Recent content by Spring

The second equation is ½mv2 but I used c (speed of light) to calculate it for light. Once again, I am unsure because I devided by m of photon which equals to zero, deviding by nothing. Also because there is a different way to calculate the energy of a photon. Ep = hf . That way it means that...

I am clearly talking about black holes. The event horizon is the limit where even a photon won't escape it. I tried to calculate it in the easy way using enegry calculation m * MG/R = mc^2 / 2 but I do not know if I am using the right equation or even if I can divide by the m because it...

I will be deviding you're question into 2, if you don't mind. If you have constant velocity Then the added forced to you will be zero, even if gavity affects you, constant velocity means something is negating this gravity. That means weightless (not massless). On the other hand you might feel...

Alright, so what you have here Is fun. in fact when you open the parentheses you will get: u2 - u > 0. Now to know when that happends you can simply solve what it should look like. since the u^2 has a positive feature you can infer it will be a "smiling" parabule. and you can clearly see the...

How does this help me? Getting from here to F = sqrt((mg)2 + 2mghk) is still vary uncleare and in fact was the main question of this form. I see how mghk is there, but the whole mg2 is a mystery for me. also I arrived at sqrt(½mghk).

True, so I did what I needed to calculate the the distance strached, also with spring like motion calculations, and after that I devided the overall spring energy by the distance to get the force at the end of the stretch sequence. I was shocked to find it was to no avail as the correct answer...

Basically. I will try not to be annoying with a big background story. Adam is a man attached to a rope, he weighs at m (kg). Adam fell from h (meters) and only has his trusty rope with spring current of k (kg/m^2) to block his fall. What force does Adam feels in Newtons. Normally I would try to...

I see, I didn't account for that. I still think you are mistaken as sqrt((mg2)) is force squered and rooted and so is sqrt(2mghk) notice they are added to each other and not multiplied. m is mass (in MKS is kg) g is acceleration (in MKS is m/s^2) h is height (in MKS is m) k is spring constant...

Which one? Also, can you explain in detail what is wrong about which? I got my new answer info from wikipedia, some inaccurations might occur.

Hello, I have been turning to this subject and tried to develop the right equation on my own. What I did was: [FONT=Georgia]mgh = ½kx2 (fall energy = spring energy) x2 = 2mgh/k (obviously devided by ½k) sqrt(2mgh/k) = x (taking out square root) F = mgh/x = mgh/sqrt(2mgh/k) = sqrt(½mghk)...