A2.00 kg pendulum bob on a string 1.50 m long is released with a velocity of 0.00 m/s when the support string makes an angle of 60.0 degrees with the vertical. what is the tension in the string at the bottom of the swing

39.2 Newtons First, calculate how high the bob is above the bottom of the swing. Since the string is making a 60 degree angle from the vertical, the angle the string makes against the horizon is 30 degrees. And the sine of 30 degrees is 0.5. So the vertical distance from the top of the pendulum is 0.5 * 1.5 m = 0.75m and therefore the distance the bob will drop is 1.5m - 0.75m = 0.75m. Now at the point of release we have all gravitational potential energy and that will be: 2 kg * 9.8 m/s^2 * 0.75 m = 14.7 kg*m^2/s^2 = 14.7 Joules At the bottom of the swing all of that potential energy will have been converted to kinetic energy. The equation for that is E = 0.5 MV^2 14.7 J = 0.5 * 2.0 kg * V^2 14.7 kg*m^2/s^2 = 1 kg * V^2 14.7 m^2/s^2 = V^2 3.834057903 m/s = V Now the tension in the string at the bottom of the swing will be the sum of the gravitational attraction and the centripetal force. The equation for centripetal force is: F = MV^2/r Where F = Force M = Mass V = velocity r = radius We already know that V^2 is 14.7 m^2/s^2 from calculating the velocity above, so we have F = M*14.7 m^2/s^2/1.5 m F = 2.0 kg * 14.7 m^2/s^2 / 1.5 m F = 29.4 kg*m^2/s^2 / 1.5 m F = 19.6 kg*m/s^2 And the gravitational attraction is F = 2 kg * 9.8 m/s^2 = 19.6 kg*m/s^2 So the total force experienced is 19.6 kg*m/s^2 + 19.6 kg*m/s^2 = 39.2 kg*m/s^2 = 39.2 N So the tension in the sting at the bottom of the swing is 39.2 newtons.

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An  atom  (ours, not carroll's) consists of a positively charged  nucleus, surrounded by one or more negatively charged particles called  electrons. the positive charges equal the negative charges, so the atom has no overall charge; it is electrically neutral. most of an atom's mass is in its nucleus; the mass of an electron is only 1/1836 the mass of the lightest nucleus, that of hydrogen. although the nucleus is heavy, it is quite small compared with the overall size of an atom. the radius of a typical atom is around 0.1 to 0.25 nanometres (nm), whereas the radius of a nucleus is about 10-6  nm.  *  if an atom were enlarged to the size of the earth, its nucleus would be only 200 feet in diameter and could easily rest inside a small football stadium.read more on -