integrated over x and then divided by the length it was integrated over.

F_s=-kx=ma, a=\frac{-kx}{m} We now need to find an equation that satisfies: \frac{d^2x}{d^2t}=\frac{kx}{m} We get (letting w^2=k/m): x(t)=A*Cos(wt+\phi) The time it takes for one cycle is:T_1=2 \pi\sqrt{\frac{m}{k} Now, let's solve for T again using a different method...

Could someone walk me through this one? I know what the answer is, but don't really see how that answer is come by. In the book, it just states the answer, so I guess its something obvious, but the answer still eludes me :( Homework Equations \int{Cos^2{(ax)}}dx

could someone explain to me what a Candella is? I am not really getting it. Okay, so it is a Watt/Steradian. Okay. Lets say that when we get a flux of 100 photons passing through a surface per second, that that is equal to one watt. So if we have, let's say 50 Candella, we have a flux...

Is there a minimum value for how small a wavelength can become?

While we are at it, could someone explain to me what a Candella is? I am not really getting it. Okay, so it is a Watt/Steradian. Okay. Lets say that when we get a flux of 100 photons passing through a surface per second, that that is equal to one watt. So if we have, let's say 50...

For a lamp that is stated to have x lumens, is that x lumens total, or x lumens per steradian?