Recent content by mircobot

Homework Statement There's an 110V source and two resisters in series. The first resister, R, is a fixed resistor. The second resister, Rx, is a variable resister (a potentiometer). The potentiometer is to be designed to adjust current I from 1A to 10A. Calculate the values of R and Rx...

Nevermind, I completely forgot that work is a scalar quantity and that obviously means it results from a dot product. I solved it... I just found the vectors of the force and direction, and found their dot product.

Homework Statement Find the work done when a force of 65 N acting at an angle of 170° from the x-axis is applied to an object that moves through a displacement of 2.2 m at an angle of 30°. Homework Equations W = Fcos\Thetad where d is the distance moved in the direction of the...

I know this seem pretty ridiculous to ask to such a large forum, but for my physics lab I have found a spring constant to have a value of 0.1368 N/cm with an uncertainty of 5x10^-4. How do I express the spring constant? So far I think it is (13.68 +/- 0.05)x10^-2 N/cm but the scientific...

Wow... speechless. That was it, off by a factor of ten. Final answer = 7.695 kg*m^2 I tell ya, sometimes you get so caught up in trying to comprehend a problem that you forget what the problem actually stated. Thank you so much, Doc Al for your patience and help.

I entered 0.07695 and I don't have to include units because webassign doesn't take them. I also enter 0.077. The problem gives me the period, like you said, which is messing me up. The first question is: (a) Assuming R = 0.90 m and m = 3.0 kg, calculate the structure's rotational inertia...

I can't seem to get it no matter what I try.

It is still wrong! My final answer is 0.07695 kg*m^2. http://www.webassign.net/hrw/hrw7_11-45.gif do I have to change the hoop's equation at all because the diameter is 2R... I don't know anymore. I thought I had it. Unless my calculations are incorrect, but I have checked them numerous...

So if left or right of the axis does not matter then according to the Parallel Axis Theorem (I = I_com + Md^2) the (3 kg)((0.09 m)/2)^2 will be added to the moment of inertia, not subtracted. The 0.09 m is divided by two in the equation because the center of mass has only shifted half of the...

Sorry, I've been confusing myself and you over this problem. I_parallelRod = (3 kg)(0.09 m)^2 I_hoop = (1/2)(3 kg)(0.09 m)^2 + (3 kg)(0.09 m)^2 I_horizontalRod = (1/12)(3 kg)(0.09 m)^2 - (3 kg)(0.09 m)^2 The horizontal rods have a moment of inertia of (1/12)(3 kg)(0.09 m)^2, but it...

Yes, it is a vertical rod. The furthest rod has a distance of 0.90 m away from axis

Sure...

Oops I meant to say what is wrong with my horizontal equation that I first posted?

Ok, so... I_parallelRod = (3 kg)(0.09 m)^2 The parallel rod is the length of the rod away from the axis. I_hoop = (1/2)(3 kg)(0.09 m)^2 + (3 kg)(0.09 m)^2 The hoop is the length of the radius away from the axis. So the summation of these moments of inertia is: I =...

I = 0, because the distance away from the axis is 0, thus I = Md^2 = 0