Recent content by JosieNutter

The reason it doesn't make any sense is probably because there is a term missing in every one. It should be: The actual text is: 1. The force of kinetic friction is smaller than that of static friction, but F_g remains the same. 2. Once the box is moving, F_g is smaller than the force...

Some basic notes-- One nice thing about projectile motion is that the horizontal motion and vertical motion are independent of one another. Once thrown, the projectile has constant velocity in the x direction (acceleration in the x direction is zero). Its velocity in the y direction is...

The only kinematic equations you need are: x_f = x_i +v_ix*t+0.5*a_x*t^2 y_f = y_i +v_iy*t+0.5*a_y*t^2 Keep in mind that horizontal motion and vertical motion are completely independent of one another (one of the convenient things about projectile motion). With your origin set as the...

Well, that makes me a little less wary... one of the things that's been making me shy away from Smith a bit is that I'm not sure math & science classes at a liberal arts class will be challenging enough. I'm hoping I'll get a chance to talk to some of the people in CS & Engineering there at...

Background: I'm currently taking classes at a local community college with the intention of transfering to a 4-year university after I get my AS (Associate of Science). I'm a non-traditional student in my 30s who has been working professionally in the tech/computer industry since high school...

Don't estimate - you can see exactly what V is at 2ms. R is constant. V = V_0 * e^(-t/RC) Solve for R an input data from the graph where t=2ms.

Er... nevermind. I finally realized it was asking for nC - not C. Big difference in numbers there. :p

I worked this one out on my own and ended up with: q = \sqrt{\frac{.003kg*9.8\frac{m}{s^2}*tan(20)*(2sin(20))^2 m}{(9*10^9\frac{Nm^2}{C^2})}} (Which, as far as I can tell, is identical to the one above.) It gives me an answer of 7.46*10^-7... which is apparently...