Recent content by jonc1258

The denominator is kind of hard to see, it's sqrt(x)-3. Thanks for all the replies, I'll see if I can figure it out

The problem statement Evaluate the indefinite integral ∫\frac{\sqrt{x}}{\sqrt{x}-3}dx The attempt at a solution My first thought was to substitute u for √(x)-3, but then du would equal \frac{1}{2\sqrt{x}}dx, and there's no multiple of du in the integrand. Next, I tried splitting up...

What if the first bearing were the trigger for a Gauss rifle? http://sci-toys.com/scitoys/scitoys/magnets/gauss_rifle/ready_to_fire.jpg If the second magnet was .1 meters in front of the first magnet and there was another bearing lying against the second magnet (like in the picture), wouldn't...

27.1 = .005a a = 5420m/s^2 v^{2}_{f} = v^{2}_{i} + 2ad v^{2}_{f} = (0) + 2(5420)(.1) vf^2 = \sqrt{1084} v=32.9m/s Yes?

Homework Statement Find the final velocity of a 5g steel ball bearing being pulled by a neodymium magnet with a pull force of 6.1 pounds (27.1 Newtons) in a straight line on a frictionless surface after being pulled for .1 meters Homework Equations F=MA The Attempt at a Solution f=ma...