Recent content by corochena

Given two people of different complexion, one short and chubby and the other one tall and thin, but same weight (180 lbs each): a> Do they have the same number of atoms in their bodies b> If not, will the percent difference be significant? c>Is there a way to know an estimate of this...

So I guess the sport tv channels use Force instead of Energy or Impulse just because Force (Pounds, ton, kg) is more common term, less people would understand if they used Joules or Kg.m/s

yungman, you are right, but implicitly there is a time in the punch, kick or tackle, just we don't know exactly its value, but you can imagine or estimate it I will know that if I get punched by a 4 kg mass at 10 m/s speed I will receive 20 J energy in about 0.1 second, so even if I don't know...

Why when a boxer, mma artist, football player, or any other athlete makes a collision such as a punch, kick or tackle we measure that using Force units (pounds or kilograms)? Would not make more sense use Energy units or Impulse units? I was just watching espn and they say a football...

No, it does not. OK, in order to yield units of kg we have to to this: Total Mass = density of area . Total Area density of area = M / A so dm = M / A . dA then dm = M / A . length of strip . dx ... and you have Kg

First we assume that the body is uniform, so its mass is proportional to its area in any section. So I am going to use area (A) instead of mass for my calculations. dA = height of vertical strip . dx height of vertical strip = 2/3 . x Then you use the formula for center of mass...

Thanks both of you. What a nice site is this! It was difficult to get someone interested in this kind of stuff, ... not anymore!

Thank all of you guys! I am going to search my papers and bring the formula so we can discuss it. The formula works well for the case of a horizontal rain (extreme case), so I have the feeling that it is correct.

I know I have to prove that x must be equal to h/2 (that is equivalent to say that the lowest CM possible must be located at the surface of contact of the two substances). There is another way to do it and it is not necessary to make the assumption x = h/2, but you need to find the derivative...

Yes, I am taking moments about the CM of the cylinder Cc. But I think m1 . x = m2 (H/2 - h/2) is wrong. The correct expression is as I wrote before (hopefully!) m1 . x = m2 . (H/2 - x). I think the new picture will explain better.

Also I did a search in the Internet and found that Mythbusters tried to solve this problem weighting the clothes during the walk and they found that it is better to walk than to run (to get wet the least). So we came to different results. But, a regular box is very different from a moving human...

I found this problem in the book Resnick and Halliday many years ago (1992) and I had it in my mind until 2005 or so when I was able to get some advance. "If you want to walk through a distance L in a rainy day, in order to get wet the little as possible what would you do, run as fast as...

Thanks a lot! And... how were you able to answer so quickly? Is there some kind of alarm or something?

It's a bit difficult because the simbology but I'll try: The center of mass of cylinder 1 and 2 (heavier and lighter) are located in the middle because they are homogeneus. So I am able to find the CM of the whole cylinder: By definition of CM: m1 x = m2 (H/2 - x), x is the distance...

Solution Well, I realized that the heavier substance must be located at the bottom of the cylinder and the ligther top. Then I computed the position of the center of mass of each sub-cylinder top and bottom. Let it be: d1: density of the heavier substance d2: density of the lighter...